Another probability distribution question

**Glitch** · February 2nd 2011, 07:06 PM

The question:

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

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

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

Then by geometric series:

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

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

What have I screwed up this time? :/

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

**mr fantastic** · February 2nd 2011, 07:19 PM

Originally Posted by Glitch

The question:

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

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

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

Then by geometric series:

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

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

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 $\displaystyle \sum_{i=0}^{k} p (1 - p)^i = p \sum_{i=0}^{k} (1 - p)^i$ . 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.

**Glitch** · February 2nd 2011, 07:28 PM

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

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