Question

The probability of a man hitting a target is $\frac{1}{4}$. How many times must he fire so that the probability of his hitting the target atleast once is greater than $\frac{2}{3}$ ?

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (A)

Here $p= \frac{1}{4}$,
so $q = 1-\frac{1}{4}=\frac{3}{4}$.
Let there be $n$ trials.
The probability of hitting the target atleast once in $n$ trials $= 1 - P$(no success) $= 1 - q^{n}$.
According to question
$1-q^{n} > \frac{2}{3}$
$\Rightarrow 1-\left(\frac{3}{4}\right)^{n} > \frac{2}{3}$
$\Rightarrow 1-\frac{2}{3} > \left(\frac{3}{4}\right)^{n}$
$\Rightarrow \left(\frac{3}{4}\right)^{n} < \frac{1}{3}$
Now, $\frac{3}{4} > \frac{1}{3}$,
$\left(\frac{3}{4}\right)^{2}= \frac{9}{16} > \frac{1}{3}$,
$\left(\frac{3}{4}\right)^{3} = \frac{27}{64} > \frac{1}{3}$
but $\left(\frac{3}{4}\right)^{4} = \frac{81}{256} < \frac{1}{3}$
Hence, the man must fire atleast $4$ times.