Geometric Distributions - Problem

Quote: A Concise Course in Advanced Statistics - Crawshaw, Pg. 277 Question 10.

Quote:

*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**

Reminding, Geometric Model:

$\displaystyle P(X=r) = q^{r-1}p$, $\displaystyle r = 1, 2, 3..\infty. $ And to the limit $\displaystyle \sum_{r=1 }^{\infty }P(X=r)=1$

Where,

**r** : rth attempt

**q** : the probability of failure

**p** : the probability of success

**p** (the probability of success) is a constant, but I have trouble understanding what **p** > 0.5 really means from the quoted question above.

Does it means a range of values of **p** then how to solve the question,

or it just simply requires one constant **p** = 0.6.

If that is so, there I am trying to solve it as follows:

$\displaystyle q^{2-1}p=0.1275$

$\displaystyle P(>2)=1-P(X\leq 2)$

Therefore,

$\displaystyle 1-P(X\leq 2)=1-\left [ P(1) +P(2)\right ]=1-\left [ p+qp \right ]=1-\left [ {\color{Red} 0.6}+0.1275 \right ]$ = **0.2725** $\displaystyle \neq 0.7225$

You are welcome to give your help. Thanks. (Hi)

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 $\displaystyle (1- p)^2p= 0.1275$

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 $\displaystyle p = (1- q)$ to choose only one value satisfying $\displaystyle P > 0.5.$

And Secondly, to Sir Plato for confirming the textbook printing error for $\displaystyle P > 0.5$ instead of $\displaystyle P < 0.5$ :)