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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
The probability distribution of a random variable X is given below:
$\text{X}$ $0$ $1$ $2$ $3$
$\text{P}(\text{X})$ $\text{k}$ $\frac{\text{k}}{2}$ $\frac{\text{k}}{4}$ $\frac{\text{k}}{8}$
  1. Determine the value of k.
  2. Determine $\text{P}(\text{X}\leq2)$ and $\text{P}(\text{X}\geq2)$
  3. Find $\text{P}(\text{X}\leq2)+\text{P}(\text{X}\geq2)$
✓

Answer

We have,
$\text{X}$ $0$ $1$ $2$ $3$
$\text{P}(\text{X})$ $\text{k}$ $\frac{\text{k}}{2}$ $\frac{\text{k}}{4}$ $\frac{\text{k}}{8}$
  1. Since, $\sum_\limits{\text{i}=1}^\text{n}\text{P}_{\text{i}}=1,\text{i}=1,2, ....,\text{n}$ and $\text{P}_{\text{i}}\geq0$
$\therefore\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}+\frac{\text{k}}{8}=1$
$\Rightarrow8\text{k}+4\text{k}+2\text{k}+\text{k}=8$
$\therefore\text{k}=\frac{8}{15}$
  1. $\text{P}(\text{X}\leq2)=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}$
$=\frac{(4\text{k}+2\text{k}+\text{k})}{4}=\frac{7\text{k}}{4}$
$=\frac{7}{4}\cdot\frac{8}{15}=\frac{14}{15}$
And $\text{P}(\text{X}\geq2)=\text{P}(3)=\frac{\text{k}}{8}$
$=\frac{1}{8}\cdot\frac{8}{15}=\frac{1}{15}$
  1. $\text{P}(\text{X}\leq2)+\text{P}(\text{X}\geq2)$
$=\frac{14}{15}+\frac{1}{15}=1$

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