Engineering
Mathematics
Conditional Probability
Question
If P(A) = 0.7, P(B) = 0.55, P(C) = 0.5, P(A⋂B) = x, P(A⋂C) = 0.45, P(B⋂C) = 0.3 and P(A⋂B⋂C) = 0.2, then
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Solution
P(AUBUC) = P(A) + P(B) + P(C) – P(A⋂B) – P(A⋂C) – P(B⋂C) + P(A⋂B⋂C)
= P(A⋂B) = P(A) + P(B) + P(C) – P(A⋂B⋂C) – P(A⋂C) – P(B⋂C) + P(A⋂B⋂C)
= P(A⋂B) = 0.7 + 0.55 + 0.5 0 – P(A⋂B⋂C) – 0.45 – 0.3 + 0.2 = 1.2 P(A⋂B⋂C)
Maximum P(AUBUC) is 1, hence, minimum P(A⋂B) = 1.2 – 1 = 0.2
Maximum P(A⋂B) could have been the minimum of P(A) and P(B) i.e. 0.55,
however, that would mean that 1.2 – P(A⋂B⋂C) = 0.55 ≤ P(A⋂B⋂C) = 0.65 i.e. less than A: This is not possible
hence, Maximum P(A⋂B) could be 1.2 – 0.7 = 0.5.
∴ 0.2 ≤ P(A⋂B) ≤ 0.5
0.2 ≤ x ≤ 0.5
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