Explanation

(c)
Let $x$ be the probability of success in each trial, then $(1 - x)$ will be the probability of failure in each trial.
Thus, probability of exactly successes in a series of three trials
$= P({\overline{E}}_{1}\ E_{2}{\ E}_{3} + E_{1}{\overline{E}}_{2}{\ E}_{3} + E_{1}\ E_{2}{\overline{E}}_{3})$
$= (1 - x)x \bullet x + x(1 - x)x + x \bullet x(1 - x)$
$= 3x^{2}(1 - x)$
and the probability of three success
$P\left( E_{1}E_{2}E_{3} \right) = x \bullet x \bullet x = x^{3}$
According to question,
$9x^{3} - 3x^{2}(1 - x)$
$\Rightarrow 3x = 1 - x$
$\Rightarrow 4x = 1$
$\Rightarrow x = \frac{1}{4}$
Hence, the probability of success in each trial is $\frac{1}{4}.$