Geometric Distributions - Problem

Apr 2012
52
2
Rose Belle, Mauritius
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
Apr 2005
20,249
7,909
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.
 
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Plato

MHF Helper
Aug 2006
22,474
8,643
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 $1-(0.85+0.1275)=0.0225$.
 
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Apr 2012
52
2
Rose Belle, Mauritius
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 \(\displaystyle P < 0.5\) :)