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8890244575, , PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , Introduction, Numerical study of chances of occurrence of, events is dealt in probability theory., The theory of probability is applied in many, diverse fields and the flexibility of the theory, provides approximate tools for so great a variety of, needs., , Definitions of various terms, (1) Sample space : The set of all possible, outcomes of a trial (random experiment) is called its, sample space. It is generally denoted by S and each, outcome of the trial is said to be a sample point., (2) Event : An event is a subset of a sample, space., (i) Simple event : An event containing only a, single sample point is called an elementary or, simple event., (ii) Compound events : Events obtained by, combining together two or more elementary events, are known as the compound events or, decomposable events., (iii) Equally likely events : Events are equally, likely if there is no reason for an event to occur in, preference to any other event., (iv) Mutually exclusive or disjoint events :, Events are said to be mutually exclusive or disjoint or, incompatible if the occurrence of any one of them, prevents the occurrence of all the others., (v) Mutually non-exclusive events : The, events which are not mutually exclusive are known, as compatible events or mutually non exclusive, events., , (vi) Independent events : Events are said to be, independent if the happening (or non-happening) of, one event is not affected by the happening (or nonhappening) of others., (vii) Dependent events : Two or more events, are said to be dependent if the happening of one, event affects (partially or totally) other event., (3) Exhaustive number of cases : The total, number of possible outcomes of a random, experiment in a trial is known as the exhaustive, number of cases., (4) Favourable number of cases : The number, of cases favourable to an event in a trial is the total, number of elementary events such that the, occurrence of any one of them ensures the, happening of the event., (5) Mutually exclusive and exhaustive system, of events : Let S be the sample space associated, with a random experiment. Let A1, A2, …..An be, subsets of S such that, (i) Ai  A j   for i  j, and (ii), A1  A2  ....  An  S, , Then the collection of events A1, A2 , ....., An is said, to form a mutually exclusive and exhaustive system, of events., If E1, E2 , ....., En are elementary events associated, with a random experiment, then, (i) Ei  E j   for i  j, and, (ii), E1  E2  ....  En  S, , So, the collection of elementary events, associated with a random experiment always form a

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8890244575, , PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, Similarly, odds against an event E, , system of mutually exclusive and exhaustive system, of events., In this system, P( A1  A2 .......  An ),  P( A1 )  P(A2 )  .....  P(An )  1 ., , Classical definition of probability, If a random experiment results in n mutually, exclusive,, equally, likely, and, exhaustive, outcomes, out of which m are favourable to the, occurrence of an event A, then the probability of, occurrence of A is given by, m Number of outcomes favourable to A, P ( A) , , n, Number of total outcomes, It is obvious that 0 ≤ m ≤ n. If an event A is, certain to happen, then m = n, thus P(A) = 1., If A is impossible to happen, then m = 0 and, so P(A) = 0. Hence we conclude that 0 ≤ P(A) ≤ 1., , , , Addition theorems on probability, Notations : (i) P( A  B) or P( A  B) = Probability, of happening of A or B, = Probability of happening of the events A or, B or both, = Probability of occurrence of at least one, event A or B, (ii) P(AB) or P(AB) = Probability of, happening of events A and B together., (1) When events are not mutually exclusive : If, A and B are two events which are not mutually, exclusive, then, P ( A  B)  P ( A)  P ( B)  P ( A  B), or P( A  B)  P( A)  P(B)  P( AB ), , Further, if A denotes negative of A i.e., event that A doesn’t happen, then for above, cases m, n; we shall have, nm, m, ,, , P( A ) , 1,  1  P( A), n, n, , For any three events A, B, C, P( A  B  C )  P( A)  P(B)  P(C )  P( A  B), ,  P(B  C)  P(C  A)  P( A  B  C), , or P( A  B  C)  P( A)  P(B)  P(C)  P( AB )  P(BC ), , P( A)  P( A )  1 ., , Notations : For two events A and B,, (i) A or A or AC stands for the non-occurrence, or negation of A., (ii) A  B stands for the occurrence of at, least one of A and B., (iii) A  B stands for the simultaneous, occurrence of A and B., (iv) A  B stands for the non-occurrence of, both A and B., (v) A  B stands for “the occurrence of A, implies occurrence of B”., , Problems based on combination and, permutation, (1) Problems based on combination or selection, : To solve such kind of problems, we use, n!, n, ., Cr , r!(n  r)!, (2) Problems based on permutation or, arrangement : To solve such kind of problems, we, n!, use n Pr , ., (n  r)!, , Odds in favour and odds against an event, As a result of an experiment if “a” of the, outcomes are favourable to an event E and “b” of, the outcomes are against it, then we say that, odds are a to b in favour of E or odds are b to a, against E., Thus odds in favour of an event E, Number of favourable cases, a a /(a  b) P(E), ,  , , Number of unfavourab le cases b b /(a  b) P(E ), ., , Number of unfavourab le cases, b P( E ), .,  , Number of favourable cases, a P ( E), ,  P(CA )  P( ABC ), , (2) When events are mutually exclusive : If A, and B are mutually exclusive events, then, n( A  B)  0  P( A  B)  0,  P ( A  B)  P ( A)  P ( B ) ., For any three events A, B, C which are, mutually exclusive,, P( A  B)  P(B  C)  P(C  A)  P( A  B  C), , =0,  P( A  B  C)  P( A)  P(B)  P(C) ., The probability of happening of any one of, several mutually exclusive events is equal to the, sum of their probabilities, i.e. if A1, A2 ..... An are, mutually exclusive events, then, P(A1  A2  ...  An )  P(A1 )  P(A2 )  .....  P( An ), ,  A )   P(A ) ., , i.e. P(, , i, , i, , (3) When events are independent : If A and B, are independent events, then P( A  B)  P( A).P(B),  P( A  B)  P( A)  P(B)  P( A).P(B) ., (4) Some other theorems, (i) Let A and B be two events associated with, a random experiment, then, (a) P( A  B)  P(B)  P( A  B), (b) P( A  B )  P( A)  P( A  B), If B  A, then, (a) P( A  B )  P( A)  P(B), , (b) P(B)  P( A), , Similarly if A  B, then, (a) ( A  B)  P(B)  P( A), , (b) P( A)  P(B), ,  Probability of occurrence of neither A nor B is, P ( A  B )  P ( A  B)  1  P ( A  B )

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PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 8890244575, , (ii) Generalization of the addition theorem : If, A1, A2 ,....., An are n events associated with a, random, experiment,, then, n, n, n,  n, , P  Ai  , P( Ai ) , P( Ai  A j ) , P( Ai  A j  Ak ) , , , i, j 1, i, j, k 1,  i1  i1, , , , , , , , , , i j, , i j  k, , that the events A1, A2 , ....., Ai1 have already, happened., (iii) Multiplication theorems for independent, events : If A and B are independent events, associated with a random experiment, then, i.e., the probability of, P( A  B)  P( A) . P(B), simultaneous occurrence of two independent, events is equal to the product of their, probabilities. By multiplication theorem, we, have P(A  B)  P(A) . P(B / A) . Since A and B are, , ...  (1)n 1 P( A1  A2  .....  An ) ., , If all the events Ai(i  1, 2..., n) are mutually, , independent events, therefore, Hence, P( A  B)  P( A) . P(B) ., , n,  n, , P( Ai ), exclusive, then P  Ai  , , ,  i 1  i 1, i.e., P( A1  A2  ....  An )  P( A1 )  P(A2 )  ....  P(An ) ., , , , , , (iv) Extension of multiplication theorem for, independent events, :, If, are, A1 , A2 , ...., An, independent events associated with a random, experiment, then, P(A1  A2  A3  ...  An )  P(A1 )P(A2 )... P(An ) ., , (iii) Booley’s inequality : If A1, A2 , .... An are n, events associated with a random experiment,, then, , By multiplication theorem, we have, , n, n,  n,  n, , , P( Ai )  (n  1) (b) P  Ai  , P( Ai ), (a) P  Ai  , , , , ,  i1  i1,  i1  i1, These results can be easily established by, using the Principle of mathematical induction., , , , , , , , , , P(A1  A2  A3  ...  An )  P(A1 )P( A2 / A1 )P(A3 / A1  A2 ), ... P(An / A1  A2  ...  An 1 ), Since, A1, A2 , ...., An 1 , An, events, therefore, , Conditional probability, Let A and B be two events associated with a, random experiment. Then, the probability of, occurrence of A under the condition that B has, already occurred and P(B)  0, is called the, conditional probability and it is denoted by, P(A/B)., Thus, P(A/B) = Probability of occurrence of, A, given that B has already happened., P( A  B) n( A  B), ., , , P(B), n(B), Similarly, P(B/A) = Probability of occurrence, of B, given that A has already happened., P( A  B) n( A  B), ., , , P( A), n( A), , Hence, P(A1  A2  ...  An )  P(A1 )P(A2 ).... P(An ) ., (2) Probability of at least one of the n, independent events : If p1 , p2 , p3 , ........, pn be the, probabilities of happening of n independent, events A1, A2 , A3 , ........, An respectively, then, (i) Probability of happening none of them,  P( A1  A2  A3 ......  An )  P( A1 ).P( A2 ).P( A3 )..... P( An ), , ., ,  (1  p1 )(1  p2 )(1  p3 )....(1  pn ), (ii) Probability of happening at least one of, them,  P( A1  A2  A3 ....  An )  1  P( A1 )P( A2 )P( A3 ).... P( An ), , ., ,  1  (1  p1 )(1  p2 )(1  p3 )...(1  pn ), (iii) Probability of happening of first event, and, not, happening, of, the, remaining,  P( A1 )P( A2 )P( A3 )..... P( An ), ,  p1(1  p2 )(1  p3 ).......( 1  pn ), , Total probability and Baye’s rule, , P( A1  A2  A3  ....  An )  P( A1 )P( A2 / A1 )P( A3 / A1  A2 ), , where P( Ai / A1  A2  ...  Ai1 ) represents the, conditional probability of the event Ai , given, , independent, , P( An / A1  A2  ...  An 1 )  P( An ), , P(A  B)  P(B) . P(A / B) , if P(B)  0., , .... P(An / A1  A2  ...  An 1 ) ,, , are, , P(A2 / A1 )  P(A2 ), P(A3 / A1  A2 )  P(A3 ), ....,, ,  Sometimes, P(A/B) is also used to denote the, probability of occurrence of A when B occurs., Similarly, P(B/A) is used to denote the probability of, occurrence of B when A occurs., (1) Multiplication theorems on probability, (i) If A and B are two events associated with, a, random, experiment,, then P(A  B)  P(A) . P(B / A) , if P(A)  0 or, (ii) Extension of multiplication theorem : If, A1 , A2 , ...., An are n events related to a random, experiment, then, , P ( B / A)  P ( B) ., , (1) The law of total probability : Let S be the, sample space and let E1 , E2 , ..... En be n mutually, exclusive and exhaustive events associated with, a random experiment. If A is any event which, occurs with E1 or E2 or …or En, then

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PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 8890244575, , P(A)  P(E1 ) P(A / E1 )  P(E2 ) P(A / E2 )  ...  P(En ) P( A / En ), ., (2) Baye’s rule : Let S be a sample space and, E1 , E2 , ..... En be n mutually exclusive events such, n, , that, , E, , i, ,  S and P(Ei )  0 for i = 1, 2, ……, n., , i 1, , We can think of (Ei’s as the causes that lead to, the outcome of an experiment. The probabilities, P(Ei), i = 1, 2, ….., n are called, prior, probabilities. Suppose the experiment results in, an outcome of event A, where P(A) > 0. We have, to find the probability that the observed event A, was due to cause Ei, that is, we seek the, conditional, probability, These, P(Ei / A) ., probabilities are called posterior probabilities,, given, by, Baye’s, rule, as, P(Ei ).P( A / Ei ), ., P(Ei / A)  n, ,  P( E ) P ( A / E ), k, , k, , k 1, , Binomial distribution, (1) Geometrical method for probability : When, the number of points in the sample space is, infinite, it becomes difficult to apply classical, definition of probability. For instance, if we are, interested to find the probability that a point, selected at random from the interval [1, 6] lies, either in the interval [1, 2] or [5, 6], we cannot, apply the classical definition of probability. In, this case we define the probability as follows:, , P{x  A} , , Now, since (X  x i ) is an event, we can talk of, , P(X  x i ) . If P(X  xi )  Pi (1  i  n) , then the system, of numbers.,  x1 x 2  x n , ,  is said to be the probability,  p1 p 2  pn , distribution of the random variable X., The expectation (mean) of the random, n, , variable X is defined as E( X ) , , Let X be the sum of numbers on the dice., Then X(12)  3, X(43)  7 , etc. Also, {X = 7} is the, event {61, 52, 43, 34, 25, 16}. In general, if X is, a random variable defined on the, sample space S and r is a real number, then {X =, r} is an event., If the random variable X takes n distinct values, x1 , x 2 , ...., x n , then {X  x1 } , {X  x 2 }, ...., {X  x n }, are mutually exclusive and exhaustive events., X = x1, X = x2, X = x4, , X = x3, , X = xn, , i i, , and the, , i 1, , variance of X is defined as, var( X ) , , n, , n, ,  p (x  E(X ))   p x, 2, , i, , 2, i i, , i, , i 1, ,  (E( X ))2 ., , i 1, , (3) Binomial probability distribution :, random variable X which takes values 0, 1, 2,, n is said to follow binomial distribution if, probability distribution function is given, P( X  r)  n Cr p r q n r , r  0, 1, 2, ....., n, , A, …,, its, by, , where p, q > 0 such that p + q = 1., The notation X ~ B(n, p) is generally used to, denote that the random variable X follows, binomial distribution with parameters n and p., We have P( X  0)  P( X  1)  ...  P( X  n) .,  n C0 p 0 q n  0  n C1 p1q n 1  ...  n Cn p n q n n  (q  p)n  1n  1, , Measure of region A, Measure of the sample space S, , where measure stands for length, area or, volume depending upon whether S is a onedimensional,, two-dimensional, or, threedimensional region., (2) Probability distribution : Let S be a sample, space. A random variable X is a function from, the set S to R, the set of real numbers., For example, the sample space for a throw of, a pair of dice is, {11, 12,  , 16, 21, 22 ,  , 26, S , , , , , 61, 62,  , 66 }, , p x, , Now probability of, (a) Occurrence of the event exactly r times, P( X  r)  n Cr q n r p r ., , (b) Occurrence of the event at least r times, P( X  r)  n Cr q n r p r  ...  p n , , n, , , , n, , CX p X qn X ., , X r, , (c) Occurrence of the event at the most r, times, P(0  X  r)  q n  n C1q n 1 p  ...  n Cr q n r p r , , r, , p, , X n X, , q, , X 0, , .,  If the probability of happening of an event in, one trial be p, then the probability of successive, happening of that event in r trials is p r .,  If n trials constitute an experiment and the, experiment is repeated N times, then the frequencies, of 0, 1, 2, …, n successes are given by, N.P(X  0), N.P(X  1), N.P(X  2), ...., N.P(X  n) ., (i) Mean, distribution, , and, , variance, , of, , the, , binomial, , The binomial probability distribution is

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PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 8890244575, X, P(X ), , 0, n, , 1, , C0 q n p 0, , n, , 2, , C1q n 1 p, , n, , ....., , envelopes, 1, 1, 1, 1, ,  ,  ...  (1)n ., 2! 3! 4 !, n!, (iv) Probability that exactly r letters are in right, envelopes, 1 1 1 1, 1 ,      .....  (1)n r, ., r!  2! 3! 4!, (n  r)!, , n, , C2q n  2 p 2 ..... nCnq 0 p n, , The mean of this distribution is, n, , n, ,  X p   X., , n, , i i, , i1, , C X q n  X p X  np ,, , X 1, , The variance of the Binomial distribution is, 2, and the standard deviation is,   npq, , , ,   (npq ) ., , probability of the occurrence of that event is, , (ii) Use of multinomial expansion : If a die has, m faces marked with the numbers 1, 2, 3, ….m, and if such n dice are thrown, then the, probability that the sum of the numbers, exhibited on the upper faces equal to p is given, by the coefficient of x p in the expansion of, , and the probability of non-occurrence of that event, , (x  x 2  x 3  ....  x m )n, ., mn, , and the probability of non-occurrence of that event, , (4) The poisson distribution : Let X be a, discrete random variable which can take on the, values 0, 1, 2,... such that the probability, function of X is given by, , , , f (x )  P( X  x ) , ,  x e , x!, , , x  0,1,2,...., , where  is a given positive constant. This, distribution is called the Poisson distribution, and a random variable having this distribution is, said to be Poisson distributed., , Independent events are always taken from, different experiments, while mutually exclusive, events are taken from a single experiment.,  Independent events can happen together while, mutually exclusive events cannot happen together.,  Independent events are connected by the word, “and” but mutually exclusive events are connected, by the word “or”.,  Number of exhaustive cases of tossing n coins, simultaneously (or of tossing a coin n times) = 2n.,  Number of exhaustive cases of throwing n dice, simultaneously (or throwing one dice n times) = 6n.,  Probability regarding n letters and their, envelopes : If n letters corresponding to n, envelopes are placed in the envelopes at random,, then, , is, , , If odds in favour of an event are a : b, then the, , b, ., ab, , If odds against an event are a : b, then the, , probability of the occurrence of that event is, is, , a, ab, , b, ab, , a, ., ab, , Let A, B, and C, Then, Verbal description of, event, , are three arbitrary events., Equivalent set theoretic, notation, , (i) Only A occurs, (ii) Both A and B, but, not C occur, (iii) All the three, events occur, (iv) At least one, occurs, (v) At least two, occur, , (i) A  B  C, (ii) A  B  C, , (vi) One and no, more occurs, , (vi), , (iii) A  B  C, (iv) A  B  C, (v), ( A  B)  ( B  C )  ( A  C ), , ( A  B  C ) ( A  B  C ), , , , (i) Probability that all letters are in right, envelopes  1 / n ! ., (ii) Probability that all letters are not in right, 1, envelopes  1  ., n!, (iii) Probability that no letter is in right, ,  ( A  B  C)  ( A  B  C), , (vii) Exactly two, of A, B and C, occur, , (vii), ( A  B  C )  ( A  B  C),  ( A  B  C)  ( A  B  C), , (viii) None occurs, (ix) Not more than, two occur, , (viii), , A  B C  A  B C, (ix), ( A  B)  (B  C)  ( A  C), ( A  B  C ), , (x) Exactly one of, A and B occurs, , (x) ( A  B )  ( A  B)

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8890244575, , PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, (c) Independent, (d) Dependent only on A, , Definition of various terms, 1., , 2., , Two coins are tossed. Let A be the event that, the first coin shows head and B be the event, that the second coin shows a tail. Two events, A and B are, [MP PET 1989], (a) Mutually exclusive, (b) Dependent, (c) Independent and mutually exclusive, (d) None of these, If P ( A1  A2 )  1  P( A1c ) P( A2c ) where c stands for, complement, then the events A1 and A 2 are, , Definition of probability, 1., , [MNR 1988; UPSEAT 2000], , 1, (a), 169, , (c), 2., , [MP PET 1989], , 3., , (a) Mutually exclusive (b) Independent, (c) Equally likely, (d) None of these, Two fair dice are tossed. Let A be the event, that the first die shows an even number and B, be the event that the second die shows an odd, number. The two event A and B are, , 3., , [IIT 1979], , 4., , 5., , 6., , (a) Mutually exclusive, (b) Independent and mutually exclusive, (c) Dependent, (d) None of these, A card is drawn from a pack of 52 cards. If A =, card is of diamond, B = card is an ace and, A  B  card is ace of diamond, then events A, and B are, (a) Independent, (b) Mutually exclusive, (c) Dependent, (d) Equally likely, If A and B are two independent events, then A, and B are, (a) Not independent, (b) Also independent, (c) Mutually exclusive (d) None of these, Let A, B, C be three mutually independent, events. Consider the two statements S 1 and S 2, S 1 : A and B  C are independent, , 4., , 3., , 4., , S 2 : A and B  C are independent, , Then, (a) Both S 1 and S 2 are true, , [IIT 1994], , (b) Only S 1 is true, (c) Only S 2 is true, (d) Neither S 1 nor S 2 is true, 7., , If P( A)  2 / 3 , P(B)  1 / 2 and P( A  B)  5 / 6 then, events A and B are, [Kerala (Engg.) 2002], (a) Mutually exclusive, (b) Independent, as, well, as, mutually, exhaustive, , Two card are drawn successively with, replacement from a pack of 52 cards. The, probability of drawing two aces is, , 5., , 1, 2652, , (b), , 1, 221, , (d), , 4, 663, , In a single throw of two dice, the probability of, getting more than 7 is, [MP PET 1991], (a), , 7, 36, , (b), , 7, 12, , (c), , 5, 12, , (d), , 5, 36, , The probability of drawing a white ball from a, bag containing 3 black balls and 4 white balls,, is, 3, 7, , (a), , 4, 7, , (b), , (c), , 1, 7, , (d) None of these, , A and B toss a coin alternatively, the first to, show a head being the winner. If A starts the, game, the chance of his winning is, (a) 5/8, (b) 1/2, (c) 1/3, (d) 2/3, If two balanced dice are tossed once, the, probability of the event, that the sum of the, integers coming on the upper sides of the two, dice is 9, is, [MP PET 1987], (a), , 7, 18, , (b), , 5, 36, , (c), , 1, 9, , (d), , 1, 6, , From a well shuffled pack of cards one card is, drawn at random. The probability that the, card drawn is an ace is, 4, 13, , (a), , 1, 13, , (b), , (c), , 3, 52, , (d) None of these, , A single letter is selected at random from the, word “PROBABILITY”. The probability that the, selected letter is a vowel is [MNR 1986; UPSEAT, 2000], , (a), , 2, 11, , (b), , 3, 11

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8890244575, (c), 6., , 4, 11, , 1, n!, , (c) 1 , , 8., , (b) 1 , 1, n, , (d) None of these, , (a), , 2, 25, , (b), , (c), , 11, 100, , (d) None of these, , 14., , 9, 100, , 15., , 1, 4, , 1, (b), 3, , 16., , If a dice is thrown twice, then the probability, of getting 1 in the first throw only is, 1, 36, , (b), , 3, 36, , (c), , 5, 36, , (d), , 1, 6, , 17., , Two cards are drawn one by one at random, from a pack of 52 cards. The probability that, both of them are king, is, [MP PET 1994], , 11., , (a), , 2, 13, , (b), , 1, 169, , (c), , 1, 221, , (d), , 30, 221, , 18., , A coin is tossed and a dice is rolled. The, probability that the coin shows the head and, the dice shows 6 is, , 12., , 1, (b), 12, , 1, (c), 2, , (d) 1, , 1, 4, , (b), , (c), , 3, 4, , (d) 1, , (c), , 1, 52, , (d) None of these, , Two dice are thrown simultaneously. The, probability of getting the sum 2 or 8 or 12 is, (a), , 5, 18, , (b), , 7, 36, , (c), , 7, 18, , (d), , 5, 36, , A dice is thrown twice. The probability of, getting 4, 5 or 6 in the first throw and 1, 2, 3, or 4 in the second throw is, , 7, 36, , (b), , 1, 3, , (d) None of these, , Two cards are drawn from a pack of 52 cards., What is the probability that at least one of the, cards drawn is an ace, (a), , 33, 221, , (b), , 188, 221, , (c), , 1, 26, , (d), , 21, 221, , One card is drawn from each of two ordinary, packs of 52 cards. The probability that at least, one of them is an ace of heart, is, (a), , 103, 2704, , (b), , 1, 2704, , (c), , 2, 52, , (d), , 2601, 2704, , A box contains 6 nails and 10 nuts. Half of the, nails and half of the nuts are rusted. If one, item is chosen at random, what is the, probability that it is rusted or is a nail, , (c), 19., , [MNR 1978], , 1, 2, , (b), , [MP PET 1992, 2000], , A coin is tossed twice. The probability of, getting, head, both, the, times, is, (a), , 1, 26, , 3, (a), 16, , [MP PET 1994; Pb. CET 2001], , 1, (a), 8, , 17, 2704, , (a), , (c), , (d) None of these, , (a), , From a pack of 52 cards two are drawn with, replacement. The probability, that the first is a, diamond and the second is a king, is, , (a) 1, , There are two childrens in a family. The, probability that both of them are boys is, , (c), , 10., , 1, n!, , From a book containing 100 pages, one page is, selected randomly. The probability that the, sum of the digits of the page number of the, selected page is 11, is, , 1, (a), 2, , 9., , 13., , (d) 0, , There are n letters and n addressed envelopes., The probability that all the letters are not kept, in the right envelope, is, (a), , 7., , PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 20., , 11, 16, , 5, (b), 16, , (d), , 14, 16, , The probability of getting at least one tail in 4, throws of a coin is [MNR 1983; Kurukshetra CEE 1998], 1, 16, , (a), , 15, 16, , (b), , (c), , 1, 4, , (d) None of these, , Three letters are to be sent to different persons, and addresses on the three envelopes are also, written. Without looking at the addresses, the

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8890244575, , PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , probability that the letters go into the right, envelope is equal to, , are 1/3, 1/4 and 1/5 respectively. The, probability that the question will be solved is, , [MNR 1982; MP PET 1990; Orissa JEE 2004], , 1, (a), 27, , 1, (b), 9, , 4, 27, , 1, (d), 6, , (c), 21., , 22., , 23., , (c), , 1, 18, , (b), , (c), , 1, 6, , (d) None of these, , The probability of getting a total of 5 or 6 in a, single throw of 2 dice is, (a), , 1, 2, , (b), , 1, 4, , (c), , 1, 3, , (d), , 1, 6, , The probability of a sure event is, (a) 0, (b) 1, (d), , (a) 1, (c), , 27., , 1, (a), 6, 1, (c), 3, , 28., , 1, 100, 1, (c), 20, , 5, 36, 1, (c), 6, , (a), , 30., , 1, 6, , 1, (b), 3, 5, (d), 6, , the probability of getting at least 2 tails, , 1, 8, 1, (c), 2, , (a), , 34., , 35., , 36., , 37., , 11, 36, 1, (d), 3, , (b), , 1, 4, , (d) None of these, , 1, 2, , (d) None of these, , Two dice are thrown simultaneously. What is, the probability of obtaining sum of the, numbers less than 11, (a), , 17, 18, , (b), , (c), , 11, 12, , (d) None of these, , 1, 12, , The probability that an ordinary or a non-leap, year has 53 sunday, is, [MP PET 1996], 1, 7, , (a), , 2, 7, , (b), , (c), , 3, 7, , (d) None of these, , A card is drawn at random from a pack of 52, cards., The probability that the drawn card is a, [AI CBSE 1981], court card i.e. a jack, a queen or a king, is, 3, 52, 4, (c), 13, , (a), , 38., , (b), , In a throw of a die, what is the probability of, getting a number less than 7, (a) 0, (b) 1, (c), , 1, 50, 1, (d), 10, , A problem of mathematics is given to three, students whose chances of solving the problem, , in, , [MP PET 1988], , (b), , Two dice are thrown simultaneously. What is, the probability of obtaining a multiple of 2 on, one of them and a multiple of 3 on the other, , 5, , 1, 1, (a), (b), 52, 26, 1, (c), (d) None of these, 18, [MP PET 1988], 33. In a simultaneous throw of three coins, what is, , (d) None of these, , From 10,000 lottery tickets numbered from 1, to 10,000, one ticket is drawn at random., What is the probability that the number, marked on the drawn ticket is divisible by 20, (a), , 29., , 1, (b), 2, , number, , A card is drawn from a well shuffled pack of, cards. The probability of getting a queen of, [MPclub, PET or, 1988], king of heart is, , [IIT 1980; Kurukshetra CEE 1998; DCE 2001], , (a) 0.936, (b) 0.784, (c) 0.904, (d) 0.216, In a single throw of two dice the probability of, obtaining an odd number is, , getting, , 32., , 1, 2, , The probability of happening an event A in one, trial is 0.4. The probability that the event A, happens at least once in three independent, trials is, , 3, 4, 3, (d), 5, , (b), , [MP PET 1988], , 1, 12, , (a), , 2, 3, 4, 5, , The probability of, throwing a dice is, , 31., , Two dice are thrown. The probability that the, sum of numbers appearing is more than 10, is, , (c) 2, 24., , (a), , (b), , 3, 13, , (d) None of these, , Two dice are thrown together. The probability, that sum of the two numbers will be a multiple, of 4 is, [MP PET 1990]

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8890244575, , 39., , (a), , 1, 9, , (b), , 1, 3, , (c), , 1, 4, , (d), , 5, 9, , 1, 5, 4, (c), 5, , (c), , 42., , 1, 3, 1, 2, , 1., , 2, 5, , 2., , 2, 3, 1, (d), 6, , (b), , 2, 13, , (b), , (c), , 4, 663, , (d) None of these, , 1, (b), 6, , (d) None of these, , 44., , (b), , (c), , 1, 4, , (d) None of these, , Three fair coins are tossed. If both heads and, tails appears, then the probability that exactly, one head appears, is, (a), (c), , 45., , 1, 12, , 1, 2, , (b), , 1, 2, , 1, (d), 3, , A bag contains 4 white, 5 black and 6 red, balls. If a ball is drawn at random, then what, is the probability that the drawn ball is either, white or red, , 52, , C6, , (d), , C3, , 52, , C6, , 1, 2, , (b), , 1, 16, , (d) None of these, , (b), , 10, 19, , (d) None of these, , A bag contains 3 red, 7 white and 4 black, balls. If three balls are drawn from the bag,, then the probability that all of them are of the, same colour is, 6, 71, 10, (c), 91, , 1, 6, , 3, 8, , C 3  26 C 3, , 26, , If out of 20 consecutive whole numbers two, are chosen at random, then the probability, that their sum is odd, is, , (a), , 6., , (b), , A man draws a card from a pack of 52 playing, cards, replaces it and shuffles the pack. He, continues this processes until he gets a card of, spade. The probability that he will fail the first, two times is, [MNR 1980], , 5, 19, 9, (c), 19, , [MNR 1988], , (a), , 26, , (a), , 5., , (d) None of these, , C6, , 9, 16, 9, (c), 64, , 4., , A man and a woman appear in an interview for, two vacancies in the same post. The, probability of man's selection is 1/4 and that, of the woman's selection is 1/3. What is the, probability that none of them will be selected, , 26, , (a), , Two dice are thrown together. If the numbers, appearing on the two dice are different, then, what is the probability that the sum is 6, , 1, 4, , Six cards are drawn simultaneously from a, cards. What is the probability, that 3 will be red and 3 black, , (c), 3., , (b), , [MPpack, PET 1988], of playing, , (a), , 2, 663, , 2, 15, , 1, 2, , Two cards are drawn at random from a pack of, 52 cards. The probability that both are the, cards of spade is, 1, 26, 1, (c), 17, , (d) None of these, , (a), , (c), , (b), , (a), , Two cards are drawn from a pack of 52 cards., What is the probability that one of them is a, queen and the other is an ace, , 5, (a), 36, , 43., , (b), , 4, 15, , Use of permutations and combinatons in probability, , The probability of getting a number greater, than 2 in throwing a die is, (a), , 41., , (a), , If in a lottary there are 5 prizes and 20 blanks,, then the probability of getting a prize is, (a), , 40., , PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , (b), , 7, 81, , (d) None of these, , If four persons are chosen at random from a, group of 3 men, 2 women and 4 children., Then the probability that exactly two of them, are children, is, [Kurukshetra CEE 1996; DCE 1999], , 10, (a), 21, 5, (c), 21, , 7., , 8, 63, 9, (d), 21, , (b), , A box contains 25 tickets numbered 1, 2, ......., 25. If two tickets are drawn at random then, the probability that the product of their, numbers is even, is

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8890244575, 11, 50, 37, (c), 50, , (a), , 8., , 10., , 11., , 12., , (d) None of these, , (c), 15., , 13, (b), 15, , (d) None of these, , 5, 11, , (b), , (c), , 4, 11, , (d) None of these, , Twenty tickets are marked the numbers 1, 2,, ..... 20. If three tickets be drawn at random,, then what is the probability that those marked, 7 and 11 are among them, , 1., , 2., , 1, 19, , (a), , 3, 190, , (b), , (c), , 1, 190, , (d) None of these, , If Mohan has 3 tickets of a lottery containing 3, prizes and 9 blanks, then his chance of, winning prize are, (a), , 34, 55, , (b), , (c), , 17, 55, , (d) None of these, , 3., , 21, 55, , 4., , A bag contains 3 white and 7 red balls. If a, ball is drawn at random, then what is the, probability that the drawn ball is either white, or red, , 7, 10, , (b), , 3, 10, , (d), , 10, 10, , 5., , A bag contains 4 white, 5 red and 6 black, balls. If two balls are drawn at random, then, the probability that one of them is white is, (a), , 44, 105, , (b), , (c), , 11, 21, , (d) None of these, , 11, 105, , 6., , A bag contains 6 red, 4 white and 8 blue balls., If three balls are drawn at random, then the, probability that 2 are white and 1 is red, is, (a), , 5, 204, , (b), , 7, 102, , 3, 68, , (d), , 1, 13, , A committee of five is to be chosen from a, group of 9 people. The probability that a, certain married couple will either serve, together or not at all, is, [CEE 1993], (a), , 1, 2, , (b), , 5, 9, , (c), , 4, 9, , (d), , 2, 9, , Odds in favour and odds against, Addition, theorem on probability, , 21, 55, , (a), , (c), , 14., , 13, 50, , A word consists of 11 letters in which there are, 7 consonants and 4 vowels. If 2 letters are, chosen at random, then the probability that all, of them are consonants, is, , (a) 0, , 13., , (b), , From a class of 12 girls and 18 boys, two, students are chosen randomly. What is the, probability that both of them are girls, 22, (a), 145, 1, (c), 18, , 9., , PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 7., , If the odds against an event be 2 : 3, then the, probability of its occurrence is, (a), , 1, 5, , (b), , (c), , 3, 5, , (d) 1, , 2, 5, , If the odds in favour of an event be 3 : 5, then, the probability of non-occurrence of the event, is, (a), , 3, 5, , (b), , 5, 3, , (c), , 3, 8, , (d), , 5, 8, , A card is drawn from a pack of 52 cards. A, gambler bets that it is a spade or an ace. What, are the odds against his winning this bet, (a) 17 : 52, (b) 52 : 17, (c) 9 : 4, (d) 4 : 9, An event has odds in favour 4 : 5, then the, probability that event occurs, is, (a), , 1, 5, , (b), , 4, 5, , (c), , 4, 9, , (d), , 5, 9, , For an event, odds against is 6 : 5. The, probability that event does not occur, is, (a), , 5, 6, , (b), , 6, 11, , (c), , 5, 11, , (d), , 1, 6, , In a horse race the odds in favour of three, horses are 1 : 2 , 1 : 3 and 1 : 4 . The probability, that one of the horse will win the race is, (a), , 37, 60, , (b), , 47, 60, , (c), , 1, 4, , (d), , 3, 4, , The odds against a certain event is 5 : 2 and, the odds in favour of another event is 6 : 5. If

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PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 8890244575, , both the events are independent, then the, probability that at least one of the events will, happen is, [RPET 1997], 50, 77, 25, (c), 88, , 52, 77, 63, (d), 88, , (a), , 8., , (b), , If odds against solving a question by three, students are 2 : 1, 5 : 2 and 5 : 3 respectively,, then probability that the question is solved, only by one student is [RPET 1999], , 18, (a), 595, 3, (c), 10, , 10., , 11., , 12., , 6, (b), 17, 2, (d), 7, , A party of 23 persons take their seats at a, round table. The odds against two persons, sitting together are, [RPET 1999], (a) 10 : 1, (b) 1 : 11, (c) 9 : 10, (d) None of these, If two events A and B are such that, P ( A  B) , , 3., , 5, ,, 6, , P ( AB ) , , If, , P ( A) , , (d) None of these, 1, 1, , P (B) , 2, 3, , P ( A  B) , , and, , 1, ,, 4, , then, , (a) 1, , (b) 0, , 1, (c), 2, , (d), , 1, 3, , If A and B are two events such that P ( A)  0 and, A, P (B)  1, then P   , B,  , , (d) None of these, , Three ships A, B and C sail from England to, India. If the ratio of their arriving safely are 2 :, 5, 3 : 7 and 6 : 11 respectively then the, probability of all the ships for arriving safely is, , 5, 6, , B, P  , A, , 24, (b), 56, , 31, (a), 56, 25, (c), 56, , 9., , 2., , (c), , [IIT 1982; RPET 1995, 2000; DCE 2000; UPSEAT, 2001], , A, (a) 1  P  , B, , A, (b) 1  P  , B, , 1 [Pb.,  P (CET, A  B2000], ), P (A), (d), P (B ), P (B ), In a single throw of two dice what is the, probability of obtaining a number greater than 7,, if 4 appears on the first dice, 1, 1, (a), (b), 3, 2, 1, (c), (d) None of these, 12, 1, If A and B are two events such that P ( A)  ,, 3, , (c), 4., , 5., , 1, 1, and P ( A )  , then the, 2, 3, , P (B ) , , 1, 4, , and P ( A  B) , , B, 1, , then P   , 5, A, 37, (b), 45, , events A and B are, 37, (a) Independent, (a), 40, (b) Mutually exclusive, 23, (c) Mutually exclusive and independent, (c), (d) None of these, 40, (d) None of these, 3, The probabilities of three mutually exclusive, 6., If A and B are two events such that P ( A)  ,, 8, events are 2/3, 1/4 and 1/6. The statement is [MNR 1987; UPSEAT 2000], 3, 5, A, (a) True, (b) Wrong, P (B ) , and P ( A  B)  , then P   , 4, 8, B, (c) Could be either, (d) Do not know, 2, 2, (b), 5, 3, 3, (c), (d) None of these, 5, If the events A and B are mutually exclusive, then, A, P  , B, , (a), , 7., , Conditional probability, Baye's theorem, 1., , Two dice are thrown. What is the probability that, the sum of the numbers appearing on the two, dice is 11, if 5 appears on the first, 1, (a), 36, , 1, (b), 6, , (a) 0, P ( A  B), (c), P ( A), 8., , (b) 1, P ( A  B), (d), P (B), , If A and B are two events such that A  B, then, B, P  , A, (a) 0, , (b) 1

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PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 8890244575, , 9., , 10., , (c) 1/2, (d) 1/3, If A and B are two independent events, then, A, P  , B, (a) 0, (b) 1, (c) P (A), (d) P (B), ,  AB , (c) P , 0,  A  B , , 16., ,  B  1, (c) P   ,  A  2, , 17., , [MP PET 2003], , 12., , (a), , 2, 5, , (b), , 3, 5, , (c), , 7, 10, , (d), , 19, 60, , For a biased die, the probabilities for different, faces to turn up are, Face :, 1, 2, 3, 4, 5, 6, Probabilit, 0.2 0.22 0.11 0.25 0.05 0.17, y:, The die is tossed and you are told that either face, 4 or face 5 has turned up. The probability that it, is face 4 is, 1, 1, (a), (b), 6, 4, (c), , 13., , 15., , (d) None of these, , 18., , 1, 3, , If A and B are two independent events such that, 1, 1, P ( A)  , P(B)  , then, 2, 5,  A 1, (a) P   , B 2, ,  A  5, (b) P , ,  AB 6, , A biased die is tossed and the respective, probabilities for various faces to turn up are, given below, Face :, 1, 2, 3, 4, 5, 6, Probabilit, 0.1 0.24 0.19 0.18 0.15 0.14, y:, If an even face has turned up, then the probability, that it is face 2 or face 4, is, (a) 0.25, (b) 0.42, (c) 0.75, (d) 0.9, If two events A and B are such that, P ( A c )  0 . 3, P ( B )  0 . 4, and, then, P( AB c )  0.5,, , [IIT 1994], , 19., , (d) None of these, , Three coins are tossed. If one of them shows tail,, then the probability that all three coins show tail,, is, 1, 1, (a), (b), 7, 8, 2, 1, (c), (d), 7, 6, , (d) All of the above, , P[B /( A  B c )] is equal to, , A pair has two children. If one of them is boy,, then the probability that other is also a boy, is, 1, 1, (a), (b), 4, 2, (c), , 14., , 5, 6, ,  A  3, P  ,  B 4, , [IIT 1989], , (a) A and B are independent (b), , c, , 11., ,  A 1, For two events A and B, if P( A)  P   , and, B 4, , B 1, P    , then,  A 2, , If E and F are independent events such that, 0  P(E)  1 and 0  P (F)  1, then, (a) E and F (the complement of the event F) are, independent, (b) E c and F c are independent,  Ec , E, (c) P    P  c   1, F, F , (d) All of the above, B, If 4 P(A)  6 P (B)  10 P (A  B)  1, then P   , A, , (d) All of the above, , 20., , (a), , 1, 2, , (b), , (c), , 1, 4, , (d) None of these, , 1, 3, , A letter is known to have come either from, LONDON or CLIFTON; on the postmark only the, two consecutive letters ON are legible. The, probability that it came from LONDON is, (a), , 5, 17, , (b), , 12, 17, , (c), , 17, 30, , (d), , 3, 5, , 0  P ( A)  1 ,, 0  P (B)  1, Let, P(A)  P(B)  P(A) P(B). Then, , and, , P( A  B) , [IIT 1995], , (a) P(B / A)  P(B)  P( A), (b) P( A c  B c )  P( A c )  P(B c ), (c) P( A  B)c  P( A c ) P(B c ), (d) P( A / B)  P( A), 21., , For a biased die the probabilities for different, faces to turn up are given below, Face :, Probabilit, y:, , 1, , 2, , 3, , 4, , 0.1, , 0.32, , 0.21, , 0.15, , 5, , 6, , 0.05 0.17, , [

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8890244575, , PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , The die is tossed and you are told that either face, 1 or 2 has turned up. Then the probability that it, is face 1, is, [IIT 1981], , 22., , (c), 27., , 5, 22, , (a), , 5, 21, , (b), , (c), , 4, 21, , (d) None of these, , 24., , 25., , (a), , 1, 5, , (b), , 3, 8, , (c), , 1, 3, , (d), , 2, 3, , There are 3 bags which are known to contain 2, white and 3 black balls; 4 white and 1 black balls, and 3 white and 7 black balls respectively. A ball, is drawn at random from one of the bags and, found to be a black ball. Then the probability that, it was drawn from the bag containing the most, black balls is, (a), , 7, 15, , (b), , (c), , 3, 4, , (d) None of these, , (c) P (E / F)  P (E / F )  1, (d) P (E / F )  P (E / F )  1, 28., , 3, 3, (b), 4, 8, 1, 1, (c), (d), 8, 2, A and B are two events such that P (A)= 0.8,, P(B)=0.6 and P( A  B)  0 .5, then the value of, P ( A / B) is, 5, 6, , (b), , 5, 8, ,  A 1, For two events A and B, if P( A)  P   , and, B 4, B 1, P    , then, A 2, , [MP PET 2003], , (a) A and B are independent (b),  B  1, (c) P   ,  A  2, , 29., , 5, 19, , In an entrance test there are multiple choice, questions. There are four possible answers to, each question of which one is correct. The, probability that a student knows the answer to a, question is 90%. If he gets the correct answer to, a question, then the probability that he was, guessing, is, 37, 1, (a), (b), 40, 37, 36, 1, (c), (d), 37, 9, A coin is tossed three times in succession. If E is, the event that there are at least two heads and F, is the event in which first throw is a head, then, E, P  , [MP PET 1996], F, , (a), , If E and F are the complementary events of, events, E, and, F, respectively, and, if, [IIT 1998], 0  P (F)  1, then, (b) P (E / F)  P (E / F )  1, , In a certain town, 40% of the people have brown, hair, 25% have brown eyes and 15% have both, brown hair and brown eyes. If a person selected, at random from the town, has brown hair, the, probability that he also has brown eyes, is, , 30., , 31., , 32., , 33., ,  A  3, P  ,  B 4, , (d) All of these, , Two cards are drawn one by one from a pack of, cards. The probability of getting first card an ace, and second a coloured one is (before drawing, second card first card is not placed again in the, pack), [UPSEAT 1999; 2003], 1, 5, (a), (b), 52, 26, 4, 5, (c), (d), 13, 221, One dice is thrown three times and the sum of the, thrown numbers is 15. The probability for which, number 4 appears in first throw, 1, 1, (a), (b), 18, 36, 1, 1, (c), (d), 9, 3, One ticket is selected at random from 100 tickets, numbered 00, 01, 02, ...... 98, 99. If X and Y, denote the sum and the product of the digits on, the tickets, then P (X  9 / Y  0) equals, 1, 2, (b), 19, 19, 3, (c), (d) None of these, 19, A man is known to speak the truth 3 out of 4, times. He throws a die and reports that it is a six., The probability that it is actually a six, is, 1, 3, (a), (b), 8, 5, 3, (c), (d) None of these, 4, A bag ‘A’ contains 2 white and 3 red balls and bag, ‘B’ contains 4 white and 5 red balls. One ball is, drawn at random from a randomly chosen bag, , (a), , (a), , 26., , (d) None of these, , (a) P (E / F)  P (E / F)  1, , [MNR 1988], , 23., , 9, 10

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PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 8890244575, , and is found to be red. The probability that it was, drawn from bag ‘B’ was, , (a), , [BIT Ranchi 1988; IIT 1976], , 34., , 35., , 5, 5, (a), (b), 14, 16, 25, 5, (c), (d), 18, 52, A bag X contains 2 white and 3 black balls and, another bag Y contains 4 white and 2 black balls., One bag is selected at random and a ball is drawn, from it. Then the probability for the ball chosen, be white is, [EAMCET 2003], 2, 7, (a), (b), 15, 15, 14, 8, (c), (d), 15, 15, Bag A contains 4 green and 3 red balls and bag B, contains 4 red and 3 green balls. One bag is taken, at random and a ball is drawn and noted it is, green. The probability that it comes bag B, 2, 2, (a), (b), 7, 3, 3, 1, (c), (d), 7, 3, , Binomial distribution, 1., , 2., , 16, 25, , 4, 8, (d), 25, 5, A fair coin is tossed n times. If the probability, that head occurs 6 times is equal to the, probability that head occurs 8 times, then n is, equal to, , 5., , [Kurukshetra CEE 1998; AMU 2000], , 6., , (a) 15, (b) 14, (c) 12, (d) 7, If three dice are thrown together, then the, probability of getting 5 on at least one of them is, 125, 215, (a), (b), 216, 216, (c), , 7., , 1, 216, , (d), , If a dice is thrown 7 times, then the probability, of obtaining, 5 exactly 4 times is, [DCE 2005], 4, , (a), , 7, , 1 5, C4    , 6 6, 4, , 1 5, (c)    , 6 6, , 8., , 91, 216, , 3, , 3, , 3, , (b), , 7, , 1 5, C4    , 6 6, 3, , 1 5, (d)    , 6 6, , 4, , 4, , If x denotes the number of sixes in four, consecutive throws of a dice, then P (x  4) is, , 8 coins are tossed simultaneously. The probability, of getting at least 6 heads is [AISSE 1985; MNR 1985; MP PET 1994] 1, 4, (a), (b), 229, 57, 1296, 6, (a), (b), 64, 256, 1295, (c) 1, (d), 37, 7, 1296, (c), (d), 64, 256, 9., A man make attempts to hit the target. The, In a box containing 100 eggs, 10 eggs are rotten., 3, . Then the, probability of hitting the target is, The probability that out of a sample of 5 eggs, 5, none is rotten if the sampling is with replacement, probability that A hit the target exactly 2 times in, is [MP PET 1991;, 5 attempts, is,  1 , (a)  ,  10 , , 5, , 1, (b)  , 5, , 5, , 5, , 5, , 9,  9 , (c)  , (d)  , 5,  ,  10 , If the probability that a student is not a swimmer, is 1/5, then the probability that out of 5 students, one is swimmer is, 4, , (a), (c), 4., , (b), , (c), , MNR 1986; RPET 1995; UPSEAT 2000], , 3., , 9, 25, , 5, , 4  1, C1    , 5 5, , 4 1,  , 5 5, , (b), , 5, , C1, , 4 1,  , 5 5, , 10., , 144, 625, , (b), , (c), , 216, 625, , (d) None of these, , If a dice is thrown 5 times, then the probability of, getting 6 exact three times, is, 125, 125, (a), (b), 3888, 388, , 4, , (c), 11., , 4, , (d) None of these, , In a box of 10 electric bulbs, two are defective., Two bulbs are selected at random one after the, other from the box. The first bulb after selection, being put back in the box before making the, second selection. The probability that both the, bulbs are without defect is [MP PET 1987], , 72, 3125, , (a), , 625, 23328, , (d), , 250, 2332, , The binomial distribution for which mean = 6 and, variance = 2, is, 2 1, (a)   , 3 3, , 6, , 2 1, (b)   , 3 3, , 9, , 1 2, (c)   , 3 3, , 6, , 1 2, (d)   , 3 3, , 9

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PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 8890244575, 12., , 13., , 14., , 15., , A dice is thrown ten times. If getting even number, is considered as a success, then the probability of, four successes is, (a), , 10, , (c), , 10, , 1, C4  , 2, , 4, , 1, C4  , 2, , 8, , (b), , 10, , (d), , 10, , 1, C4  , 2, , 6, , 1, C6  , 2, , 10, , If the mean and variance of a binomial variate X, are 2 and 1 respectively, then the probability that, X takes a value greater than 1, is, 4, 2, (a), (b), 3, 5, 15, 7, (c), (d), 16, 8, At least number of times a fair coin must be, tossed so that the probability of getting at least, one head is at least 0.8, is, (a) 7, (b) 6, (c) 5, (d) None of these, A biased coin with probability p, 0  p  1, of heads, , 20., , [DSSE 1981], , 21., , 22., , is tossed until a head appears for the first time. If, the probability that the number of tosses required, 2, is even is , then p , 5, (a), , 1, 2, , 1, (c), 4, , 16., , 17., , (b), , 1, 3, , 23., , (d) None of these, , The probability of a bomb hitting a bridge is, 1, and two direct hits are needed to destroy it., 2, The least number of bombs required so that the, probability of the bridge beeing destroyed is, greater then 0.9, is, (a) 8, (b) 7, (c) 6, (d) 9, If X follows a binomial distribution with, parameters n  6 and p. If 9 P (X  4)  P (X  2),, , 24., , then p , , 18., , 1, 3, , (b), , (c), , 1, 4, , (d) 1, , A die is tossed thrice. If getting a four is, considered a success, then the mean and variance, of the probability distribution of the number of, successes are, [DSSE 1987], 1 1, 1 5, (a) ,, (b) ,, 2 12, 6 12, 5 1, ,, (d) None of these, 6 2, A die is tossed twice. Getting a number greater, than 4 is considered a success. Then the variance, , (c), 19., , 1, 2, , (a), , of the probability distribution of the number of, successes is, [AISSE 1979], 4, 2, (a), (b), 9, 9, 1, (c), (d) None of these, 3, A die is thrown three times. Getting a 3 or a 6 is, considered success. Then the probability of at, least two successes is, , 25., , 2, 7, (a), (b), 9, 27, 1, (c), (d) None of these, 27, In a simultaneous toss of four coins, what is the, probability of getting exactly three heads, 1, 1, (a), (b), 2, 3, 1, (c), (d) None of these, 4, A coin is tossed successively three times. The, probability of getting exactly one head or 2 heads,, is [AISSE 1990], 1, 1, (a), (b), 4, 2, 3, (c), (d) None of these, 4, The items produced by a firm are supposed to, contain 5% defective items. The probability that a, sample of 8 items will contain less than 2, defective items, is, [MP PET 1993], , (a), , 27  19 , , , 20  20 , , (c), , 153, 20, , 7, ,  1 , , ,  20 , , (b), , 533  19 , , , 400  20 , , (d), , 35, 16, , 7, ,  1 , , ,  20 , , 6, , 6, , 3, ., 4, He tries 5 times. The probability that he will hit, the target at least three times is, 371, 291, (a), (b), 364, 464, 459, 471, (c), (d), 502, 512, A fair coin is tossed a fixed number of times. If, the probability of getting 7 heads is equal to that, of getting 9 heads, then the probability of getting, 3 heads is, 35, 35, (a) 12, (b) 14, 2, 2, 7, (c) 12, (d) None of these, 2, , The probability that a man can hit a target is

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PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 8890244575, , Definition of various terms, 1, , d, , 2, , b, , 6, , a, , 7, , a, , 3, , d, , 4, , c, , 5, , b, , 11, , d, , 12, , d, , 13, , d, , 14, , d, , 15, , b, , 16, , a, , 17, , c, , 18, , d, , 19, , b, , 20, , b, , 21, , c, , 22, , c, , 23, , a, , 24, , d, , 25, , a, , 26, , c, , 27, , c, , 28, , a, , 29, , d, , 30, , a, , Definition of probability, Definition of various terms, , 1, , a, , 2, , c, , 3, , a, , 4, , d, , 5, , c, , 6, , a, , 7, , c, , 8, , b, , 9, , a, , 10, , c, , 11, , c, , 12, , c, , 13, , b, , 14, , b, , 15, , c, , 16, , b, , 17, , b, , 18, , a, , 19, , a, , 20, , c, , 21, , a, , 22, , d, , 23, , b, , 24, , b, , 25, , b, , 26, , b, , 27, , b, , 28, , c, , 29, , b, , 30, , d, , 3., 4., , (d) They are mutually independent., (c) It is obvious., , 31, , c, , 32, , b, , 33, , c, , 34, , b, , 35, , c, , 5., , 36, , b, , 37, , b, , 38, , c, , 39, , a, , 40, , b, , (b) Since A  B and A  B are mutually exclusive, events such that A  ( A  B )  ( A  B), , 41, , d, , 42, , c, , 43, , a, , 44, , c, , 45, , d, , 1., 2., , (d) They are basically independent., (b) P(A1  A2 )  1  [1  P(A1 )][1  P(A2 )], ,  P( A1 )  P( A2 )  P( A1 ). P( A2 ) ., ,  P( A)  P( A  B )  P( A  B),  P( A  B )  P( A)  P( A  B)  P( A)  P( A)P(B), , Use of permutations and combinatons in probability, 1, , c, , 2, , c, , 3, , c, , 4, , b, , 5, , c, , 6, , a, , 7, , c, , 8, , a, , 9, , b, , 10, , a, , 11, , a, , 12, , d, , 13, , a, , 14, , c, , 15, , c, , Odds in favour and odds against, Addition, theorem on probability, 1, , c, , 2, , d, , 3, , c, , 4, , c, , 5, , b, , 6, , b, , 7, , b, , 8, , c, , 9, , a, , 10, , a, , 11, , a, , 12, , b, , 13, , c, , 14, , d, , 15, , d, , ( A, B are independent),  P( A  B )  P( A)(1  P(B))  P( A)P(B ), ,  A and B are also independent., 6., , (a) B  C is independent to A, so S 1 is true, , B  C is also independent to A, so S 2 is true., 7., , (a) P( A  B)  P(A)  P(B)  P( A  B), 5 2 1,    P( A  B)  P( A  B)  0, 6 3 2,  Events A and B are mutually exclusive., , Definition of probability, 2, , 1., , Conditional probability, Baye's theorem, 2., 1, , b, , 2, , c, , 3, , c, , 4, , b, , 5, , a, , 6, , a, , 7, , a, , 8, , b, , 9, , c, , 10, , d, , 11, , a, , 12, , c, , 13, , c, , 14, , a, , 15, , d, , 16, , d, , 17, , c, , 18, , c, , 19, , b, , 20, , cd, , 21, , a, , 22, , b, , 23, , a, , 24, , b, , 25, , a, , 26, , a, , 27, , ad, , 28, , d, , 29, , c, , 30, , a, , 31, , b, , 32, , a, , 33, , d, , 34, , c, , 35, , c, , Binomial distribution, , 1,  4 , (a) Required probability    , ., 169,  52 , , (c) Required probability is, P(getting 8)  P(9)  P(10)  P(11)  P(12), , , 5, 4, 3, 2, 1, 15, 5, , , , , , , ., 36 36 36 36 36 36 12, , 3., , (a) Required probability , , 4., , (d) The chance of head , , d, , 2, , d, , 3, , b, , 4, , b, , 5, , b, , 6, , d, , 7, , a, , 8, , a, , 9, , a, , 10, , b, , 1, 1, and not of head , 2, 2, , Since A has first throw, he can win in the, first, third, …,  Probability of A’s winning, 2, , 1, , 4, ., 7, , , , 4, , 1 1 1 1 1,    .    .  ........, 2 2 2 2 2

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PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 8890244575, , 3, , , , sum twelve can be found in one way i.e.,, , 5, , 1 1, 2, 1,        .........  ., 2 2, 3, 2, , (6, 6)., , (c) Required probability , , 4, 1,  ., 36 9, , 6., , (a) Required probability , , 4, 1, ., , 52 13, , 7., , c) Since there are one A, two I and one O,, hence, the, required, probability, , 17., , (b) Let P(A) and P(B) be the probability of the, events then P( A and B)  P( A). P(B) , , 18., , 1, , (b) Required probability is 1  P (All letters in, , 19., , 1, right envelope)  1 , n!, , 9., , {As there are total number of n ! ways in, which letters can take envelopes and just, one way in which they have corresponding, envelopes}., (a) Favourable ways {29, 92, 38 , 83, 47, 74 , 56, 65}, Hence required probability , , 10., , 8, 2, , ., 100 25, , (c) To be both boys the probability, 11 1,     ., 22 4, , 11., , (c) Probability of getting 1 in first throw , , 663  564, 99, 33, 48 47, ., , , , ., 52 51, 13 . 51, 13 . 51 221, , (a) Required probability is 1  P (no ace of, heart), 1, , 20., , 51 51 (52  51), 103, ., , , ., 52 52, 52 . 52, 2704, , (c) Total rusted items, nails  3 ., , Both, , independent, , required probability , 12., , events,, , so, , 1, 2, , 22., , (d) Total no. of ways placing 3 letters in three, envelops  3 !, out of these ways only one, way is correct., , (b) Required probability , , 24., , (b) Total number of ways  36, Favourable numbers of cases are, (4, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)  9, , 4, 52, , Hence the required probability , 25., , (b) It is obvious., , 26., , (b) Here P( A)  0 . 4 and P( A )  0 .6, , 4, 3, 1, , , Hence required probability , ., 52 51 221, , (b) Required probability      , , 14., , 11 1, (b) Required probability       ., 22 4, , 15., , (c) Required probability  P(Diamond ). P(king), ,  (0 .6)3, , Thus required Probability  1  (0.6)3  0.784 ., (b) Required probability , , 28., , (c) Number of tickets numbered such that it is, divisible by 20 are, , 10000,  500 ., 20, , Hence required probability =, , (b) The sum 2 can be found in one way i.e., The sum 8 can be found in five ways i.e., (6, 2), (5, 3), (4, 4),(3, 5),(2, 6) . Similarly the, , 1, ., 2, , 27., , 13 4, 1, ., , ., 52 52 52, , (1, 1), , 9, 1,  ., 36 4, , Probability that A does not happen at all, , 1, ., 12, , 13., , 16., , 1, 1,  ., 3! 6, , 2 1 1, , ., 36, 12, , 23., , and probability of also second to be a king, , , , 15, ., 16, , (1 4), (2, 3), (3, 2),, , 3, , 51, , 11, 26, , 4, , (a) Required probability  1    , , 1 5, 5,  , ., 6 6 36, , (c) Probability of first card to be a king , , 38, 11, , ., 6  10 16, , Hence the required probability , , the, , unrusted, , 21., , 1, 6, , 5, 6, , are, ,  3  5  8;, ,  Required probability , , Probability of not getting 1 in second, throws , , 1 2 1,   ., 2 3 3, , (a) Required probability is 1  P (no ace), , 1  2 1, 4, , , ., 11, 11, , 8., , 7, ., 36, , Hence required probability , , 5., , 29., , 500, 1, , ., 10000, 20, , (b) Favourable cases for one are three i.e. 2, 4, and 6 and for other are two i.e. 3, 6.

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PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 8890244575, Hence, , required, , probability, ,  3  2 , 1  11,  , 2, , 36, 36, , ,  36, , {As same way happen when dice changes, numbers among themselves}, 30. (d) The probability of students not solving the, problem, , are, , 1, , 1 2,  ,, 3 3, , 1, , 1 3, , 4 4, , Therefore the probability that the problem, is not solved by any one of them, , , (d) Required probability , , 42., , I, ,  5,, (c) Obviously numbers will be  4 ,,  2,, ,  1,, , and, , 1 4, 1 , 5 5, , required probability , 43., , 2 3 4 2,   , 3 4 5 5, , Hence the, , 1, ., 6, , (c) Required probability , , 32., , (b) The probability of card to be queen of club, , 44., , 36., , 1, 1, 2, 1, , , , ., 52 52 52 26, , S  HHH , HHT ....... TTT ., , 4 1,  ., 8 2, , (b) It is obvious., (c) Favourable cases to get the sum not less, than 11 are {(5, 6), (6, 6), (6, 5)}  3, Hence favourable cases to get the sum less, than 11 are (36  3)  33 . So required, 33 11, , ., 36 12, , (b) In a non-leap year, we have 365 days i.e.,, 52 weeks and one day. So, we may have, any day of seven days. Therefore, 53, , Use of permutations and combinations in probability, 13, , C2, , 52, , C2, , 26, , C3 ., , 13 . 12, 1, , ., 52 . 51 17, , 1., , (c) Required probability , , 2., , (c) Required probability , , 3., , (c) The required probability is given by, 39, 52, , C1, C1, , , , 39, 52, , C1, C1, , , , 13, , C1, , 52, , C1, , , , 52, , , , 26, , C3, , C6, , ., , 3 3 1, 9,   , ., 4 4 4 64, , Hence required probability , , (b) The total number of ways in which 2, integers can be chosen from the given 20, integers 20 C 2 ., , Hence required probability , , The sum of the selected numbers is odd if, exactly one of them is given and one is odd.,  Favourable, number, of, outcomes, , 37., , (b) Court cards are king, queen and jack, , 38., , 12, 3, , ., 52 13, (c) S  {(3,1), (2, 2), (1, 3), (6, 2), (5, 3), (4, 4), (3, 5), (2, 6), (6, 6)}, , 40., , 45., , (c) Total ways are 8 and favourable ways are 4, , 1, Sunday, required probability = ., 7, , 39., , 3 1,  ., 6 2, 4, (d) Probability for white ball P(W ) , 15, 6, Probability for red ball P(R ) , 15, Probability (white or red ball)  P(W )  P(R), 4, 6, 10 2, , , ,  ., 15 15 15 3, , Hence required probability , , the required probability , , probability , , (c) Since both heads and tails appears, so, n(E)  {HTT , THT , TTH }, , 1, ., 52, , Hence probability , , 3 2 1,   ., 4 3 2, , n(S )  {HHT , HTH , THH , HTT , THT , TTH }, , Both are mutually exclusive events, hence, , 34., 35., , (a) Let E1 be the event that man will be, selected and E 2 the event that woman will, be selected. Then, , So, P(E1  E 2 )  P(E1 )  P(E 2 ) , , and also probability of card to be a, , king of heart is, , 33., , 4, 2, ., , 6 . 5 15, , 1 3, 1, 1, so P(E1 )  1   and P(E 2 ) , 3, 4, 4 4, 2, So P(E 2 ) , 3, Clearly E1 and E 2 are independent events., , probability that problem is, , 31., , 1, 52, , II, , 1, 2  . Hence, , 4, , 5, , P ( E1 ) , , 2 3, solved  1   ., 5 5, , is, , 4 .4, 8, 2, ., 52 . 51, 663, , 41., , 5, 1,  ., 25 5, 4 2, (b) Required probability   ., 6 3, , (a) Required probability , , 9, 1,  ., 36 4, , 4., , 10 C1 10 C1, ,  Required probability , , 10, , C1 10 C1, 20, , C2, , , , 10, ., 19

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PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 8890244575, 5., , (c) Required probability , , , 6., , 3, , C3 7 C3  4 C3, 14, , ace , , C3, , 1  35  4, 40, 10, , , ., 14 . 13 . 2, 14 . 26 91, , (a) Total number of ways  9 C 4 , 2 children are, chosen in, , 4, , 4, , Hence required probability =, , C 2 5 C 2, 9, , C4, , , , 10, ., 21, , 13, 25, , C2, 13 . 12 37, 1, , ., 25, . 24 50, C2, , (c) Required probability , , 4, 4,  ., 4 5 9, , 5., , (b) Required probability , , 6, 6, , ., 6  5 11, , 6., , (b) Probabilities of winning the race by three, horses are, , 12, , C2, , 30, , C2, , 9., , (b) Required probability , , 10., , (a) 7, 11 have always to be in that group of, three, therefore 3rd ticket may be chosen, in 18 ways., Hence, required, probability, is, , 7, , , , against A are 5 : 2 , therefore P( A) , , therefore P(B) , , C2, 7 .6, 21, , , ., C2 11 . 10 55, ,  9 C 3  84, , The required probability  1  P( A ) P(B ), 2 , 6  52, ,  1  1   1   , ., 7 , 11  77, , , (c) The probability of solving the question by, , 8., , P( A) , , 13., 14., 15., , (d) Required probability , (a) Required probability , (c) Required probability , (c) Required probability , , C1  C1, 7, , 10, , 4, , C1 . 11, 15, , 4, , C1, , , C2, , C 2  6 C1, 18, , 7, , C3, , 9, , C5, , C3, , , , , , , 7, , C5, , 9, , C5, , 3, ., 68, , , 56, 4,  ., 126 9, , favour and odds against, Addition theorem on, probability, (c) Required probability , , 2., , (d) Required probability  1 , , or A B C or A B C ), ,  P( A) P(B ) P(C )  P( A ) P(B) P(C )  P( A ) P(B ) P (C), , , , , , 1 5 5 2 2 5 2 5 3, 25  20  30, 25, . .  . .  . . , , ., 3 7 8 3 7 8 3 7 8, 168, 56, , (a) We have ratio of the ships A, B and C for, arriving safely are 2 : 5, 3 : 7 and 6 : 11, respectively., The probability of ship A for arriving safely, , 9., , 2, 2, , 25 7, , Similarly, for B , , 3, 3, , 3  7 10, , and for C =, , 6, 6, , 6  11 17, ,  Probability of all the ships for arriving, , 3, ., 5, , 1., , 3, 8, , Then probability of question solved by only, one student, , 10, ., 10, , 4 . 11 . 2, 44, , ., 15 . 14, 105, , and, , 2, 1, 3, ; P(B)  ; P(C ) , 7, 8, 3, ,  P( A B C, , 3, , 1 2, ,, 3 7, , respectively., , 84, 34, , ., Hence required probability  1 , 220 55, , 12., , 6, ., 11, , these three students are, , (a) Mohan can gets one prize, 2 prizes or 3, prizes and his chance of failure means he, get no prize., Number of total ways 12 C 3  220, Favourable number of ways to be failure, , 2, ., 7, , The odds in favour of B are 6 : 5 ,, , 11, , 18 . 3 . 2, 18, 3, ., , , 20, 20 .19 .18 190, C3, , 1 1 1 47,   , ., 3 4 5 60, , (b) Let A and B be two given events. The odds, , 12  11, 22, , ., 30  29 145, , (a) Required probability =, , 1 1, 1, ,, and ., 3 4, 5, , Hence required probability , 7., , 8., , 11., , 4., , (c) Required probability is 1  P, (Both odd numbers are chosen), 1, , 16, 4, . Hence odds in favour is 4 : 9., , 52 13, , So the odds against his winning is 9 : 4 ., , C 2 ways and other 2 persons, , are chosen in 5 C 2 ways., , 7., , (c) Probability of the card being a spade or an, , 3., , safely, , =, , 3 5,  ., 8 8, , 10., , 2 3, 6, 18, , , , ., 7 10 17 595, , (a) Required probability , , (21) ! 2 ! 1, 1, , , ., (22) !, 11 1  10

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PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 8890244575, , 11., ,  Odds against = 10 : 1., (a) We have P( A  B)  P( A)  P(B)  P( AB ), ,  P(E / F)  P(E c / F c )  1., , 5 1, 1, 4 2,    P(B)   P(B)  , 6 2, 3, 6 3, 1 2 1, Thus, P( A). P(B)     P( AB ), 2 3 3, , 12., , (a) P , , 12., , (c) Let A be the event that face 4 turns up and, B be the event that face 5 turns up then, P( A)  0 .25 , P(B)  0 . 05 . Since A and B are, mutually, exclusive,, so, P( A  B)  P( A)  P(B)  0 .25  0 .05  0 .30 ., , Hence events A and B are independent., (b) Since here P( A  B  C)  P( A)  P(B)  P(C), , , , A , , which is equal to,  AB, , 2 1 1 13, , which is greater than 1.,   , 3 4 6 12, , , , We have to find P , , Hence the statement is wrong., , P, , Conditional probability, Baye's theorem, 1., , (b) Since we are given that 5 appears on first, die so to get sum 11, six must be on the, second and hence, the required probability, , , (c) P  , B, , 4., , (b) Since 4 has appeared on the first, so we are, required 4 or 5 or 6 on second dice., , probability of a boy in two, , The probability that the second child is also, boy is, , P( A  B) P( A  B) 1  P( A  B), , , ., P(B ), P(B ), P(B ), , Hence required probability , , 3 1,  ., 6 2, , P ( A  B) , , 14., , 23,  B  1  P( A  B) 1  60, 37 3 37, (a) P   , , ,  , ., 1, 60, 2 40, P( A ), A, 1, 3, P( A  B) (3 / 8 )  (5 / 8)  (3 / 4 ) 2, (a) P( A / B) , ,  ., P(B), (5 / 8), 5,  A  P ( A  B), , P (B), B, , (a) P , , 1, 4, , (a) Let E be the event in which all three coins, shows tail and F be the event in which a, coin shows tail.,  F  {HHT , HTH , THH , HTT , THT , TTH , TTT }, and E  {TTT } ., P(E  F) 1,  ., P(E), 7, 1, (d) P( A / B)  P( A) as independent event  ., 2, P[ A  ( A  B)], P{ A /( A  B)} , P( A  B), , Required probability  P(E / F) , , 15., , 0,  A, 0 ., ,  B  P (B), , So, P( A  B)  0 . Hence P , , {Since A  ( A  B)  A  [ A  B  A  B],  A  A  B  A  B  a}, , (b) Since A  B  A  B  B  A  A, , 1, 1, P ( A), 5,  A , 2,  P, ,  2 , , 6, 6,  A  B  P ( A  B) 1  1  1, 2 5 10, 10,  AB , and similarly P , .,  A  B , ,  B  P (B  A ) P ( A ), , 1., , P( A), P ( A),  A, , Hence P , ,  A  P ( A  B ) P ( A). P (B ), ,  P ( A) ., , P (B ), P(B), B, , 9., , (c) P , , 10., , (d) P(E  F)  P(E). P(F), , P( A  B) 1 / 4 1, ,  ., P( A), 3/4 3, , We have to find P(B / A) , , Since A and B are mutually exclusive., , 8., , P (A ) , , Favourable cases are BB, BG, GB  3, , 3., , 7., , (c) Let, , 3, children , 4, , P( A  B) 1 / 4 1, ,  ., P( A), 1/2 2, , (c) P(B / A) , , 6., , [ A  ( A  B)], P( A), 0.25 5, , ,  ., P( A  B), P( A  B) 0.30 6, , Because cases are BB, BG, GB, GG  4, , 2., , 5., , 13., , 1, ., 6, ,  A, ,  B  P( A  B) (1 / 10 ) 2, ,  ., , P ( A), (1 / 4 ), 5,  A, , 11., , Now,, , 8., ,  A, , B, , (d) A, B are independent as P( A)  P , , P(E  F )  P(E)  P(E  F)  P(E)[1  P(F)]  P(E).P(F ), c, , c, , and, P(E c  F c )  1  P(E  F)  1  [P(E)  P(F)  P(E  F),  [1  P(E)][1  P(F)]  P(E c ) P(F c ), , Also P(E / F)  P(E) and P(E c / F c )  P(E c ), , 1 3,  A , P   1  , as A, B are independent, B, 4, 4,  , ,  A, B are independent., 1 1,  B , P    P(B )  1   ., , A, 2 2,  

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PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 8890244575, 9., , (c) Let A be the event that even face turns up, and B be the event that it is 2 or 4., Then, , E 2  Bag II is chosen and E 3  Bag III is, , chosen., 1, , P(B)  P(2)  P(4 )  0 .24  0 . 18  0 . 42, , So, P(B / A) , 10., , , , , , 11., ,  A, P ,  E2, , P(B  A) P(B) 0.42, , ,  0.75 ., P( A), P( A) 0.56, , (c) P[B /( A  B c )] , , , , 16., , (b) We define the following events :, A1 : Selecting a pair of consecutive letter, from the word LONDON., A2 : Selecting a pair of consecutive letters, from the word CLIFTON., E : Selecting a pair of letters ‘ON’., ,  7, ,  10, , , P(E3 )P( A / E3 ), 7, ., , P(E1 )P( A / E1 )  P(E2 )P( A / E2 )  P(E3 )P( A / E3 ) 15, , (b) We define the following events :, A1 : He knows the answer., E : He gets the correct answer., , Then P( A1 ) , ,  E, 9, 9, 1, , P( A 2 )  1 , , , P , 10, 10 10,  A1, , Then P( A1  E)  ; as there are 5 pairs of, ,  E  1, , P , ,  A2  4, , consecutive letters out of which 2 are ON., ,  Required probability, , there, , are, , 6, , pairs, , 17., , It means A and B are independent events, so A c and B c will also be independent., Hence, P( A  B)c  P( A c  B c )  P( A c ) P(B c ) (Demorgan’s, law), As A is independent of B, hence, P ( A / B)  P ( A) ,, , (a) S  {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT, n (E)  4, n (F)  4 and n (E  F)  3, , 2, P( A1  E), 12,  A1 , 5, P,  , , , ., 2, 1, E, P, (, A, , E, ), , P, (, A, , E, ), 17, , , 1, 2, , 5 6, , (c, d) Since P( A  B)  P(A) P(B), ,  E  P ( E  F) 3 / 8 3,  P  , ,  ., P(F), 4 /8 4, F,  A  P( A  B) 0 . 5 5, ,  ., , P (B), 0 .6 6, B, , 18., , (a) P , , 19., , (a, d) P(E / F)  P(E / F) , , , (a) Required probability , , 14., , (b) P( A) , , [ E  F and E  F are disjoint], , , 0 .1, 0 .1, 5, , , ., 0 .1  0 .32 0 .42 21, , P{(E  E )  F} P(F), , 1, P(F), P (F ), , Similarly we can show that (b) and (c) are, not true while (d) is true., , 40, 25, 15, , P (B ) , and P( A  B) , 100, 100, 100, ,  E  P(E  F ) P(E  F ) P(F ), E, P   P  , , , 1, F, P(F), P(F), F, P(F ),  ,  , , 3,  B  P( A  B) 15 / 100, ,  ., , A, P, (, A, ), 40, /, 100, 8,  , , So P , , (a) Consider the following events :, A  Ball drawn is black; E1  Bag I is, chosen;, , P ( E  F)  P ( E  F), P (F), , P{(E  F)  (E  F)}, P (F), , { P(A  B)  P(B)P(A / B)} ., , 13., , ,   1,, , , , P( A 2 )P(E / A 2 ), 1, A ,  P 2  , , ., E, P, (, A, ), P, (, E, /, A, ), , P, (, A, ), P, (, E, /, A, ), 37, , , 1, 1, 2, 2, , of, , consecutive letters of which one is ON.,  The required probability is, , 15., ,  A, P ,  E3, , A2 : He does not know the answer., , 2, 5, , 12., ,  1,  ,,  5, , , Required probability  P  3 ,  A , , 0 .7  0 .5 1, P( A)  P( A  B c ), ,  ., 0 .8, 4, P( A)  P(B c )  P( A  B c ), , as, , 3, , E , , P(B  ( A  B c )), P( A  B c ), , P ( A  B), P( A)  P(B c )  P( A  B c ), , 1, P ( A 2  E)  ;, 6, ,  A , , Then P(E1 )  (E 2 )  P(E 3 )  , P   ., 3  E1  5, , P( A)  P(2)  P(4 )  P(6)  0 .24  0 .18  0 .14  0 .56, , 20., , 1, P(B  A) 1, B 1,   P ( B  A) ,  , 8, P( A), 2,  A 2, , (d) P , , 1, P( A  B) 1,  A 1, P  ,  P (B) , , , 2, P(B), 4, B 4

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PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 8890244575, , P( A  B ) , , 1,  P( A). P(B), 8, , 1 3, , 3, 6 4, ,  ., 1 3 5 1, 8,   , 6 4 6 4, ,  Events A and B are independent.,  A   P( A   B) P( A ) P(B) 3, , , , P (B), P(B), 4, B, , Now, P , , 25., ,  B'  P(B' A' ) P(B' ) P( A' ) 1, ,  ., , P( A ' ), P( A ' ), 2,  A' , , and P , 21., , (c) P(E1 ) , , E, 4, 1, , P 2, , 52 13,  E1, ,  15, 5, ,  51  17, , , E, P(E1  E 2 )  P(E1 ).P  2,  E1, , 22., , Since both the bags are equally likely to be, selected, we have P(E1 )  P(E2 ) , ,  1 5, 5, ,  13 . 17  221 ., , , Hence by Bay’s theorem, we have, P ( E 2 / E) , ,  A  n ( A  B), , n(B), B, , But P , , 26., , When n ( A  B) and n(B) respectively denote, the number of digits in A  B and B., Now n (B)  36 , because first throw is of 4., So another two throws stop by 6  6  36, types. Three dices have only two throws,, which starts from 4 and give sum 15 i.e.,, (4, 5, 6) and (4, 6, 5)., , 27., , n (B)  36 ;, , (b) Event (Y  0 ) is {00, 01, 09, 10, 20, .......... 90}, , 24., , P(Y  0), , 1, 5, 3, , P(E )  , P( A / E) , 6, 6, 4, 1, P( A / E ) , 4, , P(E).P( A / E), P(E).P( A / E)  P(E ).P( A / E ), , Probability of picked bag B P(B) , , 1, 2, , and, , Total probability of green ball =, , 2 3, 1, , , 7 14, 2, , Probability of fact that green ball is drawn, from bag B, G, P  , B, , From Baye’s theorem,, P(E / A) , , 1, 2, , 3, G 1 3,  P(B).P     ,  B  2 7 14, , 2, ., , 19, , (a) Let E denote the event that a six occurs, and A the event that the man reports that, it is a ‘6’, we have, P( E) , , Probability of picked bag A P( A) , , Probability of green ball picked from bag B, , Hence required probability,  P(X  9) /(Y, , (c) It is based on Baye’s theorem., , G 1 4 2,  P( A).P     ,  A 2 7 7, , 19, 2, P(Y  0) , and P( X  9)  (Y  0) , 100, 100, ,  0), , 1 2 1 2, 8,    , ., 2 5 2 3 15, , Probability of green ball picked from bag A, , Also (X  9)  (Y  0)  09, 90, we have, , P(X  9)  (Y,  0) , , (c) Let A be the event of selecting bag X, B be, the event of selecting bag Y and E be the, event of drawing a white ball, then, P ( E / A)  2 / 5, P( A)  1 / 2, P(B)  1 / 2 ,, P ( E / B)  4 / 6  2 / 3 ., P(E)  P( A)P(E / A)  P(B)P(E / B) , , 2, 1,  A, ,  P , .,  B  36 18, , 23., , P( E 2 ) P( E / E 2 ), P(E1 ) P(E / E1 )  P(E 2 ) P(E / E 2 ), , 1 5, ., 25, 2 9, , , ., 1 3 1 5 52, .  ., 2 5 2 9, , A, ., B, , B. So we have to find P , , n ( A  B)  2 ,, , 1, 2, , Also P(E / E1 )  3 / 5 and P(E / E2 )  5 / 9., , (a) We have to find the bounded probability to, get sum 15 when 4 appears first. Let the, event of getting sum 15 of three thrown, number is A and the event of apearing 4 is, , So,, , (d) Let E1 be the event that the ball is drawn, from bag A, E 2 the event that it is drawn, from bag B and E that the ball is red.We, have to find P(E2 / E) ., , G, 1 3, P(B)P  , , 3, B, 2 7, ,  ., 1, 4, 1, 3, 7, G, G,   , P( A)P    P(B)P  ,  A, B 2 7 2 7, , Binomial distribution, 1., , (d) The required probability

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PHOTON INSTITUTE UJJAIN, NDA, X-GROUP Probability, , 8890244575, 6, , 2, , 7, , 8, , 3, , 37, 1 1, 1 1, 1, ., 8 C6   .   8 C7   .   8 C8   , 256, 2 2, 2 2, 2, , 2., , (d) Let P(fresh egg ) , , 90, 9, , p, 100 10, , 1 5, 6 6, , (b) Required probability  5 C 3     , , 11., , (d) np  6, npq  2  q , , 10, 1, P(rotten egg ) , ,  q ; n  5, r  5, 100 10, , 0, , 5, , 4 1 , 5 5 , , 12., , 4, , 13., , 1 1, n C6    , 2 2, , head, , 6, , 14., , parameters n and p , , times, , n 8, , 8, , n 8, ,  1 n C0 P 0 (1  P)n  0 .8, n, , 6, , n, , 6., , n 6, , 1 1, n C8    , 2 2, , 15., , This shows that the least value of n is 3., (b) Let X denotes the number of tosses, required. Then P(X  r)  (1  p)r 1 . p, for, , C 6  n C 8  (n  6)(n  7)  56  n  14 ., , r  1, 2, 3 ......, , Let E denote the event that the number of, tosses required is even., Then P(E)  P[(X  2)  ( X  4 )  ( X  6)  ........], , (d) Required probability, 2, , 2, , 3, , 0, , 91, 15, 1 5, 1 5,  3 C1      3 C 2     3 C3     , ., 6, 6, 6, 6, 6, 6, 216,   ,    ,    , 4, , P(E)  P( X  2)  P( X  4 )  P( X  6)  ......, , 3, , 7., , 1 5, (a) Required probability  7 C 4     ., 6 6, , 8., , (a) Probability of coming ‘six’ in one throw is, , P(E)  (1  p)p  (1  p)3 p  (1  p)5 p  ....... , , But we are given that P( E) , , 1, 6, , p, , Hence required probability is given by, 4, , 0, , 1, 1 5,  4 C 4   .  , ., 1296, 6 6, , 9., , (a) Probability of success ( p) , , 16., 3, 2, q 1 p , 5, 5, , Hence the probability of 2 hits in 5, attempts, 2, , 3, , 144, 3 2, 5 C 2     , ., 5, 5, 625,    , , n, , 1, 1, 1,     0.2      2 n  5, 5, 2, 2, , occurs 8 times  n C 8    , 1 1,  nC6    , 2 2, , 1, ., 2, , We have, P( X  1)  0 .8  1  P( X  0)  0 .8, , and probability that head, 8, , 15, ., 16, , (d) Suppose the coin tossed n times. Let X be, the number of heads obtained. Then X, follows a binomial distribution with, , n 6, , 1 1, 2 2, , 4, , Thus p( X  1)  1  p( X  0)  14 C0   , , 0, , occurs, , 1, 1, or p , 2, 2, , 1, 2, , 2, 1,   q and n  2, r  2, 10 5, , that, , 6, , 10, , 1, 10 C 6   ., 2, , and n  4, , 16, 4  1,  2 C 2   .  , ., 25, 5 5, , (b) Probability, , 10, , and variance ( X )  npq  1  q , , Hence required probability  n C r p r .q n  r, , 5., , 9, , (d) We have mean ( X )  np  2, , 8, 4,  p, (b) Here P (without defected) , 10 5, , 2, , 6, , 1 1, 1, 10 C 4     10 C 4  , 2 2, 2, , 4, , (b) Required probability = 5 C1    , , P (defected) , , 2, 3, , (d) Required probability, , {Here strictly one is swimmer}, 4., , 1, 2, and n  9., ,p, 3, 3, , 1, 3, ,  9   1 ,  9 ,  5 C 5   .     .,  10   10 ,  10 , , 3., , 125, ., 3888, , Hence the binomial distribution is    ., , So the probability that none egg is rotten, 5, , 2, , 10., , 1 p, 2p, , 2, , then we get, 5, , 1, ., 3, , (a) Let n be the least number of bombs, required and X the number of bombs that, hit the bridge. Then X follows a binomial, distribution with parameter n and p , , 1, ., 2, , Now P( X  2)  0 .9  1  P( X  2)  0 .9,  P( X  0)  P( X  1)  0 .1, n, , 1, 1, n C0    nC1  , 2, 2, , n 1, , 1,    0 .1  10 (n  1)  2 n, 2

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PHOTON INSTITUTE UJJAIN NDA, X-GROUP Probability, , 8890244575, , 17., , This gives n  8., (c) 9 . 6 C 4 p 4 q 2  6 C 2 p 2 q 4, Putting q  1  p, we get required result., , 18., , (d) We are given that n  3, p , Mean  np  3 , , 1 1, , 6 2, , 1, 6, , Variance  nqp  3  , 19., , (b) Obviously, p , also, , 1, 5, , q, 6, 6, , 5, 5, , ., 6 12, , 2 1, 1 2,   q 1  ,, 6 3, 3 3, , n  2. Therefore,, , variance, , 1 2 4,  npq  2    ., 3 3 9, , 20., , (b) Required, probability  P (exactly, two) P (exactly three), 2, , 3, , 2 1, 7, 2 4, 2, ., 3 C2 .    3C3    , , 9 27 27, 6 6, 6, 3, , 1 1, 2 2, , 21., , (c) Required probability  4 C 3 .    , , 22., , (c) Required probability, 1, , 2, , 2, , 1, ., 4, , 1, , 3 3 6 3, 1 1, 1 1,  3 C 1   .   3 C 2   .      ., 8 8 8 4, 2 2, 2 2, , 23., , (a) Required probability, 1, , 7, , 0, , 8, , 27  19 ,  1   19  8  1   19 ,  8 C1 ,  ,   C0 ,  ,  , , , 20  20 ,  20   20 ,  20   20 , , 24., , (d) We have p , , 7, , 3, 1, q, and n  5, 4, 4, , Therefore required probability, 3, , 2, , , , 5 . 81, , 4, , 3 1, 3 1, 3,  5 C 3      5 C 4     5 C 5  , 4 4, 4 4, 4, , , 25., , 10 . 27, 45, , 45, , , , 5, , 243 270  405  243 459, , , ., 1024, 512, 45, , (a) Let the coin be tossed n times, 7, , 1 1, P (7 heads)  n C 7    , 2 2, , n 7, , 1,  n C7  , 2, , 9, , 1 1, 2 2, , and P (9 heads)  n C 9    , , n 9, , n, , 1, n C9  , 2, , n, , P (7 heads)  P (9 heads)  n C 7  n C 9  n  16,  P (3, 3, , 1 1, 16 C 3    , 2 2, , heads), 16  3, , 1, 16 C 3  , 2, , 16, , , , 35, ., 2 12