Geometric Distributions - Problem

Re: Geometric Distributions - Problem

In order that "the first success occurs on the second trial", there must be failure on the first trial and a success on the second. The probability of failure on the first attempt is 1- p and the probability of success on the second trial is (1- p)p. So the probability of "first success on the second trial is

Solve p- p^3= 0.1275 or p^3- p- 0.1275. You don't say how you got p= 0.6. It certainly does not satisfy that. I get between 9.0 and 9.1.

Re: Geometric Distributions - Problem

Quote:

Originally Posted by

**zikcau25** X~Geo(p) and the probability that the first success is obtained on the second attempt is 0.1275.

If p > 0.5, find P(X > 2).

Textbook ANS: 0.7225

I think that the given answer should be $0.0225$. It is just a typo.

$\mathcal{P}(X=2)=(1-p)p=0.1275$ gives $p=0.85$ (that is p>.5).

So $1-(0.85+0.1275)=0.0225$.

Re: Geometric Distributions - Problem

Thanks to Sir HallsofIvy, for reminding me to evaluate, quadratically, two values of p from the relationship to choose only one value satisfying

And Secondly, to Sir Plato for confirming the textbook printing error for instead of :)