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

Answer 1

The correct option is B $\frac{32}{625}$

${}^5{C}_1{p}^1{q}^4=0.4096\cdots \left(1\right){}^5{C}_2{p}^2{q}^3=0.2048\cdots \left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow \frac{q}{2p}=2\Rightarrow q=4p$
We know
$p+q=1\Rightarrow p=\frac{1}{5},q=\frac{4}{5}P\left(\text{exactly}3\right)={}^5{C}_3\left(p{)}^3\right(q{)}^2\Rightarrow P\left(\text{exactly}3\right)=10\times \frac{4^2}{5^5}\therefore P\left(\text{exactly}3\right)=\frac{32}{625}$