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 hit exactly by one of them

#### Solution

Let event A: A can hit the target,

event B: B can hit the target,

event C: C can hit the target.

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

∴ P(A') = 1 – P(A) = `1 - 3/4 = 1/4`

P(B') = 1 – P(B) = `1 - 1/2 = 1/2`

P(C') = 1 – P(C) = `1 - 5/8 = 3/8`

Since A, B, C are independent events,

A', B', C' are also independent events.

Let event W: Target is hit exactly by one of them.

P(W) = 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 xx 1/2 xx 3/8) + (1/4 xx 1/2 xx 3/8) + (1/4 xx 1/2 xx 5/8)`

= `9/64 + 3/64 + 5/64`

= `17/64`