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Question
If P(B) = `3/5`, P(A|B) = `1/2` and P(A∪ B) = `4/5`, then P(A∪ B)′ + P( A′ ∪ B) = ______.
Options
`1/5`
`4/5`
`1/2`
1
Solution
If P(B) = `3/5`, P(A|B) = `1/2` and P(A∪ B) = `4/5`, then P(A∪ B)′ + P( A′ ∪ B) = 1.
Explanation:
Given that: P(B) = `3/5`, P(A|B) = `1/2` and P(A∪ B) = `4/5`
`"P"("A"/"B") = ("P"("A" ∩ "B"))/("P"("B"))`
⇒ `1/2 = ("P"("A" ∩ "B"))/(3/5)`
⇒ P(A ∩ B) = `3/10`
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
`4/5 = "P"("A") + 3/5 - 3/10`
∴ P(A) = `4/5 - 3/5 + 3/10`
= `1/5 + 3/10`
= `5/10`
= `1/2`
Now P(A ∪ B)' + P(A' ∪ B)
= 1 – P(A ∪ B) + 1 – P(A ∩ B')
= `2 - 4/5 - "P"("A")*"P"("B'")`
= `6/5 - 1/2 * (1 - 3/5)`
= `6/5 - 1/2 xx 2/5`
= `6/5 - 1/5`
= `5/5`
= 1
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