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If the p.m.f. of a r.v. $\mathrm{X}$ is given by $\mathrm{P}(\mathrm{X}=x)=\frac{\left(\begin{array}{l}5 \\ x\end{array}\right)}{2^{5}}$ if $x=0,1,2, \ldots \ldots 5$ $=0$ otherwise then which of the following is not true?
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Verified Answer
The correct answer is:
$P(X \leq 3) \leq P(X \geq 3)$
(B)
$\begin{aligned}
P(X \leq 2) &=P(X=0)+P(X=1)+P(X=2) \\
&=\frac{{ }^{5} C_{0}}{2^{5}}+\frac{{ }^{5} C_{1}}{2^{5}}+\frac{{ }^{5} C_{2}}{2^{5}}=\frac{1+5+10}{2^{5}}=\frac{16}{2^{5}} \\
P(X \geq 4) &=P(X=4)+P(X=5) \\
&=\frac{C_{4}}{2^{4}}+\frac{C_{6}}{2^{6}}=\frac{5+1}{2^{5}}=\frac{6}{2^{5}}
\end{aligned}$
$\begin{aligned} \mathrm{P}(\mathrm{X} \leq 3) &=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ &=\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{2}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}=\frac{1+5+10+10}{2^{5}}=\frac{26}{2^{5}} \\ \mathrm{P}(\mathrm{X} \geq 3) &=\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\ &=\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{4}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{5}}{2^{5}}=\frac{10+5+1}{2^{5}}=\frac{16}{2^{5}} \\ \mathrm{P}(\mathrm{X} \leq 1) &=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\ &=\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}=\frac{1+5}{2^{5}}=\frac{6}{2^{5}} \end{aligned}$
$\begin{aligned}
P(X \leq 2) &=P(X=0)+P(X=1)+P(X=2) \\
&=\frac{{ }^{5} C_{0}}{2^{5}}+\frac{{ }^{5} C_{1}}{2^{5}}+\frac{{ }^{5} C_{2}}{2^{5}}=\frac{1+5+10}{2^{5}}=\frac{16}{2^{5}} \\
P(X \geq 4) &=P(X=4)+P(X=5) \\
&=\frac{C_{4}}{2^{4}}+\frac{C_{6}}{2^{6}}=\frac{5+1}{2^{5}}=\frac{6}{2^{5}}
\end{aligned}$
$\begin{aligned} \mathrm{P}(\mathrm{X} \leq 3) &=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ &=\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{2}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}=\frac{1+5+10+10}{2^{5}}=\frac{26}{2^{5}} \\ \mathrm{P}(\mathrm{X} \geq 3) &=\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\ &=\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{4}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{5}}{2^{5}}=\frac{10+5+1}{2^{5}}=\frac{16}{2^{5}} \\ \mathrm{P}(\mathrm{X} \leq 1) &=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\ &=\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}=\frac{1+5}{2^{5}}=\frac{6}{2^{5}} \end{aligned}$
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