Question
In binomial distribution with five Bernoulli’s trials, the probability of one and two success are 0.4096 and 0.2048 respectively. Find the probability of success.

Answer

Given: $X \sim B(n=5, p)$
The probability of $X$ success is
$
P(X=x)={ }^n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots, n
$
i.e. $P(X=x)={ }^5 \mathrm{C}_x p^x q^{5-x}, x=0,1,2,3,4,5$
Probabilities of one and two successes are
$
P(X=1)={ }^5 C_1 p^1 q^{5-1}
$
and $P(X=2)={ }^5 \mathrm{C}_2 p^2 q^{5-2}$ respectively
Given: $P(X=1)=0.4096$ and $P(X=2)=0.2048$
$\therefore \frac{P(X=2)}{P(X=1)}=\frac{0.2048}{0.4096}$
i.e. $\frac{{ }^5 \mathrm{C}_2 p^2 q^{5-2}}{{ }^5 \mathrm{C}_1 p^2 q^{5-1}}=\frac{1}{2}$
i.e. $2 \times{ }^5 C_2 p^2 q^3=1 \times{ }^5 C_1 p q^4$
i.e. $2 \times \frac{5 \times 4}{1 \times 2} \times p^2 q^3=1 \times 5 \times p q^4$
i.e. $20 p^2 q^3=5 p q^4$
i.e. $4 p=q$
i.e. $4 p=1-p$
i.e. $5 p=1$
$
\therefore p=\frac{1}{5}
$
Hence, the probability of success is $\frac{1}{5}$.

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