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.