Question

An artillery target may be either at point I with probability $\frac{8}{9}$ or at point II with probability $\frac{1}{9}$. We have 55 shells, each of which can be fired either at point I or II. Each shell may hit the target, independent of the other shells, with probability $\frac{1}{2}$. Maximum number of shells that must be fired at point I to have maximum probability is

• A. $20$
• B. $25$
• C. $29$
• D. $35$

Answer 1

The correct option is C $29$

Let A denote the event that target is hit when $x$ shells are fired at point $I$. Let $P_1$ and $P_2$, denote the event that the target is at point $I$ and $II$, respectively.
We have $P\left(P_1\right)=\frac{8}{9},P\left(P_2\right)=\frac{1}{9}$,
$P\left(A/P_1\right)=1-(1/2{)}^x,P(A/P_2)=1-(1/2{)}^{55-x}$.
Now from total probability theorem,
$P\left(A\right)=P\left(P_1\right)P\left(A/P_1\right)+P\left(P_2\right)P\left(A/P_2\right)$
$=\frac{1}{9}(8-8{\left(\frac{1}{2}\right)}^x+1-{\left(\frac{1}{2}\right)}^{55-x})$
$=\frac{1}{9}(9-8{\left(\frac{1}{2}\right)}^x-{\left(\frac{1}{2}\right)}^{55-x})$
Now,
$\frac{dP\left(A\right)}{dx}=\frac{1}{9}(-8{\left(\frac{1}{2}\right)}^x\ln \left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^{55-x}\ln \left(\frac{1}{2}\right))$
Where, $x\in R^{+}$
$=\frac{1}{9}\ln \left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^{55-x}(1-{\left(\frac{1}{2}\right)}^{2x-58})$
$>0\text{if}x<29$
$<0\text{if}x>29$
Therefore, $P\left(A\right)$ is maximum at $x=29$.
Thus, $29$ shells must be fired at point $I$