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The probability of hitting a target by three marks men is $\frac{1}{2} , \frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability that exactly two of them will hit the target is $\lambda$ and that at least two of them hit the target is $\mu$ then $\lambda + \mu$ is equal to :-
$\frac{13}{24}$
$\frac{6}{24}$
$\frac{7}{24}$
None
Solution
$\lambda = \overline {\rm{A}} {\rm{BC}} + \overline {\rm{B}} {\rm{CA}} + \overline {\rm{C}} {\rm{AB}}$
$=\left(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right)+\left(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{2}\right)+\left(\frac{3}{4} \times \frac{1}{2} \times \frac{1}{2}\right)=\frac{6}{24}=\frac{1}{4}$
$\mu=\lambda+\mathrm{ABC}=\frac{1}{4}+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right)=\frac{7}{24}$
$\lambda+\mu=\frac{1}{4}+\frac{7}{24}=\frac{13}{24}$
