Question

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

1. is hit exactly by one of them
2. is not hit by any one of them
3. is hit
4. is exactly hit by two of them

Answer 1

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.

(a) 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`

(b) Let event X: Target is not hit by any one of them. 

P(X) = P(A' ∩ B' ∩ C')

= P(A') · P(B') · P(C')

`= 1/4 xx 1/2 xx 3/8`

`= 3/64`

(c) Let event Y: Target is hit.

P(Y) = 1 - P (target is not hit by any one of them)

`= 1 - 3/64`

`= 61/64`

(d) Let event Z: Target is hit by exactly two of them.

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

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

`= 29/64`