A, B, and C try to hit a target simultaneously but independently. Their respective probabilities of hitting the target are `3/4, 1/2` and `5/8`. Find the probability that the target is exactly hit by two of them

#### Solution

Let A ≡ the event that A hits the target

B ≡ the event that B hits the target

C ≡ the event that C hits the target.

It is given that,

P(A) = `3/4`, P(B) = `1/2`, P(C) = `5/8`

P(A') = 1 – P(A) = `1/4`

Similarly, P(B') = `1/2`, P(C) = `3/8`

Now, A, B, C are independent events

∴ A', B', C' are independent events

Let F ≡ the event that target is hit by exactly two of them

∴ F = (A ∩ B ∩ C') ∪ (A ∩ B' ∩ C) ∪ (A' ∩ B ∩ C)

Events in the brackets are mutually exclusive.

Also A, B, C, A', B', C' are independent

∴ P(F) = P(A ∩ B ∩ C') + P(A ∩ B' ∩ C) + P(A' ∩ B ∩ C)

= P(A)·P(B)·P(C') + P(A)·P(B')·P(C) + P(A')·P(B)·P(C)

= `3/4*1/2*3/8 + 3/4*1/2*5/8 + 1/4*1/2*5/8`

= `9/64 + 15/64 + 5/64`

= `29/64`.