# Geometric Distributions - Problem

#### zikcau25

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

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

[HR][/HR]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
[HR][/HR]
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)

Last edited:

#### HallsofIvy

MHF Helper
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.

• 1 person

#### Plato

MHF Helper
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 $1-(0.85+0.1275)=0.0225$.

• 1 person

#### zikcau25

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 $$\displaystyle P < 0.5$$ 