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  • The probability distribution of a random variable X is given below: (i) Determine the value of k.

The probability distribution of a random variable X is given below:
\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}) & \mathrm{K} & \frac{\mathrm{k}}{2} & \frac{\mathrm{k}}{4} & \frac{\mathrm{k}}{8} \\ \hline \end{array}
(i) Determine the value of k.
(ii) Determine P (X ≤ 2) and P (X > 2)
(iii) Find P (X ≤ 2) + P (X > 2).

Answers (1)

Solution

Given-
\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}) & \mathrm{K} & \frac{\mathrm{k}}{2} & \frac{\mathrm{k}}{4} & \frac{\mathrm{k}}{8} \\ \hline \end{array}

(i) since,

\\\sum_{i=1}^{n} P_{i}=1,2, \ldots, n$ and $P_{i} \geq 0 \\\\\therefore \mathrm{k}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{4}+\frac{\mathrm{k}}{8}=1 \\\\\Rightarrow 8 k+4 k+2 k+k=8 \\\\\therefore 15 k=8 \\\\k=\frac{8}{15}
(ii)\\\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2) \\=\mathrm{k}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{4} \\=\left(\frac{4 \mathrm{k}+2 \mathrm{k}+\mathrm{k}}{4}\right)

\\ =\frac{7 \mathrm{k}}{4}=\frac{7}{4} \times \frac{8}{15}=\frac{14}{15} \\\\\mathrm{P}(\mathrm{X}>2)=\mathrm{P}(3) \\ =\frac{\mathrm{k}}{8} \\ =\frac{1}{8} \times \frac{8}{15} \\ =\frac{1}{15} \\ \text { (iii) } \mathrm{P}(\mathrm{X} \leq 2)+\mathrm{P}(\mathrm{X}>2)=\frac{14}{15}+\frac{1}{15}=1

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