Question

Let in a Binomial distribution, consisting of $5$ independent trials, probabilities of exactly $1$ and $2$ successes be $0.4096$ and $0.2048$ respectively. Then the probability of getting exactly $3$ successes is equal to:

• A. $\frac{32}{625}$
• B. $\frac{80}{243}$
• C. $\frac{40}{243}$
• D. $\frac{128}{625}$

Answer 1

Answer: (A)

$P(X=1)={}^5{C}_1\cdot p\cdot q^4=0.4096$
$P(X=2)={}^5{C}_2\cdot p^2\cdot q^3=0.2048$
$\Rightarrow \frac{q}{2p}=2$
$\Rightarrow q=4p$ and $p+q=1$
$\Rightarrow p=\frac{1}{5}$ and $q=\frac{4}{5}$
Now
$P(X=3)={}^5{C}_3\cdot {\left(\frac{1}{5}\right)}^3\cdot {\left(\frac{4}{5}\right)}^2$
$=\frac{10\times 16}{125\times 25}=\frac{32}{625}$