Question

If the range of a random variable $X$ is $\{0,1,2,3,\dots \dots \}$ with $P(X=k)=\frac{(k+1)\left(a\right)}{3^k}$ for sample of $4$ items is drawn at random without $k\ge 0$, then $a=$

• A. $2/3$
• B. $4/9$
• C. $8/27$
• D. $16/81$

Answer 1

Answer: (B)

$4/9$

Use the concept of probability distribution function,

${\sum }_{k=0}^{\infty }\left(P\right(X=k\left)\right)=1$

So, ${\sum }_{k=0}^{\infty }\frac{(k+1)a}{3^k}=a{\sum }_{k=0}^{\infty }(\frac{k}{3^k}+\frac{1}{3^k})$

${\sum }_{k=0}^{\infty }\frac{k}{3^k}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.......=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.......=\frac{1}{3}(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+.....)$

Use the Taylor series

$\frac{1}{(1-x{)}^2}=1+2x+3x^2+4x^3+...........$

Let us assume that $x=\frac{1}{3}$

${\sum }_{k=0}^{\infty }\frac{k}{3^k}=\frac{\frac{1}{3}}{(1-\frac{1}{3}{)}^2}=\frac{3}{4}$

The second Summation will be infinite Geometric Sequence where $a=\frac{1}{3}$ and $r=\frac{1}{3}$

${\sum }_{k=0}^{\infty }\frac{1}{3^k}=1+\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{3}{2}$

Therefore,

${\sum }_{k=0}^{\infty }\left(P\right(X=k\left)\right)=1$

$(\frac{3}{4}+\frac{3}{2})\times a=1$

Hence, $a=\frac{4}{9}$