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Question:
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The p.m.f of a random variable $\mathrm{X}$ is
$P(X=x)=\frac{1}{2^5}\left(\begin{array}{l}5 \\ x\end{array}\right), \quad x=0,12345$ then $=0 \quad$ otherwise,
Options:
$P(X=x)=\frac{1}{2^5}\left(\begin{array}{l}5 \\ x\end{array}\right), \quad x=0,12345$ then $=0 \quad$ otherwise,
Solution:
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Verified Answer
The correct answer is:
$\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X} \geq 3)$
$$
\begin{aligned}
& \mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{1}{32}{ }^5 \mathrm{C}_{\mathrm{x}} \text {, where } \mathrm{x}=0,1,2,3,4,5 \\
& =0, \text { otherwise } \\
& \therefore \mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\
& =\frac{1}{32}\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_1+{ }^5 \mathrm{C}_2\right]=\frac{1}{32}(1+5+10)=\frac{16}{32}=\frac{1}{2} \\
& \mathrm{P}(\mathrm{X} \geq 3)=\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\
& =\frac{1}{32}\left[{ }^5 \mathrm{C}_3+{ }^5 \mathrm{C}_4+{ }^5 \mathrm{C}_5\right]=\frac{1}{32}(10+5+1)=\frac{16}{32}=\frac{1}{2}
\end{aligned}
$$
\begin{aligned}
& \mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{1}{32}{ }^5 \mathrm{C}_{\mathrm{x}} \text {, where } \mathrm{x}=0,1,2,3,4,5 \\
& =0, \text { otherwise } \\
& \therefore \mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\
& =\frac{1}{32}\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_1+{ }^5 \mathrm{C}_2\right]=\frac{1}{32}(1+5+10)=\frac{16}{32}=\frac{1}{2} \\
& \mathrm{P}(\mathrm{X} \geq 3)=\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\
& =\frac{1}{32}\left[{ }^5 \mathrm{C}_3+{ }^5 \mathrm{C}_4+{ }^5 \mathrm{C}_5\right]=\frac{1}{32}(10+5+1)=\frac{16}{32}=\frac{1}{2}
\end{aligned}
$$
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