The random variable X has pmf pk 052k k 01 Find the probab

The random variable X has pmf p(k) = 0.5/(2^k), k = 0,1,... Find the probability that X is even.

Solution

P(even) = P(0) + P(2) + P(4) + P(6) ...

= 0.5/(0^2) + 0.5/(2^2) + 0.5/(2^4) + 0.5/(2^6) ...

As we can see, this is an infinite geometric series with first term a1 = 0.5/2^0 = 0.5, and a common ratio of 1/2^2 = 0.25.

Thus, as the sum of an infinite geometric series is S = a1 / (1-r),

P(even) = P(0) + P(2) + P(4) + P(6) ...

= 0.5/(2^0) + 0.5/(2^2) + 0.5/(2^4) + 0.5/(2^6) ...

= 0.5 / (1-0.25)

= 2/3 OR 0.66666667 [ANSWER]