Q.136. Probability

Question 136.
Suppose that for some experiment, we have three independent events A, B and C.
Suppose that: i) P(AUB) = 0.7 ii) P(C/A) = 0.3 iii) P(B/C) = 0.5.
Find the probability that exactly one of A, B or C occurs.

Answer 136.
As A, B and C are independent events, from ii) and iii),
P(C/A) = P(C) = 0.3 and
P(B/C) = P(B) = 0.5

From i) P(AUB) = 0.7
=> P(A) + P(B) - P(A∩B) = 0.7
=> P(A) + P(B) - P(A)*P(B) = 0.7
......[because A and B are independent events]
=> P(A) + 0.5 - P(A) * (0.5) = 0.7
=> (0.5)P(A) = 0.2
=> P(A) = 0.4
=>
P(A') = 1 - P(A) = 1 - 0.4 = 0.6
P(B') = 1 - P(B) = 1 - 0.5 = 0.5
P(C') = 1 - P(C) = 1 - 0.3 = 0.7

Probability that exactly one of A, B or C occurs is given by
P(A∩B'∩C') + P(B∩A'∩C') + P(C∩A'∩B')
... [because (A∩B'∩C'), (B∩C'∩A') and (C∩A'∩B') are
... mutually exclusive events.]
= P(A)*P(B')*P(C') + P(B)*P(C')*P(A') + P(C)*P(A')*P(B')
... [because A, B and C are independent events]
= (0.4)*(0.5)*(0.7) + (0.5)*(0.7)*(0.6) + (0.3)*(0.6)*(0.5)
= 0.14 + 0.21 + 0.09
= 0.44.

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