A rifleman if firing at a distant target and has only 10% chance of hitting it. The least number of rounds, he must fire in order to have more than 50% chance of hitting it at least once, is

7

Let p be the probability that the rifleman hits the target.

Then, $p = \frac{10}{100}=\frac{1}{10}$ and $q=\frac{9}{10}$

Suppose n rounds are fired. Let X be the number of times the rifleman hits the target in n trials. Then,

$P(X=r)= {^nC}_r \left(\frac{1}{10}\right)^r \left(\frac{9}{10}\right)^{n-r}, r= 0, 1, 2, ..., n$

Now,

$P(X ≥ 1)≥\frac{1}{2}$

$⇒ 1-p(X=0)≥\frac{1}{2}$

$⇒ P(X=0)≤ \frac{1}{2}⇒ \left(\frac{9}{10}\right)^{n}\left(\frac{9}{10}\right)^{n-r}\frac{1}{2}⇒n=7,8,9,...$

Hence, the rifleman must fire at least 7 rounds.