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. $80/243$
• B. $32/625$
• C. $128/625$
• D. $40/243$

Answer 1

The correct option is B

$32/625$

Detrmine the probability of getting exactly $3$ successes

Let probability of one success be denoted by $\mathrm{P}(\mathrm{X}=1)$ and probability of two successes be denoted by $\mathrm{P}(\mathrm{X}=2)$. Therefore,

$\mathrm{P}(\mathrm{X}=1)={}^5\mathrm{C}_1{\mathrm{p}}^1{\mathrm{q}}^4=0.4096\left(1\right)$

$\mathrm{P}(\mathrm{X}=2)={}^5\mathrm{C}_2{\mathrm{p}}^2{\mathrm{q}}^3=0.2048\left(2\right)$

Here, $\mathrm{p}=$probability of success for any trial

$\mathrm{q}=$probability of failure for any trial

Total number of independent trials$=5$

Now divide equation $\left(1\right)$ by $\left(2\right)$ we get,

$$\begin{gathered} \frac{{}^5\mathrm{C}_1{\mathrm{p}}^1{\mathrm{q}}^4}{{}^5\mathrm{C}_2{\mathrm{p}}^2{\mathrm{q}}^3}=\frac{0.4096}{0.2048} \\ \Rightarrow \frac{5\mathrm{q}}{10\mathrm{p}}=2 \\ \Rightarrow \frac{\mathrm{q}}{2\mathrm{p}}=2 \\ \Rightarrow \mathrm{q}=4\mathrm{p} \end{gathered}$$

We also know that $\mathrm{p}+\mathrm{q}=1$. Therefore,

$$\begin{gathered} \mathrm{p}+4\mathrm{p}=1 \\ \Rightarrow 5\mathrm{p}=1 \\ \Rightarrow \mathrm{p}=\frac{1}{5} \end{gathered}$$

Therefore, $\mathrm{q}=4\mathrm{p}=4\times \frac{1}{5}=\frac{4}{5}$

Now probability of three successes can be calculated as follows,

$$\begin{gathered} \mathrm{P}(\mathrm{X}=3)={}^5\mathrm{C}_3{\mathrm{p}}^3{\mathrm{q}}^2 \\ =10\times {\left(\frac{1}{5}\right)}^3{\left(\frac{4}{5}\right)}^2 \\ =\frac{32}{625} \end{gathered}$$

Therefore the probability of getting exactly $3$ successes is equal to $\frac{32}{625}$.

Hence, option B is the correct answer.