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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions2 Marks
MCQ
If the probability mass function of a discre random variable X is
$\begin{aligned}\mathrm{P}(x) & =\frac{\mathrm{C}}{x^3} ; \quad x=1,2,3 \\& =0 ; \quad \text { otherwise }\end{aligned}$
Then $\mathrm{E}(\mathrm{X})=$
  • A
    $\frac{343}{297}$
  • ✓
    $\frac{294}{251}$
  • C
    $\frac{297}{294}$
  • D
    $\frac{251}{294}$
✓

Answer

Correct option: B.
$\frac{294}{251}$
(B)
$\mathrm{P}(1)=\frac{\mathrm{C}}{1^3}, \mathrm{P}(2)=\frac{\mathrm{C}}{2^3}, \mathrm{P}(3)=\frac{\mathrm{C}}{3^3}$
$\text { Since, } \mathrm{P}(1)+\mathrm{P}(2)+\mathrm{P}(3)=1$
$\therefore \frac{\mathrm{C}}{1^3}+\frac{\mathrm{C}}{2^3}+\frac{\mathrm{C}}{3^3}=1$
$\Rightarrow \mathrm{C}\left(\frac{1}{1}+\frac{1}{8}+\frac{1}{27}\right)=1$
$\Rightarrow \mathrm{C}\left(\frac{216+27+8}{216}\right)=1$
$\Rightarrow \mathrm{C}=\frac{216}{251}$
$\therefore E ( X )=\sum x_{ i } \cdot P \left(x_{ i }\right)$
$=(1) \frac{\mathrm{C}}{1^3}+(2) \frac{\mathrm{C}}{2^3}+(3) \frac{\mathrm{C}}{3^3}$
$=\mathrm{C}\left(1+\frac{1}{4}+\frac{1}{9}\right)=\mathrm{C}\left(\frac{36+9+4}{36}\right) $
$=\frac{216}{251} \times \frac{49}{36}$
$=\frac{294}{251}$

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