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 $21$ shells each of which can be fired at point $I$ or $II$. Each shell may hit the target independently of the other shell with probability $\frac{1}{2}$. How many shells must be fired at point $I$ to hit the target with maximum probability?

• A. $P\left(A\right)$ is maximum where $x=11$.
• B. $P\left(A\right)$ is maximum where $x=12$.
• C. $P\left(A\right)$ is maximum where $x=14$.
• D. $P\left(A\right)$ is maximum where $x=15$.

Answer 1

The correct option is B $P\left(A\right)$ is maximum where $x=12$.

Let $A$ denote the event that the target is hit when $x$ shells are fired at point $I$.
Let $E_1$ and $E_2$ denote the events hitting $I$ and $II$, respectively
$\therefore P\left(E_1\right)=\frac{8}{9},P\left(E_2\right)=\frac{1}{9}$
Now $P\left(\frac{A}{E_1}\right)=1-{\left(\frac{1}{2}\right)}^x$ and $P\left(\frac{A}{E_2}\right)=1-{\left(\frac{1}{2}\right)}^{21-x}$
Hence $P\left(A\right)=\frac{8}{9}[1-{\left(\frac{1}{2}\right)}^x]+\frac{1}{9}[1-{\left(\frac{1}{2}\right)}^{21-x}]$
$\therefore \frac{dP\left(A\right)}{dx}=\frac{8}{9}\left[{\left(\frac{1}{2}\right)}^x\log 2\right]+\frac{1}{9}[-{\left(\frac{1}{2}\right)}^{21-x}\log 2]$
For maximum probability $\frac{dP\left(A\right)}{dx}=0$
$\therefore x=12$ $[\because 2^{3-x}=2^{x-21}\Rightarrow 3-x=x-21]$
Since $\frac{d^{2}P\left(A\right)}{d{x}^2}<0$ for $x=12$
$\therefore P\left(A\right)$ is maximum for $x=12$