Question ID: #1177
If the probability that the random variable $X$ takes the value $x$ is given by $P(X=x)=k(x+1)3^{-x}$, $x=0,1,2,3,\dots$ where $k$ is a constant, then $P(X\ge3)$ is equal to
- (1) $\frac{7}{27}$
- (2) $\frac{4}{9}$
- (3) $\frac{8}{27}$
- (4) $\frac{1}{9}$
Solution:
$$\sum_{x=0}^{\infty} P(X=x) = 1$$
$$\sum_{x=0}^{\infty} k(x+1)3^{-x} = 1$$
$$k \left( 1 + 2(3^{-1}) + 3(3^{-2}) + 4(3^{-3}) + \dots \right) = 1$$
$$\frac{1}{k} = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \dots \quad \text{— (i)}$$
$$\frac{1}{3k} = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \dots \quad \text{— (ii)}$$
$$\text{Subtract (ii) from (i):}$$
$$\frac{1}{k} – \frac{1}{3k} = 1 + \left(\frac{2}{3} – \frac{1}{3}\right) + \left(\frac{3}{3^2} – \frac{2}{3^2}\right) + \dots$$
$$\frac{2}{3k} = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots$$
$$S_{\infty} = \frac{a}{1-r}$$
$$\frac{2}{3k} = \frac{1}{1 – \frac{1}{3}}$$
$$\frac{2}{3k} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$$
$$k = \frac{4}{9}$$
$$P(X \ge 3) = 1 – (P(X=0) + P(X=1) + P(X=2))$$
$$P(X \ge 3) = 1 – k \left( (0+1)3^{-0} + (1+1)3^{-1} + (2+1)3^{-2} \right)$$
$$P(X \ge 3) = 1 – \frac{4}{9} \left( 1 + \frac{2}{3} + \frac{3}{9} \right)$$
$$P(X \ge 3) = 1 – \frac{4}{9} \left( \frac{9 + 6 + 3}{9} \right)$$
$$P(X \ge 3) = 1 – \frac{4}{9} \left( \frac{18}{9} \right)$$
$$P(X \ge 3) = 1 – \frac{4}{9} (2)$$
$$P(X \ge 3) = 1 – \frac{8}{9}$$
$$P(X \ge 3) = \frac{1}{9}$$
Ans. (4)
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