Given that the probability of a man hitting a target with a gun is $\frac{1}{3}$. If he fires 8 times, then…
Given that the probability of a man hitting a target with a gun is $\frac{1}{3}$. If he fires 8 times, then the probability of his hitting the target atleast twice is
$5\left(\frac{2}{3}\right)^8$
$1-5\left(\frac{2}{3}\right)^8$
$\left(\frac{2}{3}\right)^8$
$\left(\frac{3}{8}\right)^4$
Solution
Probability to hit the target $=\frac{1}{3}$
Probability to not hit the target $=1-\frac{1}{3}=\frac{2}{3}$
Probability to hit the target atleast twice is
$
\begin{aligned}
& =P(2)+P(3)+P(4)+\ldots+P(8) \\
& =1-[P(0)+P(1)] \\
& =1-\left[\left(\frac{2}{3}\right)^8+{ }^8 C_1 \cdot\left(\frac{2}{3}\right)^7 \cdot \frac{1}{3}\right] \\
& =1-\left(\frac{2}{3}\right)^7 \cdot\left[\frac{2}{3}+8 \cdot \frac{1}{3}\right] \\
& =1-\left(\frac{2}{3}\right)^7 \cdot \frac{10}{3}=1-5 \cdot\left(\frac{2}{3}\right)^8
\end{aligned}
$