Question

A die is thrown a fixed number of times. If probability of getting even number $3$ times is same as the probability of getting even number $4$ times, then the probability of getting even number exactly once is

• A. $\frac{1}{6}$
• B. $\frac{1}{9}$
• C. $\frac{5}{36}$
• D. $\frac{7}{128}$

Answer 1

The correct option is D $\frac{7}{128}$

According to the given condition
${}^n{C}_3{\left(\frac{1}{2}\right)}^n={}^n{C}_4{\left(\frac{1}{2}\right)}^n$
Where $n$ is the number of times die is thrown
${\therefore }^n{C}_3{=}^n{C}_4\Rightarrow n=7$
Thus, the required probability is
${}^7{C}_1{\left(\frac{1}{2}\right)}^7=\frac{7}{2^7}=\frac{7}{128}$