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The random variable $\mathbf{X}$ has a probability distribution $\mathbf{P}$ (X) of the following form, where $k$ is some number:
$$
P(X)=\left\{\begin{array}{c}
k, \text { if } x=0 \\
2 k, \text { if } x=1 \\
3 k, \text { if } x=2 \\
0, \text { otherwise }
\end{array}\right.
$$
(a) Determine the value of $\mathbf{k}$
(b) Find $P(X < 2), P(X \leq 2), P(X \geq 2)$
$$
P(X)=\left\{\begin{array}{c}
k, \text { if } x=0 \\
2 k, \text { if } x=1 \\
3 k, \text { if } x=2 \\
0, \text { otherwise }
\end{array}\right.
$$
(a) Determine the value of $\mathbf{k}$
(b) Find $P(X < 2), P(X \leq 2), P(X \geq 2)$
Solution:
1963 Upvotes
Verified Answer
(a) Sum of probabilities $=1$
$$
\mathrm{k}+2 \mathrm{k}+3 \mathrm{k}=1 \text { or } 6 \mathrm{k}=1, \mathrm{k}=\frac{1}{6}
$$
The probability distribution is as given below
(b) (i) $\mathrm{P}(\mathrm{X} < 2)=\mathrm{P}(0)+\mathrm{P}(1)$
$$
=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}
$$
(ii) $\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)$
$$
=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}=\frac{6}{6}=1
$$
(iii) $\mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(2)+\mathrm{P}(3)+\mathrm{P}(4)+\ldots \ldots$
$$
=\frac{3}{6}+0+0+\ldots \ldots=\frac{3}{6}=\frac{1}{2}
$$
$$
\mathrm{k}+2 \mathrm{k}+3 \mathrm{k}=1 \text { or } 6 \mathrm{k}=1, \mathrm{k}=\frac{1}{6}
$$
The probability distribution is as given below
(b) (i) $\mathrm{P}(\mathrm{X} < 2)=\mathrm{P}(0)+\mathrm{P}(1)$
$$
=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}
$$
(ii) $\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)$
$$
=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}=\frac{6}{6}=1
$$
(iii) $\mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(2)+\mathrm{P}(3)+\mathrm{P}(4)+\ldots \ldots$
$$
=\frac{3}{6}+0+0+\ldots \ldots=\frac{3}{6}=\frac{1}{2}
$$
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