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A random variable $\mathrm{X}$ has the range $\{0,1,2, \ldots\}$. If $\mathrm{P}(X=r)=k(1+r) 3^{-r}$, for $r=0,1,2, \ldots$. where $k>0$ is a real number, then $P(X=0)+P(X=1)+P(X=2)=$
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Verified Answer
The correct answer is:
$\frac{8}{9}$
Given 9 robability function $\mathrm{P}(\mathrm{X}=r)=\mathrm{K}(1+r) 3^{-r}$. The sum of all probabilities is equal to 1 .
So, $\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+$ $=1$ $\mathrm{K}\left[1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \ldots ..\right]=1$
Let $S=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+$
Multiply both side by $\frac{1}{3}$.
$$
\frac{1}{3} S=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\ldots
$$
Subtract (ii) from (i).
$$
\begin{gathered}
\mathrm{S}=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \ldots \ldots . . \\
\frac{\frac{1}{3} \mathrm{~S}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots \ldots \ldots \ldots .}{\frac{2}{3} \mathrm{~S}=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \ldots \ldots} \\
\frac{2}{3} \mathrm{~S}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}
\end{gathered}
$$
$$
\mathrm{S}=\frac{9}{4}
$$
Put in eq. (a), $\mathrm{K} \times \frac{9}{4}=1$.
$$
\mathrm{K}=\frac{4}{9}
$$
Now $\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{4}{9}\left[1+\frac{2}{3}+\frac{3}{9}\right]$
$$
=\frac{4}{9} \times\left[\frac{9+6+3}{9}\right]=\frac{4}{9} \times \frac{18}{9}=\frac{8}{9}
$$
Therefore, correct option is (b).
So, $\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+$ $=1$ $\mathrm{K}\left[1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \ldots ..\right]=1$
Let $S=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+$
Multiply both side by $\frac{1}{3}$.
$$
\frac{1}{3} S=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\ldots
$$
Subtract (ii) from (i).
$$
\begin{gathered}
\mathrm{S}=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \ldots \ldots . . \\
\frac{\frac{1}{3} \mathrm{~S}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots \ldots \ldots \ldots .}{\frac{2}{3} \mathrm{~S}=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \ldots \ldots} \\
\frac{2}{3} \mathrm{~S}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}
\end{gathered}
$$
$$
\mathrm{S}=\frac{9}{4}
$$
Put in eq. (a), $\mathrm{K} \times \frac{9}{4}=1$.
$$
\mathrm{K}=\frac{4}{9}
$$
Now $\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{4}{9}\left[1+\frac{2}{3}+\frac{3}{9}\right]$
$$
=\frac{4}{9} \times\left[\frac{9+6+3}{9}\right]=\frac{4}{9} \times \frac{18}{9}=\frac{8}{9}
$$
Therefore, correct option is (b).
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