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Question
If A and B are any two events having P(A ∪ B) = `1/2` and P`(barA) = 2/3`, then the probability of `barA ∩ B` is ______.
Options
`1/2`
`2/3`
`1/6`
`1/3`
Solution
If A and B are any two events having P(A ∪ B) = `1/2` and P`(barA) = 2/3`, then the probability of `barA ∩ B` is `1/6`.
Explanation:
We have P(A ∪ B) = `1/2`
⇒ P(A ∪ (B – A)) = `1/2`
⇒ P(A) + P(B – A) = `1/2` ......(Since A and B – A are mutually exclusive)
⇒ `1 - P(barA) + P(B - A) = 1/2`
⇒ `1 - 2/3 + P(B - A) = 1/2`
⇒ P(B – A) = `1/6`
⇒ `P(barA ∩ B) = 1/6` ....(Since `barA` ∩ B = B - A)
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| (b) If E1 and E2 are the mutually exclusive and exhaustive events | (ii) (E1 – E2) ∪ (E1 ∩ E2) = E1 |
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| (d) If E1 and E2 are two events such that E1 ⊂ E2 | (iv) E1 ∩ E2 = Φ |
