Question

Three shots are fired at a target in succession. The probabilities of a hit in the first shot is $\frac{1}{2}$, in the second $\frac{2}{3}$ and in the third shot is $\frac{3}{4}$, In case of exactly one hit, the probability of destroying the target is $\frac{1}{3}$ and in the case of exactly two hits $\frac{7}{11}$ an in the case of three hits is $1.0$. Find the probability of destroying the target in three shots

• A. $\frac{3}{4}$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{4}{5}$

Answer 1

The correct option is C $\frac{5}{8}$

$A$ : Target hit in 1st shot

$B$ : Target hit in 2nd shot

$C$ : Target hit in 3rd shot

$E_1$: destroyed in exactly one shot

$E_2$: destroyed in exactly two shot

$E_3$: destroyed in exactly three shot

$P\left(E_1\right)=P(E_{1}A\overline{B}\overline{C}\cup E_1\overline{A}\overline{B}C\cup E_1\overline{A}B\overline{C})$

$=\frac{1}{3}[\frac{1}{2}.\frac{1}{3}.\frac{1}{4}+\frac{1}{2}.\frac{1}{3}.\frac{3}{4}+\frac{1}{2}.\frac{2}{3}.\frac{1}{4}]=\frac{1+3+2}{3.24}=\frac{1}{12}$

$P\left(E_2\right)=P(E_2\overline{A}BC\cup E_{2}AB\overline{C}\cup E_{3}A\overline{B}C)$

$=\frac{7}{11}[\frac{1}{2}.\frac{2}{3}.\frac{3}{4}+\frac{1}{2}.\frac{2}{3}.\frac{1}{4}+\frac{1}{2}.\frac{1}{3}.\frac{3}{4}]=\frac{7.11}{11.24}=\frac{7}{24}$

$P\left(E_3\right)=P\left(E_{3}ABC\right)=1.\frac{1}{2}.\frac{2}{3}.\frac{3}{4}=\frac{1}{4}$

$P(E_1\cup E_2\cup E_3)=P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)$

$=\frac{1}{12}+\frac{7}{24}+\frac{1}{4}=\frac{2+7+6}{24}=\frac{15}{24}=\frac{5}{8}$