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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions2 Marks
MCQ
A random variable $X$ has the following probability distribution:
$\mathrm{X}=\boldsymbol{x}$1234
P($\mathrm{X}=\boldsymbol{x}$)$\frac{1}{8}$$\frac{1}{2}$$\frac{1}{4}$k
Then the value of k is
  • A
    $\frac{1}{4}$
  • ✓
    $\frac{1}{8}$
  • C
    $\frac{3}{8}$
  • D
    $\frac{1}{2}$
✓

Answer

Correct option: B.
$\frac{1}{8}$
(B)
$Since \sum_{x=1}^4 \mathrm{P}(\mathrm{X}=x)=1,$
$\begin{aligned} & \frac{1}{8}+\frac{1}{2}+\frac{1}{4}+\mathrm{k}=1 \\ & \Rightarrow \mathrm{k}+\frac{1+4+2}{8}=1 \\ & \Rightarrow \mathrm{k}=1-\frac{7}{8}=\frac{1}{8}\end{aligned}$

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