Engineering
Mathematics
Conditional Probability
Question
Let A, B, C be three events such that P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A⋂B) = 0.08, P(A⋂C) = 0.28, P(A⋂B⋂C) = 0.09. If P(AUBUC) ≥ 0.75. Show that P(B⋂C) lies in the interval [0.23, 0.48].
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Solution
Given:
P(A) = 0.3 P(B) = 0.4 P(C) = 0.8 P(A⋂B) = 0.08
P(A⋂C) = 0.28 P(A⋂B⋂C) 0.09
P(AUBUC) ≥ 0.75
Consider P(AUBUC) ≥ 0.75
⇒ 0.75 ≤ P(AUBUC) ≤ 1
⇒ 0.75 ≤ P(A) + P(B) + P(C) – P(A⋂B) – P(B⋂C) – P(C⋂A) + P(A⋂B⋂C) ≤ 1
⇒ 0.75 ≤ 0.3 + 0.4 + 0.8 – 0.08 – 0.28 – P(B⋂C) + 0.09 ≤ 1
⇒ 0.75 ≤ 1.23 – P(B⋂C) ≤ 1
– 0.75 ≥ P(B⋂C) – 1.23 ≥ – 1
– 0.75 + 1.23 ≥ P(B⋂C) ≥ – 1 + 1.23
0.48 ≥ P(B⋂C) ≥ 0.23
0.23 ≤ P(B⋂C) ≤ 0.48
∴ P(B⋂C) lies on interval [0.23, 0.48]
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