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

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

$P\left(A\right)=\frac{4}{5},P\left(B\right)=\frac{3}{4}$,$P\left(C\right)=\frac{2}{3}$

$\therefore P\left(C^c/E\right)$

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