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Question: Three shooters A, B, C can shoot at the target successful with probabilities \(\frac { 1 } { 2 } , \...

Three shooters A, B, C can shoot at the target successful with probabilities 12,13,14\frac { 1 } { 2 } , \frac { 1 } { 3 } , \frac { 1 } { 4 }respectively. If they aim at the same target simultaneously, then the probability that the target is destroyed, when two shots are required for it

A

1/24

B

5/24

C

7/24

D

2/24

Answer

7/24

Explanation

Solution

P(A∩B) + P(B∩C) + P(C∩A) – 2P(A∩B∩C)

= 12×13+13×14+12×14−2×12×13×14\frac { 1 } { 2 } \times \frac { 1 } { 3 } + \frac { 1 } { 3 } \times \frac { 1 } { 4 } + \frac { 1 } { 2 } \times \frac { 1 } { 4 } - 2 \times \frac { 1 } { 2 } \times \frac { 1 } { 3 } \times \frac { 1 } { 4 }

= 18+112+124+124=18+16=1448=724\frac { 1 } { 8 } + \frac { 1 } { 12 } + \frac { 1 } { 24 } + \frac { 1 } { 24 } = \frac { 1 } { 8 } + \frac { 1 } { 6 } = \frac { 14 } { 48 } = \frac { 7 } { 24 }.