Mathematics › Probability › Random Variable and its Probability Distribution
The p.m.f. of a random variable $X$ is given by $\begin{aligned} \mathrm{P}[\mathrm{X}=x] &…
The p.m.f. of a random variable $X$ is given by
$\begin{aligned}
\mathrm{P}[\mathrm{X}=x] & =\frac{\binom{5}{x}}{2^5}, \text { if } x=0,1,2,3,4,5 \\
& =0, \text { otherwise }
\end{aligned}$
Then which of the following is not correct?
$\mathrm{P}[\mathrm{X}=0]=\mathrm{P}[\mathrm{X}=5]$ $\mathrm{P}[\mathrm{X} \leq 1]=\mathrm{P}[\mathrm{X} \geq 4]$ $\mathrm{P}[\mathrm{X} \leq 2]=\mathrm{P}[\mathrm{X} \geq 3]$ $\mathrm{P}[\mathrm{X} \leq 2]\gt\mathrm{P}[\mathrm{X} \geq 3]$ Solution
$\begin{aligned} \mathrm{P}(\mathrm{X} \leq 1) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\ & =\frac{{ }^5 \mathrm{C}_0}{2^5}+\frac{{ }^5 \mathrm{C}_1}{2^5}=\frac{6}{2^5} \\ \mathrm{P}(\mathrm{X} \leq 2) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\ & =\frac{{ }^5 \mathrm{C}_0}{2^5}+\frac{{ }^5 \mathrm{C}_1}{2^5}+\frac{{ }^5 \mathrm{C}_2}{2^5}=\frac{16}{2^5}\end{aligned}$
$\begin{aligned}
\mathrm{P}(\mathrm{X} \geq 3) & =\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\
& =\frac{{ }^5 \mathrm{C}_3}{2^5}+\frac{{ }^5 \mathrm{C}_4}{2^5}+\frac{{ }^5 \mathrm{C}_5}{2^5}=\frac{16}{2^5} \\
\mathrm{P}(\mathrm{X} \leq 3) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\
& +\mathrm{P}(\mathrm{X}=3) \\
& =\frac{{ }^5 \mathrm{C}_0}{2^5}+\frac{{ }^5 \mathrm{C}_1}{2^5}+\frac{{ }^5 \mathrm{C}_2}{2^5}+\frac{{ }^5 \mathrm{C}_3}{2^5}=\frac{26}{2^5} \\
\mathrm{P}(\mathrm{X} \geq 4) & =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\
& =\frac{{ }^5 \mathrm{C}_4}{2^5}+\frac{{ }^5 \mathrm{C}_5}{2^5}=\frac{6}{2^5}
\end{aligned}$
$\therefore \quad \mathrm{P}(\mathrm{X} \leq 2)\gt\mathrm{P}(\mathrm{X} \geq 3)$ is not true.
Asked in: MHT CET 2024 (11 May Shift 1)
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