A man firing at a distant target has 10% chance of hitting the target in one shot. The number of times he must fire at the target to have about 50% chance of hitting it at least once is
(a) 5
(b) 7
(c) 9
(d) 11

Hint: First, it is clearly mentioned in the question that the hitting percentage is 10% which can be represented in the form of probability as $ \dfrac{1}{10} $ . Then, by using the rule of probability, the remaining condition he will not hit the target has the probability as $ \dfrac{9}{10} $ . Then we want to find the same for the n number of cases in which he fires n shots and to get the probability of hitting at least once is as $ 1-{{\left( \dfrac{9}{10} \right)}^{n}} $ . Then, by equating it with $ \dfrac{1}{2} $ and solving for the value for n, we get our answer.

Complete step-by-step answer:
In this question, we are supposed to find the number of times he must fire at the target to have about a 50% chance of hitting it at least once.
Now, it is clearly mentioned in the question that the hitting percentage is 10% which can be represented in form of probability as:
 $ \dfrac{10}{100}=\dfrac{1}{10} $
Now, by using the rule of the probability, the remaining condition he will not hit the target has the probability as:
 $ \begin{align}
  & 1-\dfrac{1}{10}=\dfrac{10-1}{10} \\
 & \Rightarrow \dfrac{9}{10} \\
\end{align} $
Now, we want to find the same for the n number of cases in which he fires n shots and to get the probability of hitting at least once is as:
 $ 1-{{\left( 1-\dfrac{1}{10} \right)}^{n}}=1-{{\left( \dfrac{9}{10} \right)}^{n}} $
Now, in the question we are given that the probability of the above condition is 50% which is equal to $ \dfrac{1}{2} $ .
So, by using the above condition, we get:
\[\begin{align}
  & 1-{{\left( \dfrac{9}{10} \right)}^{n}}=\dfrac{1}{2} \\
 & \Rightarrow 1-\dfrac{1}{2}={{\left( \dfrac{9}{10} \right)}^{n}} \\
 & \Rightarrow \dfrac{1}{2}={{\left( \dfrac{9}{10} \right)}^{n}} \\
\end{align}\]
So, now we need to use the logarithm on the sides to get the answer as:
 $ \log 1-\log 2=n\left( \log 9-\log 10 \right) $
Now, by using the value of the log 1, log 2, log 9 and log 10, we get:
 $ \begin{align}
  & 0-0.3010=n\left( 0.9542-1 \right) \\
 & \Rightarrow -0.3010=-n\left( 0.0458 \right) \\
 & \Rightarrow n=\dfrac{0.3010}{0.0458} \\
 & \Rightarrow n=6.572 \\
 & \Rightarrow n\approx 7 \\
\end{align} $
So, the number of times he must fire at the target to have about 50% chance of hitting it at least once is 7.
So, the correct answer is “Option B”.

Note: Now, to solve these types of questions we need to know some of the basic conversions for the log functions. So, the basic conversion is if we have function as $ \left( \dfrac{a}{b} \right) $ and we take log of the same, so we get $ \log a-\operatorname{logb} $ . Moreover, we should know the fact that probability is the ratio of the favourable outcomes to the total outcomes.