Question

The probability of hitting a target by three 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 atleast two of them hit the target is $\mu$ then $\lambda +\mu$ is -

• A. $\frac{13}{24}$
• B. $\frac{6}{24}$
• C. $\frac{7}{24}$
• D. $\frac{12}{24}$

Answer 1

The correct option is A $\frac{13}{24}$

$\lambda =\overline{A}BC+A\overline{B}C+AB\overline{C}$

$=(\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4})+(\frac{2}{3}\times \frac{1}{4}\times \frac{1}{2})+(\frac{3}{4}\times \frac{1}{2}\times \frac{1}{2})=\frac{6}{24}=\frac{1}{4}$

$\mu =\lambda +ABC=\frac{1}{4}+(\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4})=\frac{7}{24}$

$\lambda +\mu =\frac{1}{4}+\frac{7}{24}=\frac{13}{24}$