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If the probability function of a random variable $X$ is defined by $P(X=k)=a\left(\frac{k+1}{2^k}\right)$ for $k=0,1,2,3,4,5$, then the probability that $X$ takes a prime value is
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Verified Answer
The correct answer is:
$\frac{23}{60}$
We have,
$$
P(x=k)=a\left(\frac{k+1}{2^k}\right)
$$
We know that,
$$
\begin{gathered}
\Sigma P(x=k)=1 \\
\Rightarrow \quad P(x=0)+P(x=1) \\
+P(x=2)+P(x=3)+P(x=4)+P(x=5)=1
\end{gathered}
$$
$\begin{aligned} & \Rightarrow \quad a+\frac{2 a}{2}+\frac{3}{4} a+\frac{4}{8} a+\frac{5}{16} a+\frac{6}{32} a=1 \\ & \Rightarrow \quad a+a+\frac{3}{4} a+\frac{1}{2} a+\frac{5}{16} a+\frac{3}{16} a=1 \\ & \Rightarrow \quad \frac{16 a+16 a+12 a+8 a+5 a+3 a}{16}=1 \\ & \Rightarrow \quad 60 a=16 \\ & \Rightarrow \quad a=\frac{16}{60} \\ & \Rightarrow \quad a=\frac{4}{15} \\ & \therefore \text { Required probability } \\ & =P(x=2)+P(x=3)+P(x=4)=\frac{3 a}{4}+\frac{1 a}{2}+\frac{3 a}{16} \\ & =\frac{12 a+8 a+3 a}{16}=\frac{23 a}{16}=\frac{23}{16} \times \frac{4}{15}=\frac{23}{60} . \\ & \end{aligned}$
$$
P(x=k)=a\left(\frac{k+1}{2^k}\right)
$$
We know that,
$$
\begin{gathered}
\Sigma P(x=k)=1 \\
\Rightarrow \quad P(x=0)+P(x=1) \\
+P(x=2)+P(x=3)+P(x=4)+P(x=5)=1
\end{gathered}
$$
$\begin{aligned} & \Rightarrow \quad a+\frac{2 a}{2}+\frac{3}{4} a+\frac{4}{8} a+\frac{5}{16} a+\frac{6}{32} a=1 \\ & \Rightarrow \quad a+a+\frac{3}{4} a+\frac{1}{2} a+\frac{5}{16} a+\frac{3}{16} a=1 \\ & \Rightarrow \quad \frac{16 a+16 a+12 a+8 a+5 a+3 a}{16}=1 \\ & \Rightarrow \quad 60 a=16 \\ & \Rightarrow \quad a=\frac{16}{60} \\ & \Rightarrow \quad a=\frac{4}{15} \\ & \therefore \text { Required probability } \\ & =P(x=2)+P(x=3)+P(x=4)=\frac{3 a}{4}+\frac{1 a}{2}+\frac{3 a}{16} \\ & =\frac{12 a+8 a+3 a}{16}=\frac{23 a}{16}=\frac{23}{16} \times \frac{4}{15}=\frac{23}{60} . \\ & \end{aligned}$
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