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.

Ifp > 0.5, findP(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,

: rth attemptr

: the probability of failureq

: the probability of successp

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

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

or it just simply requires one constant= 0.6.p

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.