The probability that bomb will miss the target is $0 \cdot 2$. Then the probability that out of 10 bombs…
The probability that bomb will miss the target is $0 \cdot 2$. Then the probability that out of 10 bombs dropped exactly 2 will hit the target is
$\frac{288}{5^{10}}$
$\frac{144}{5^{9}}$
$\frac{144}{5^{10}}$
$\frac{288}{5^{9}}$
Solution
$(\mathrm{C})$
We have probability of missing the target $0.2$
Let $\mathrm{p}=1-0.2=0.8, \mathrm{q}=0.2, \mathrm{n}=10$ and $\mathrm{r}=2$
Hence required probability $={ }^{10} \mathrm{C}_{2}(0.8)^{2}(0.2)^{8}$
$=\frac{10 !}{2 ! 8 !} \times\left(\frac{8}{10}\right)^{2} \times\left(\frac{2}{10}\right)^{8}=\frac{45 \times 64 \times 2^{8}}{10^{10}}=\frac{9 \times 5 \times 2^{14}}{2^{10} \times 5^{10}}=\frac{9 \times 2^{4}}{5^{9}}=\frac{144}{5^{9}}$