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

[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,

[HR][/HR]

Does it means a range of values of

or it just simply requires one constant

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,

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

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

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

[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,

*: rth attempt***r***: the probability of failure***q***: the probability of success***p**[HR][/HR]

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

*then 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. (Hi)

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