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According to addition theorem of probabuity, , P(AUB) = P(A) + P(B) - P(ANB), , , , 3,2 1.4, = 242.124, 6*s 6 6, Verification :, n(AUB) 4, (AUB) = {2, 3, 4, 6} and P(AUB) = ~ nig) = 5, Example : 15., , A die is thrown once. What is the probability of getting an odd, number or an even number?, , Solution :, , Let O be an event of getting an odd number, £ be an event of getting, an even number and (OUE) is an event of getting an odd number or an, even number., , Here, S={ 1, 2,3,4,5,6}, O={1,3,5}, E={2, 4,6}, , and (OvE) (i¢., © and B are mutually exclusive), , +P(0) = 3, P(E) = 2 and P(OUE)=P(0) + P(E) = 2 +, , , , 2, é, Note: Here, (OUE) = {1, 2,3, 4, 5,6}=S., Verification, P(S) = 1, Example : 16., A card is drawn from a pack of 52 playing cards. What is the, probability that it is a Club or a King ?, Solution :, Let C be an event of getting a club(#), K be an event of getting a, king and CK is an event of getting a club king ([K)) ., , , , , , , , 302 1 PUC - Statistics Text Book, Here, PIC)= E,P(k)= Sand P(ck)= “P= 5, We know that, P(CUK) = P(C) + P(K) - P(CK), = 844-42 %-03077, , ~ $2 52 52 52, Example : 17., , In a town 15% of the population read newspaper A, 13% of the, population read newspaper B and 8% of the population read newspapers, A and B, Find the probability that a person selected at random shall, read either paper A or paper B., , Solution :, Here, P(A) = 15% = 0.15, P(B) = 13% = 0.13 and P(ANB) = 8% = 0.08, According to addition theorem of probability,, P(AUB) = P(A) + P{B) - P(ANB) = 0.15 + 0.13 - 0.08 = 0.2, Example : 18, A box contains 200 bolts and 300 nuts. 20% of bolts and half of the, , nuts are rusted. If one item is selected at random, what is the probability, that it is a rusted item or a bolt ?, , Scintinn, , , , faabis:. uiiaas’ puiuais isles wal deiansia aa Ok a RY ws

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wv, = P(A) + P(B) - P(ANB) (ANB), Note: If A, B and C are three events, then, P(AUBUC}=P{A)+P(B)+P(C)-P(ANB)-P(BNC)-P(CNA)+P(ANBNC), , Example : 14., , A die is thrown once. What is the probability of getting a multiple of, 2or3?, Solution :, , Let A be an event of getting a multiple of 2, B be an event of getting, a multiple of 3, (ANB) is an event of getting a multiple of 2 and 3, (e., both) and (AUB) is an event of getting a multiple of 2 or 3 (i.e., at, least one)., , Here, S = {1, 2, 3, 4, 5, 6}, A= (2,4, 6}, B, , , , (3, 6} and ae (6), , , , , , , , , , , , , , , , : =a 23 =e TD 5A B, APA) Soy 7 a P= Sa 142), Unit - IX: Probability Thoery 301, , _ nan) _ 1, and PANB)= "= 5, , According to addition theorem of probability, P(AUB) = P(A) + P(B) - P(AMB), , , , 2,2 4, e2422he 3, ae 8, Verification :, n(AUB) 4, ». (AUB) = {2, 3, 4, 6} and P(AUB) = ~ hig) = 5, Example : 15., , A die is thrown once. What is the probability of getting an odd, number or an even number?, Solution :, , Let O be an event of getting an odd number, E be an event of getting, an even number and (QUE) is an event of getting an odd number or an, even number., , Here, S=(1,2,3, 4, 5,6}, O=(1,3,5}, E=(2,4,6}, , and (0UE) (i.e., 0 and E are mutually exclusive), , , , 8 P(0) = 2, P(E)= 2 and P(OUE) =P(0) + P(E), Note: Here, (OUE) = {1, 2,3, 4, 5,6}=S., Verification, P(S) = 1, , Example : 16., , A card is drawn from a pack of 52 playing cards. What is the, probability that it is a Club or a King ?, Solution, , Let C be an event of getting a club(4), K be an event of getting a, king and CK is an event of getting a club king ([@K)) .

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According to addition theorem of probability,, P(AUB) = P(A) + P(B) - P(AMB) = 0.15 + 0.13 - 0.08 = 0.2, Example : 18, A box contains 200 bolts and 300 nuts. 20% of bolts and half of the, nuts are rusted. If one item is selected at random, what is the probability, that it is a rusted item or a bolt ?, , , , Solution : Here, total number of itmes = 200 + 300 = 500, Number of rusted bolts = 20% of 200 = a x 200 = 40, , Number of rusted nuts = = x 300= 150, , :, Number of rusted items = 40 + 150 = 190, , Let R be an event of getting rusted item, B be an event of getting, bolt, and RB be an event of getting rusted bolt., , 190 200 40, PIR) = Sqr PIB) = Soo and. PIRNB) = 555, , Unit - IX : Probability Thoery 303, , We know that, P(RUB) =P(R) + P(B)- P(RNB)= Ze + Ze - SO =, , , , Example :19. If P(A) = = PB) = : and P(ANB) = = then, find P(AUB)., , Solution : We know that, P(AUB) = P(A) + P(B) - P(ANB), , 1,1 15, =+2-—-=— or 0:, 105 20-20% 5, , Example : 20., It P(A)= 5, PIB)= > and (ANB) =, , 2 then, find P(AUB)., , Solution : We know that, P(AUB) = P(A) + P(B) - P(ANB), , 1,1 1.2, =1,1_1.2 or 0.6667, 5 tg 7g 5 oF 0.666, Example : 21., If P(A)= e P(B) = : and P(AUB) = 3 then, find P(ANB}., , Solution : We know that, P(AUB) = P(A) + P(B) - P(ANB), , , , , , , , aty1 121 ooaz, 8 6 4 24, Example : 22., If P(AUB)= 4, P(A) = 2 and P(ANB)= = then, find PIB)., , Solution : We know that, P(AUB) = P(A) + P(B) - P(ANB), , , , 3, , 1 1, =2+PB)-=, 6 PPG, , > pip)-2-1,421 oroas, 3.6 12 4

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SEER NN RO OR eS TNR TS Te Te EOS AT gyn ae Tee, Verification, P(S) = 1, Example : 16., A card is drawn from a pack of 52 playing cards, What is the, probability that it is a Club or a King ?, Solution :, Let C be an event of getting a club(a), K be an event of getting a, king and CK is an event of getting a club king ([R)) ., , , , , , , , 302 1 PUC- Statistics Text Book, - 38 4 = Mee), Here, P(C)= =, P(K)= = and P(CK) wo se, We know that, P(CUK) = P(C) + P(K) - P(CK), 34 41 16, =5¢+ 3: mt 709077, , Example : 17., , In a town 15% of the population read newspaper A, 13% of the, population read newspaper B and 8% of the population read newspapers, A and B. Find the probability that a person selected at random shall, read either paper A or paper B., , Solution :, Here, P(A) = 15% = 0.15, P(B) = 13% = 0.13 and P(ANB) = 8% = 0.08, According to addition theorem of probability,, P(AUB) = P(A) + P(B) - P(ANB) = 0.15 + 0.13 - 0.08 = 0.2, Example : 18, A box contains 200 bolts and 300 nuts. 20% of bolts and half of the, , nuts are rusted. If one item is selected at random, what is the probability, that it is a rusted item or a bolt ?, , Solution : Here, total number of itmes = 200 + 300 = 500, , 20, Number of rusted bolts = 20% of 200 = 75, * 200 = 40, , 1, Number of rusted nuts = 5 * 300 = 150, , :. Number of rusted items = 40 + 150 = 190, , Let R be an event of getting rusted item, B be an event of getting, bolt, and RB be an event of getting rusted bolt., , 190 200 40, PIR) = S55 PIB) = 35g and P(RNB) = 25, , 0

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VS soa) sop SSE 500, , Unit - IX: Probability Thoery 303, , 00 403, We know that, P(RUB) = P(R) + P(B) - P(RAB) = 2 + ” : 2 B07, , Example :19. If P(A) =<, P(B)= and P(ANB)= =; then, find P(AUB)., , Solution : We know that, P(AUB) = P(A) + P(B) - P(ANB), , oe, 10°5 20 20, , , , , , or 0.25, Example : 20., , If P(A) = i, P{B) = 5 and P(ANB) = 2 then, find P(AUB)., , Solution : We know that, P(AUB) = P(A) + P(B) - P(ANB), , 2, 3, , 11d, eee 0.666", eae 7, , Example : 21., , If P(A)= * P(B) = ; and P(AUB) = ; then, find P(ANB)., , Solution : We know that, P(AUB) = P(A) + P(B) - P(ANB), , =1,1 1.1 9 o0417, 86 4 24, , , , Example : 22., , If P(AUB) = 4, P(A)= 2 and PIANB) = + then, find PIB)., , Solution : We know that, P(AUB) = P(A) + P(B) - P(ANB), , , , 304 | PUC- Statistics Text Book