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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Aicharya · Accepted Answer

We have, $ P(x=k)=a\left(\frac{k+1}{2^k}ight) $ 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} $ $\begin{aligned} & \Rightarrow \quad a+\frac{2 a}{2}+\frac{3}{4} a+\frac{4}{8} a+\frac{5}{16} a+\frac{6}{32} a=1 \ & \Rightarrow \quad a+a+\frac{3}{4} a+\frac{1}{2} a+\frac{5}{16} a+\frac{3}{16} a=1 \ & \Rightarrow \quad \frac{16 a+16 a+12 a+8 a+5 a+3 a}{16}=1 \ & \Rightarrow \quad 60 a=16 \ & \Rightarrow \quad a=\frac{16}{60} \ & \Rightarrow \quad a=\frac{4}{15} \ & 	herefore 	ext { Required probability } \ & =P(x=2)+P(x=3)+P(x=4)=\frac{3 a}{4}+\frac{1 a}{2}+\frac{3 a}{16} \ & =\frac{12 a+8 a+3 a}{16}=\frac{23 a}{16}=\frac{23}{16} 	imes \frac{4}{15}=\frac{23}{60} . \ & \end{aligned}$

If the probability function of a random variable $X$ is defined by $P(X=k)=a\left(\frac{k+1}{2^k}\right)$…

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