Q78 / 150 · MHT CET 2024 (10 May Shift 1) · Mathematics
In a Binomial distribution consisting of 5 independent trials, probabilities of exactly 1 and 2 successes are 0.4096 and 0.2048 respectively, then the probability, of getting exactly 4 successes, is
- A$\frac{80}{243}$
- B$\frac{40}{243}$
- C$\frac{32}{625}$
- Correct answerD$\frac{4}{625}$
Solution
Let $\mathrm{P}(\mathrm{X}=1)$ be probability of one success and $\mathrm{P}(\mathrm{X}=2)$ be probability of two success
$\begin{array}{ll}
\therefore \quad & P(X=1)={ }^5 \mathrm{C}_1 \mathrm{p}^1 \mathrm{q}^4=0.4096 ...(i)\\
& \mathrm{P}(\mathrm{X}=2)={ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3=0.2048...(ii)
\end{array}$
Where $p=$ probability of success $\mathrm{q}=$ probability of failure $\therefore \quad$ Dividing (i) by (ii), we get $\begin{aligned} & \frac{{ }^5 \mathrm{C}_1 \mathrm{pq}^4}{{ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3}=\frac{0.4096}{0.2048} \\ & \frac{5 \mathrm{q}}{10 \mathrm{p}}=2 \\ & \Rightarrow \mathrm{q}=4 \mathrm{p} \\ \therefore \quad & \text { Also, } \mathrm{p}+\mathrm{q}=1 \\ \therefore \quad & p+4 p=1 \end{aligned}$ $\begin{array}{ll}\therefore & p=\frac{1}{5} \\ & \therefore \\ & q=\frac{4}{5}\end{array}$ Now, Probability of getting 4 successes $\begin{aligned} & =\mathrm{P}(\mathrm{X}=4) \\ & ={ }^5 \mathrm{C}_4 \mathrm{p}^4 \mathrm{q} \\ & =5 \times\left(\frac{1}{5}\right)^4 \times \frac{4}{5} \\ & =\frac{4}{625} \end{aligned}$
Where $p=$ probability of success $\mathrm{q}=$ probability of failure $\therefore \quad$ Dividing (i) by (ii), we get $\begin{aligned} & \frac{{ }^5 \mathrm{C}_1 \mathrm{pq}^4}{{ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3}=\frac{0.4096}{0.2048} \\ & \frac{5 \mathrm{q}}{10 \mathrm{p}}=2 \\ & \Rightarrow \mathrm{q}=4 \mathrm{p} \\ \therefore \quad & \text { Also, } \mathrm{p}+\mathrm{q}=1 \\ \therefore \quad & p+4 p=1 \end{aligned}$ $\begin{array}{ll}\therefore & p=\frac{1}{5} \\ & \therefore \\ & q=\frac{4}{5}\end{array}$ Now, Probability of getting 4 successes $\begin{aligned} & =\mathrm{P}(\mathrm{X}=4) \\ & ={ }^5 \mathrm{C}_4 \mathrm{p}^4 \mathrm{q} \\ & =5 \times\left(\frac{1}{5}\right)^4 \times \frac{4}{5} \\ & =\frac{4}{625} \end{aligned}$
Hence, the correct answer is option: 'D'.