# Thread: help with geometric probability

1. ## help with geometric probability

hey

I'm stuck on one of the question in our online test in statistics:

Suppose that X ε [1,2,..., ] is discrete random variable that follows a geometric distribution with mean 3.74.
What is the probability that X is an even number?

Give your solution accurate to 4 decimal places.

my idea was:

from geometric sequence:

x=1/theta

so:

3.14=1/theta -> theta=0.26737968

and then use P(X=x)=(1-theta)^x-1 * theta

to get P=0.1373945

however don't know if that's what the question is asking about (i.e what is the 'successs' in this geometric distribution, number being even?) another idea was to put:

P(X=2x), if that makes any sense, to account for the number being even however don't know how to go from there

any thoughts?
would appreciate any help

2. ## Re: help with geometric probability

Originally Posted by bart123456
hey

I'm stuck on one of the question in our online test in statistics:

Suppose that X ε [1,2,..., ] is discrete random variable that follows a geometric distribution with mean 3.74.
What is the probability that X is an even number?

Give your solution accurate to 4 decimal places.

my idea was:

from geometric sequence:

x=1/theta

so:

3.14=1/theta -> theta=0.26737968

and then use P(X=x)=(1-theta)^x-1 * theta

to get P=0.1373945

however don't know if that's what the question is asking about (i.e what is the 'successs' in this geometric distribution, number being even?) another idea was to put:

P(X=2x), if that makes any sense, to account for the number being even however don't know how to go from there

any thoughts?
would appreciate any help
Mean = 1/p => p = 1/3.75 = 4/15.

Pr(X = even) = Pr(X = 2) + Pr(X = 4) + Pr(X = 6) + .... = 1 - p using the pmf of the geometric distribution and the formula for the sum of an infinite series.

Details are left for you to do.

3. ## Re: help with geometric probability

Thank you so much!
Maybe I don't understand subject so well, but we only did infinite series sums if
-1<r<1 so don't know how to compute the sum with r=2.

I guessed to use formula

((1-θ)^2i-1)θ

and sum it from i=1 up to infinity
but I don't know how to obtain a numerical answer not in terms of ratio of i.

4. ## Re: help with geometric probability

Originally Posted by bart123456
Thank you so much!
Maybe I don't understand subject so well, but we only did infinite series sums if
-1<r<1 so don't know how to compute the sum with r=2.

I guessed to use formula

((1-θ)^2i-1)θ

and sum it from i=1 up to infinity
but I don't know how to obtain a numerical answer not in terms of ratio of i.
The common ratio r is NOT 2. I don't know how you got that since you haven't shown your working. The common ratio is a number between 0 and 1 (otherwise the infinite series would not sum to a finite number!).