Question

Three persons A, B and C shoot to hit a target. Their probabilities of hitting the target are `5/6, 4/5` and `3/4` respectively. Find the probability that:

1. Exactly two persons hit the target.
2. At least one person hits the target.

Answer 1

Given: `P(A) = 5/6, P(B) = 4/5` and `P(C) = 3/4`

So, `P(bar(A)) = 1 - P(A)`

= `1 - 5/6`

= `1/6`

`P(bar(B)) = 1 - P(B)`

= `1 - 4/5`

= `1/5`

`P(bar(C)) = 1 - P(C)`

= `1 - 3/4`

= `1/4`

i. Probability (exactly two persons hit the target)

= `P(A) xx P(B) xx P(bar(C)) + P(A) xx P(bar(B)) xx P(C) + P(bar(A)) xx P(B) xx P(C)`

= `5/6 xx 4/5 xx 1/4 + 5/6 xx 1/5 xx 3/4 + 1/6 xx 4/5 xx 3/4`

= `1/6 + 1/8 + 1/10`

= `(20 + 15 + 12)/120`

= `47/120`

ii. Probability (at least one person hits the target)

= 1 – No one will hit the target

= `1 - P(bar(A)bar(B)bar(C))`

= `1 - (1/6 xx 1/5 xx 1/4)`

= `1 - 1/120`

= `(120 - 1)/120`

= `119/120`