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About LinkBacksThe random variable X has pmf pX(x) = (1/2)^x+1, for x = 0,1,···. Find the probability that X is odd.
This is the question. Please kindly help. I have no idea as to how to go about it.
Thank you very much.
URGENT!!! PLEASE HELP!!
The random variable X has pmf pX(x) = (1/2)^x+1, for x = 0,1,···. Find the probability that X is odd.
This is the question. Please kindly help. I have no idea as to how to go about it.
Thank you very much.
URGENT!!! PLEASE HELP!!
The probability that X is odd is the sum of the mass function over all odd integers which is:Originally Posted by Ruichan
sum(k=0..infty) pX(2k+1) = sum(k=0..infty) (1/2)^(2k+2)
..................................= sum(k=0..infty) (1/4)^(k+1)
which is a geometric series which sums to 1/3.
RonL
Thank you very much.
I'm just wondering if there's a way to solve it without using the geometric series? My prof hasn't gone into solving it via geometric series hence he doesn't want us to solve it any other ways except the very basic steps he taught us.
I'm kind of slow when it comes to Math. Sorry if I sounded as if I don't know what I'm talking about.
Look at the prbability the X is even:
Pr(X even) = sum(k=0..infty) pX(2k+1) = 2 sum(k=0..infty) (1/2)^(2k+2) = 2 Pr(X odd)
So we can see that Pr(X even)=2 Pr(X odd), but these two must sum to 1,
so Pr(X even)=2/3 and Pr(X odd)=1/3.
RonL
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