Question

In a series of three trials, the probability of exactly two successes in nine times is as large as the probability of three successes. Then, the probability of success in each trial is :

• A. $ \frac{1}{2} $
• B. $ \frac{1}{3} $
• C. $ \frac{1}{4} $
• D. $ \frac{3}{4} $

Answer 1

Answer: (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 two successes in a series of trials
$= P\left(\bar{E}_{1} E_{2} E_{3}+E_{1} \bar{E}_{2} E_{3}+E_{1} E_{2} \bar{E}_{3}\right) $
$=(1-x) x \cdot x+x(1-x) x+x \cdot x(1-x) $
$=3 x^{2}(1-x)$
and the probability of three successes
$P\left(E_{1} E_{2} E_{3}\right)=x \cdot x \cdot x=x^{3}$
According to question
$\Rightarrow 9 x^{3} =3 x^{2}(1-x) $
$3 x =1-x $
$\Rightarrow 4 x=1 $
$\Rightarrow x=\frac{1}{4} $
Hence, the probability of success in each trial is $\frac{1}{4}$