Two persons $A$ and $B$ take part in a shooting competition. $A$ can hit the target with a probability of 0…
Two persons $A$ and $B$ take part in a shooting competition. $A$ can hit the target with a probability of 0.6 . $B$ can hit the target with a probability of 0.8 . $A$ has the first shoot, past which they strike alternatively. Then, the probability that $A$ wins the competition is
$\frac{7}{10}$
$\frac{15}{23}$
$\frac{2}{3}$
$\frac{11}{17}$
Solution
Given that, $P(A)=0.6$ and $P(B)=0.8$
$\Rightarrow P\left(A^{\prime}\right)=0.4$ and $\left(P\left(B^{\prime}\right)=0.2\right.$
$\therefore$ Probability that $A$ wins $=P(A)+P\left(A^{\prime}\right) P\left(B^{\prime}\right) P(A)$
$+P\left(A^{\prime}\right) P\left(B^{\prime}\right) P\left(A^{\prime}\right) P\left(B^{\prime}\right) P(A)+\ldots$
$\begin{aligned} \Rightarrow P(A \text { wins })=0.6+(0.4)(0.2)(0.6)+( & (0.4)^2 \\ & (0.2)^2(0.6)+\ldots .\end{aligned}$
$\Rightarrow P(A$ wins $)=\frac{0.6}{1-(0.4 \times 0.2)}[$ above series is an infinite GP series with $a=0.6$ and $r=0.4 \times 0.2$ ]
$\Rightarrow P(A$ wins $)=\frac{0.6}{1-0.08}=\frac{60}{92}$
$\Rightarrow P(A$ wins $)=\frac{15}{23}$