The discreate random variables $\mathrm{X}$ and $\mathrm{Y}$ are independent from one another and are…

The discreate random variables $\mathrm{X}$ and $\mathrm{Y}$ are independent from one another and are defined as $X \sim B\left(n_1, 0.5\right)$ and $\mathrm{Y} \sim \mathrm{B}\left(\mathrm{n}_2, 0.4\right)$. If the variance of both $\mathrm{X}$ and $\mathrm{Y}$ is 6 then $\sqrt{n_1+n_2}=$
  1. $7$
  2. $6$
  3. $5$
  4. $4$

Solution

$\because$ We are given that $\mathrm{X} \sim \mathrm{B}\left(\mathrm{x}_1, 0.5\right)$, $\begin{aligned} & \mathrm{Y} \sim \mathrm{B}\left(\mathrm{x}_2, 0.4\right) \\ & \text { Here } \mathrm{p}_1=0.5, \mathrm{a}_1=0.5 \\ & \mathrm{p}_2=0.4, \mathrm{a}_2=0.6 \\ & \text { Now } 6=\mathrm{x}_1 \times \mathrm{p}_1 \times \mathrm{a}_1\end{aligned}$ $\Rightarrow 6=x_1 \times 0.5 \times 0.5 \Rightarrow \frac{600}{0.5 \times 0.5} \Rightarrow x=14$ Now $6=x_2 \times p_2 \times a_2 \Rightarrow x_2 \times 0.4 \times 0.6$ $\begin{aligned} & \Rightarrow \quad \mathrm{x}_2=\frac{600}{0.4 \times 0.6} \Rightarrow \mathrm{x}_2=25 \\ & \therefore \sqrt{\mathrm{x}_1+\mathrm{x}_2}=\sqrt{24+25}=\sqrt{49}=7\end{aligned}$

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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