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(B) Given that $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \in \{0, 1, 2, \ldots, \infty\}$.We know that the sum of all probabilities in a probability distri

Vedclass · Accepted Answer

$\frac{4}{9}$

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(B) Given that $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \in \{0, 1, 2, \ldots, \infty\}$.We know that the sum of all probabilities in a probability distribution is $1$, i.e.,$\sum_{k=0}^{\infty} P(X=k) = 1$.Let $S = \sum_{k=0}^{\infty} \frac{(k+1)a}{3^k} = a \left( 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \infty ight) = 1$.Let $S = a \left( 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \ldots ight)$.Multiplying by $\frac{1}{3}$, we get $\frac{1}{3}S = a \left( \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \ldots ight)$.Subtracting the two equations:$S - \frac{1}{3}S = a \left( 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{3}{9} - \frac{2}{9}) + (\frac{4}{27} - \frac{3}{27}) + \ldots ight)$$\frac{2}{3}S = a \left( 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots ight)$.The term in the bracket is an infinite geometric series with first term $1$ and common ratio $r = \frac{1}{3}$.Sum $= \frac{a}{1-r} = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2}$.So, $\frac{2}{3}S = a \left( \frac{3}{2} ight) = \frac{3a}{2}$.Thus, $S = \frac{3a}{2} 	imes \frac{3}{2} = \frac{9a}{4}$.Since $S = 1$, we have $\frac{9a}{4} = 1$, which implies $a = \frac{4}{9}$.

If the range of a random variable $X$ is $\{0, 1, 2, 3, 4, \ldots\}$ with $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \geq 0$, then $a$ is equal to

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