# Thread: Another probability distribution question

1. ## Another probability distribution question

The question:

Calculate $\displaystyle P(X \le k)$when X has the probability distribution

$\displaystyle P(X = k) = p(1 - p)^k$, for k = 0, 1, 2, ... and 0 < p < 1.

My attempt:
$\displaystyle \sum\limits_{k = 0}^{\infty} p(1 - p)^k$

Then by geometric series:

= $\displaystyle p(1 - p)\frac{1 - (1 - p)^k }{1 - (1 - p)}$
= $\displaystyle (1 - p)-(1 - p)^{k + 1}$

The answer is supposed to be $\displaystyle 1 - (1 - p)^{k+1}$

What have I screwed up this time? :/

EDIT: It has occurred to me that I've done the geometric series wrong. >_<

2. Originally Posted by Glitch
The question:

Calculate $\displaystyle P(X \le k)$when X has the probability distribution

$\displaystyle P(X = k) = p(1 - p)^k$, for k = 0, 1, 2, ... and 0 < p < 1.

My attempt:
$\displaystyle \sum\limits_{k = 0}^{\infty} p(1 - p)^k$

Then by geometric series:

= $\displaystyle p(1 - p)\frac{1 - (1 - p)^k }{1 - (1 - p)}$
= $\displaystyle (1 - p)-(1 - p)^{k + 1}$

The answer is supposed to be $\displaystyle 1 - (1 - p)^{k+1}$

What have I screwed up this time? :/

EDIT: It has occurred to me that I've done the geometric series wrong. >_<
To start with, you should be calculating $\displaystyle \displaystyle \sum_{i=0}^{k} p (1 - p)^i = p \sum_{i=0}^{k} (1 - p)^i$. And yes, you have done the geometric series wrong. I suggest you go back and review the appropriate formula and then use the formula more carefully.

3. Yeah, once I applied it properly I got the correct answer. Thanks.