Question
A rifleman is 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
11
9
7
5
Solution
7
Let p=chance of hitting a distant target
\[\Rightarrow\] p =10% or p= 0.1
\[\Rightarrow q = 1 - 0 . 1 = 0 . 9\]
\[\text{ Let n be the least number of rounds } . \]
\[P(\text{ hitting atleast once} ) = P(X \geq 1) \]
\[ \Rightarrow 1 - P(X = 0) \geq 50 \% \]
\[ \Rightarrow 1 - P(X = 0) \geq 0 . 5\]
\[P(X = 0) \leq 0 . 5\]
\[ \Rightarrow (0 . 9 )^n \leq 0 . 5\]
\[\text{ Taking } \text{ log on both the sides, we get} \]
\[ n \text{ log } 0 . 9 \leq \log 0 . 5 \]
\[ \Rightarrow n \leq \frac{\log 0 . 5}{\log 0 . 9}\]
\[ \Rightarrow n \leq 7 . 2 \]
\[\text{ Therefore, 7 is the least number of rounds that he must fire in order } \]
\[ \text{ to have more than 50 % chance of hitting the target at least once } . \]
