Mutual information - Joint distribution of unfair coin flips I was solving a question on mutual information but got a very different answer compared to the solutions. Im finding the probability distribution of a biased coin flip and was wondering why the joint distribution is as stated below in the picture. Shouldn't it be 
$P(Y_1 = H, Y_2 = H) = p^2 + (1-p)^2 \\ 
P(Y_1 = H, Y_2 = T) = 2(p(1-p)) \\ 
P(Y_1 = T, Y_2 = H) = 2(p(1-p) \\
P(Y_1 = T, Y_2 = T) = p^2 + (1-p)^2$
Why is there a half in the solutions ?

EDIT: So I totally forgot that I was supposed to marginalize across X which has probability $p(X=1) = p(x=2) = \frac{1}{2}$ Thus 
$p(Y_1=H, Y_2=H) \\ 
= \sum_X p(Y_1=H, Y_2=H, X) \\
= p(X=1)p(Y_1=H, Y_2=H) + p(X=2)p(Y_1=H, Y_2=H) \\
= \frac{1}{2}(p^2 + (1-p)^2) $

since the probability given in the table are conditional probabilities. for example, the first one is prob of two heads conditioning x is 1
