Thread: 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

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.

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.

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!).