The p.m.f of a random variable $\mathrm{X}$ is $P(X=x)=\frac{1}{2^5}\left(\begin{array}{l}5 \ x\end{array} ight), \quad x=0,12345$ then $=0 \quad$ otherwise,

Question

The p.m.f of a random variable $\mathrm{X}$ is $P(X=x)=\frac{1}{2^5}\left(\begin{array}{l}5 \ x\end{array}ight), \quad x=0,12345$ then $=0 \quad$ otherwise,

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{1}{32}{ }^5 \mathrm{C}_{\mathrm{x}} 	ext {, where } \mathrm{x}=0,1,2,3,4,5 \ & =0, 	ext { otherwise } \ & 	herefore \mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \ & =\frac{1}{32}\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_1+{ }^5 \mathrm{C}_2ight]=\frac{1}{32}(1+5+10)=\frac{16}{32}=\frac{1}{2} \ & \mathrm{P}(\mathrm{X} \geq 3)=\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \ & =\frac{1}{32}\left[{ }^5 \mathrm{C}_3+{ }^5 \mathrm{C}_4+{ }^5 \mathrm{C}_5ight]=\frac{1}{32}(10+5+1)=\frac{16}{32}=\frac{1}{2} \end{aligned}$

The p.m.f of a random variable $\mathrm{X}$ is $P(X=x)=\frac{1}{2^5}\left(\begin{array}{l}5 \\…

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