Probability Help

[Guest]

Hi. I am having trouble with the following problem:
Pick a subset of the integers (1,2,...,100), such that the probability of picking a subset of size k is proportional to k. What is the probability that we pick the subset containing all even numbers?

I got (1/101)* (1/2^99), but the back of the book says
(1/2)*(1/2^99). Could someone explain it please?
Thank you!

[shyamjayakannan]

since it is proportional, let p(subset of size k) = ak, where a is a constant. Now, [tex]\sum_{k=1}^{100}[/tex]p(subset of size k) = 1:

[tex]\Rightarrow\sum_{k=1}^{100}ak\frac{{}^{100}C_k}{2^{100}}=1\Rightarrow\frac{a}{2^{100}}\sum_{k=1}^{100}k\times\frac{100}{k}{}^{99}C_{k-1}=1\Rightarrow\frac{100a}{2^{100}}\sum_{k=0}^{99}{}^{99}C_k=1\Rightarrow\frac{100a}{2^{100}}\times2^{99}=1\Rightarrow a=\frac{1}{50}[/tex]

So, p(subset of size k) = [tex]\frac{k}{50}[/tex]. Now, p(subset of all even numbers) = [tex]\frac{50}{50}\times\frac{1}{2^{100}}=\boxed{\frac{1}{2^{100}}}[/tex]