Q78 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics

If $\mathrm{x}$ is a random variable with p.m.f. as follows. $$ \begin{aligned} & P(X=x)=\frac{5}{16}, x=0,1 \\ & =\frac{k x}{48}, x=2, \text { then } E(x)= \\ & =\frac{1}{4}, x=3 \end{aligned} $$

  • A
    1.1875
  • Correct answer
    B
    1.3125
  • C
    1.5625
  • D
    0.5625

Solution

From given data, we write When $\mathrm{x}=0, \mathrm{P}=\frac{5}{16}=\frac{15}{48}$ $$ \begin{aligned} & x=1, P=\frac{5}{16}=\frac{15}{48} \\ & x=2, P=\frac{2 k}{48} \\ & x=3, P=\frac{1}{4}=\frac{12}{48} \end{aligned} $$ Here $\sum \mathrm{Pi}=1$ $$ \begin{aligned} & \therefore \frac{15}{48}+\frac{15}{48}+\frac{2 \mathrm{k}}{48}+\frac{12}{48}=1 \Rightarrow \mathrm{k}=3 \\ & \therefore \text { When } \mathrm{x}=2, \mathrm{P}=\frac{6}{48}=\frac{1}{8} \end{aligned} $$ $$ \begin{aligned} & \text { Now } \mathrm{E}=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \\ & =\left[\left(\frac{5}{16}\right)(0)\right]+\left[\left(\frac{5}{16}\right)(1)\right]+\left[\left(\frac{1}{8}\right)(2)\right]+\left[\left(\frac{1}{4}\right)(3)\right]=0+\frac{5}{16}+\frac{1}{4}+\frac{3}{4}=\frac{21}{16}=1.3125 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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