Question

The probability of a man hitting a target in one trial is $\frac{1}{4}$.The chances of hitting the target at least once in $n$ trials exceeds $\frac{2}{3}$, then the value of $n$ equals

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is B $4$

$p=\frac{1}{2},q=\frac{3}{4},n=?$

Now using, $p(X=r){=}^n{C}_r{p}^r{q}^{n-r}$

Given $P(X\ge 1)>\frac{2}{3}$

$\Rightarrow 1-P\left(0\right)>\frac{2}{3}\Rightarrow 1-{\left(\frac{3}{4}\right)}^n>\frac{2}{3}\Rightarrow \left(\frac{1}{3}\right)>{\left(\frac{3}{4}\right)}^n$

For $n=1,\frac{1}{3}\ngtr \frac{3}{4}$

For $n=2,\frac{1}{3}\ngtr \frac{9}{16}$

For $n=3,\frac{1}{3}\ngtr \frac{27}{64}$

For $n=4,\frac{1}{3}>\frac{81}{256}$

$\therefore$ Minimum value of $n=4$