Question

A missile target may be at a point P with probability $\frac{9}{10}$ or at a point Q with probability $\frac{1}{10}$ we have 20 shells each of which can be fired either at point P or Q Each shell may hit the target independently of the other shoot with probability $\frac{2}{3}$ Then number of shells must be fired at point P to hit any target with maximum probability is

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

Answer: (B)

$11$

Let A be the event that target is hit
$E_1,E_2$ be the events to select target P and Q respectively
Then $P\left(E_1\right)=\frac{9}{10},P\left(E_2\right)=\frac{1}{10}$
$P\left(\frac{A}{E_1}\right)=1-{\left(\frac{1}{3}\right)}^x,P\left(\frac{A}{E_2}\right)=1-{\left(\frac{1}{3}\right)}^{20-x}$ where x number of shells used to select target P
Then $P\left(A\right)=P\left(E_1\right).P\left(\frac{A}{E_1}\right)+P\left(E_2\right).P\left(\frac{A}{E_2}\right)=\frac{9}{10}(1-{\left(\frac{1}{3}\right)}^x)+\frac{1}{10}(1-{\left(\frac{1}{3}\right)}^{20-x})$
$\frac{dP\left(A\right)}{dx}=\frac{9}{10}\times -{\left(\frac{1}{3}\right)}^x\times -{\log }_{e}3+\frac{1}{10}\times {\left(\frac{1}{3}\right)}^{20-x}\times 1\times -{\log }_{e}3=0$
$\frac{{\log }_{e}3}{10}({9.3}^{-x}-3^{x-20})=0$
$3^{2-x}=3^{x-20},x=11$