Question

A can hit a target 4 times in 5 shots, B three times in 4 shots and C twice in 3 shots. They fire a target if exactly two of them hit the target then the chance that it is C who has missed is

• A. $\frac { 6 } { 13 }$
• B. $\frac { 1 } { 5 }$
• C. $\frac { 4 } { 5 }$
• D. $\frac { 4 } { 15 }$

Answer 1

Answer: (A)

$\frac { 6 } { 13 }$

Answer 2

Let A represents the event 'A hits the target', B represents the event 'B hits the target', C represents the event 'C hits the target' and E be the event that exactly two of A, B and C hit the target.

Then P(1) = $\frac { 4 } { 5 }$ , P(2) = $\frac { 3 } { 4 }$ and P(3) = $\frac { 2 } { 3 }$

\ P (C^c/E)

= $\frac { \mathrm { P } ( \mathrm { A } ) \mathrm { P } ( \mathrm { B } ) \mathrm { P } \left( \mathrm { C } ^ { \mathrm { c } } \right) } { \mathrm { P } ( \mathrm { A } ) \mathrm { P } ( \mathrm { B } ) \mathrm { P } \left( \mathrm { C } ^ { \mathrm { c } } \right) + \mathrm { P } ( \mathrm { A } ) \mathrm { P } \left( \mathrm { B } ^ { \mathrm { c } } \right) \mathrm { P } ( \mathrm { C } ) + \mathrm { P } \left( \mathrm { A } ^ { \mathrm { c } } \right) \mathrm { P } ( \mathrm { B } ) \mathrm { P } ( \mathrm { C } ) }$ = $\frac { 6 } { 13 }$