Question

If $x$ is a random variable with probability distribution $p(x=k)=\frac{(k+1)C}{2^k},k=0,1,2,3,............$, then find $C.$

Answer 1

Given:- $P(x=k)=\frac{(k+1)C}{2^k}$

To find:- $C=?$

$P(x=0)=\frac{(0+1)C}{2^0}=C$

$P(x=1)=\frac{(1+1)C}{2^1}=\frac{2C}{2^1}$

$P(x=2)=\frac{(2+1)C}{2^2}=\frac{3C}{2^2}$

$P(x=0)=\frac{(3+1)C}{2^3}=\frac{4C}{2^3}$

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As we know,

$P(x=0)+P(x=1)+P(x=2)+.....=1$

$\therefore 1=\frac{C}{2^0}+\frac{2C}{2^1}+\frac{3C}{2^2}+..........\left(1\right)$

Dividing the above equation by $2$, we get

$\frac{1}{2}=\frac{C}{2^1}+\frac{2C}{2^2}+\frac{3C}{2^3}+..........\left(2\right)$

Subtracting equation $\left(2\right)$ from $\left(1\right)$, we have

$1-\frac{1}{2}=(\frac{C}{2^0}+\frac{2C}{2^1}+\frac{3C}{2^2}+.....)-(\frac{C}{2^1}+\frac{2C}{2^2}+\frac{3C}{2^3}+.....)$

$\frac{1}{2}=C+(\frac{2C}{2^1}-\frac{C}{2^1})+(\frac{2C}{2^2}-\frac{C}{2^2})+(\frac{2C}{2^3}-\frac{C}{2^3})+.....$

$\Rightarrow \frac{1}{2}=C+\frac{C}{2^1}+\frac{C}{2^2}+\frac{C}{2^3}+.....$

$\Rightarrow \frac{1}{2}=\frac{C}{(1-\frac{1}{2})}$

$\Rightarrow 2C=\frac{1}{2}$

$\Rightarrow C=\frac{1}{4}$

Hence the value of $C$ is $\frac{1}{4}$.