If the probability that the random variable X takes the value x is given by P(X=x)=k(x+1) …

Question

If the probability that the random variable $X$ takes the value $x$ is given by
$P(X=x)=k(x+1) 3^{-x}, x=0,1,2,3 \ldots$, where $k$ is a constant, then $P(X \geq 3)$ is equal to

Accepted Answer

To find $ P(X \geq 3) $, we first determine the constant $ k $ using the total probability for $ X $.

The probability $ P(X = x) $ is given by:

$ P(X = x) = k(x+1) \cdot 3^{-x}, \quad x = 0, 1, 2, 3, \ldots $

The total probability must equal 1:

$ s = \sum_{x = 0}^{\infty} k(x+1) \cdot 3^{-x} $

Calculating that series:

$ s = \frac{k}{3^0} + \frac{2k}{3} + \frac{3k}{3^2} + \ldots $

Therefore, dividing the series by 3:

$ \frac{s}{3} = \frac{k}{3} + \frac{2k}{3^2} + \ldots $

Subtracting these:

$ s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \ldots $

The resulting series is a geometric series:

$ \frac{2s}{3} = k \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \ldots \right) $

The sum of the infinite geometric series is:

$ \frac{2s}{3} = k \cdot \frac{1}{1-\frac{1}{3}} = \frac{3k}{2} $

Equating:

$ s = \frac{9k}{4} = 1 $

Thus, solving for $ k $:

$ k = \frac{4}{9} $

Next, compute $ P(X \geq 3) $:

$ P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) $

Calculating these:

$ P(X = 0) = k = \frac{4}{9} $

$ P(X = 1) = \frac{2k}{3} = \frac{8}{27} $

$ P(X = 2) = \frac{3k}{9} = \frac{4}{27} $

Adding these probabilities:

$ P(X = 0) + P(X = 1) + P(X = 2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{8}{9} $

Finally, calculate $ P(X \geq 3) $:

$ P(X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9} $

Solutions · 1