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

Answer 1

1

Explanation:

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 . 3/5 = 3/10`

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

P(A) = `4/5 - 3/10 = 1/2 .` P(A') = 1 − P(A) = `1/2`

We know, P(A ∩ B) + P(A' ∩ B) = P(B) [as A ∩ B and A' ∩ B are mutually exclusive events]

= `3/10 + "P (A' ∩ B)" = 3/5`

= `"P (A' ∩ B)" = 3/5 - 3/10 = 3/10`

Now, P(A' ∪ B) = P(A') + P(B) − P(A' ∩ B)

= `1/2 + 3/5 - 3/10 = (5 + 6 - 3)/10 = 4/5`

P((A ∪ B)') = 1 − P(A ∪ B) = `1 - 4/5 = 1/5`

∴ P((A ∪ B)') + P(A' ∪ B) = `1/5 + 4/5 = 1`