In a Binomial distribution with $\mathrm{n}=4$, if $2 \mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=2)$,…

In a Binomial distribution with $\mathrm{n}=4$, if $2 \mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=2)$, then the variance is
  1. $\frac {36}{169}$
  2. $\frac {144}{169}$
  3. $\frac {9}{169}$
  4. $\frac {16}{169}$

Solution

$\begin{aligned} & P(X=3)={ }^4 C_3 p^3(1-p)=4 p^3(1-p) \\ & P(X=2)={ }^4 C_2 p^2(1-p)^2=6 p^2(1-p)^2 \\ & \text { Given } 2 \mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=2) \\ & \therefore \quad 8 \mathrm{p}^3(1-\mathrm{p})=18 \mathrm{p}^2(1-\mathrm{p})^2 \\ & \therefore \quad 8 \mathrm{p}=18(1-\mathrm{p}) \\ & \therefore \quad \mathrm{p}=\frac{9}{13} \\ & \text { Variance }=n p(1-p) \\ & =4 \times \frac{9}{13}\left(1-\frac{9}{13}\right) \\ & =\frac{144}{169} \\ & \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 1)

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