# Math Help - Probability pmf

1. ## Probability pmf

Let X be the number of accidents in a factory per week. , It has the pmf

f(x) = 1 / (x+1)(x+2) ; x=0,1,2,...

Find the conditional probability of x > or = 4, given that x >or = 1

2. Originally Posted by amor_vincit_omnia
Let X be the number of accidents in a factory per week. , It has the pmf

f(x) = 1 / (x+1)(x+2) ; x=0,1,2,...

Find the conditional probability of x > or = 4, given that x >or = 1
$\Pr(X \geq 4 | X \geq 1) = \frac{\Pr(X \geq 4) ~ \text{and} ~ \Pr(X \geq 1)}{\Pr(X \geq 1)} = \frac{\Pr(X \geq 4)}{\Pr(X \geq 1)}$.

Do you know how to get these two probabilities from the pmf?

3. Originally Posted by mr fantastic
$\Pr(X \geq 4 | X \geq 1) = \frac{\Pr(X \geq 4) ~ \text{and} ~ \Pr(X \geq 1)}{\Pr(X \geq 1)} = \frac{\Pr(X \geq 4)}{\Pr(X \geq 1)}$.

Do you know how to get these two probabilities from the pmf?
$f(x) = \frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}$.

Therefore $\Pr(x \geq 4) = \sum_{x=4}^{\infty} \left( \frac{1}{x+1} - \frac{1}{x+2} \right) = \left( \frac{1}{5} - \frac{1}{6} \right) + \left( \frac{1}{6} - \frac{1}{7} \right) + ...... = \frac{1}{5}$.

(This is an example of a telescoping series)

$\Pr(x \geq 1)$ is got in a similar way ....

4. Originally Posted by mr fantastic
$\Pr(x \geq 1)$ is got in a similar way ....
Also, note that $\Pr(x \geq 1) = 1 - \Pr(x < 1) \Rightarrow 1 - \Pr(x = 0)$