The problem asks us to find the probability $P(X\geq 3)$ for a discrete random variable X, whose probability mass function (PMF) is given by $P(X= x) = k(x + 1)3^{-x}$ for $x = 0, 1, 2, 3\dots$. Here, $k$ is a constant that needs to be determined first.
The sum of probabilities for all possible values of a random variable must equal 1. Therefore, we have:
$ \sum_{x=0}^{\infty} P(X=x) = 1 $
Substituting the given PMF:
$ \sum_{x=0}^{\infty} k(x + 1)3^{-x} = 1 $
We can factor out the constant $k$:
$ k \sum_{x=0}^{\infty} (x + 1)\left(\frac{1}{3}\right)^x = 1 $
The sum $\sum_{x=0}^{\infty} (x + 1)r^x$ is known to be equal to $\frac{1}{(1-r)^2}$ for $|r|<1$. In this case, $r = 1/3$.
So, the sum evaluates to:
$ \sum_{x=0}^{\infty} (x + 1)\left(\frac{1}{3}\right)^x = \frac{1}{\left(1 - \frac{1}{3}\right)^2} = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4} $
Now, substitute this back into the equation for $k$:
$ k \left(\frac{9}{4}\right) = 1 $
Solving for $k$:
$ k = \frac{4}{9} $
The PMF is therefore $P(X=x) = \frac{4}{9}(x + 1)3^{-x}$.
We want to find $P(X\geq 3)$. It is often easier to calculate this using the complement rule: $P(X\geq 3) = 1 - P(X < 3)$.
$P(X < 3)$ is the sum of probabilities for $x=0, 1, 2$:
$ P(X < 3) = P(X=0) + P(X=1) + P(X=2) $
Let's calculate each term:
Now, sum these probabilities:
$ P(X < 3) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} $
To add these fractions, find a common denominator, which is 27:
$ P(X < 3) = \frac{4 \times 3}{9 \times 3} + \frac{8}{27} + \frac{4}{27} = \frac{12}{27} + \frac{8}{27} + \frac{4}{27} = \frac{12 + 8 + 4}{27} = \frac{24}{27} $
Finally, calculate $P(X\geq 3)$:
$ P(X\geq 3) = 1 - P(X < 3) = 1 - \frac{24}{27} = \frac{27 - 24}{27} = \frac{3}{27} $
Simplify the fraction:
$ P(X\geq 3) = \frac{1}{9} $
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