Question

A can hit a target 4 times in 5 shots, B can hit the same 3

times in 4 shots and C twice in 3 shots. If they fire one each,

the probability that at least of them hit the target is-

• A. 5/6
• B. 11/12
• C. 9/10
• D. None of these

Answer 1

Answer: (A)

5/6

Answer 2

Required probability

= P (A Ē B Ē$\bar { C }$) + P (A Ē $\bar { B }$ Ē C) + P ( $\bar { A }$ Ē B Ē C) + P(A Ē B Ē C)

= P(1) P(2) P ( $\bar { C }$ ) + P(1) P( $\bar { B }$ ) P (3) + P( $\bar { A }$ ) P (2) P (C ) + P(1) P(2) P(3)

= $\frac { 4 } { 5 }$ ×$\frac { 1 } { 5 }$× $\frac { 3 } { 4 }$ × $\frac { 2 } { 3 }$

+ $\frac { 4 } { 5 }$ ×$\frac { 2 } { 3 }$ = $\frac { 50 } { 60 }$ = $\frac { 5 } { 6 }$