Probability to win a series of three games, with a given advantage for the home team 
In a game with two equal teams, the home team wins any game with probability $p>\frac12$. In a best of three playoff series, a team with a home advantage has play at home, followed by a game away, followed by a home play if necessary. The series is over as soon as one team wins two games.
What is $P[T]$, the probability that the team with home advantage wins the series? Prove that $P[T]>p$ for all $p>\frac12$.

In the first sentence, I think there's a typo. I think it meant to say $p=\frac12$ (I don't know how you can get a number as your answer if its not equal to $\frac12$)
I drew a tree diagram as shown below, and calculated the probability that Team A wins.

$.5^2 + .5^3 + .5^3 = .5$
 A: This is not a typo. $p$ is a variable and you are supposed to include it in your answer.
There are three sequences of results leading to the home advantage team winning, as you found out: WW ($p(1-p)$), WLW ($p^3$) and LWW ($(1-p)^2p$). Thus
$$P[T]=p(1-p)+p(1-p)^2+p^3=(p-p^2)(2-p)+p^3=2p-3p^2+2p^3=p(2-3p+2p^2)$$
and if $p=\frac12+d$ with $d>0$
$$P[T]=p\left(2-3\left(\frac12+d\right)+2\left(\frac12+d\right)^2\right)$$
$$=p\left(2d^2-d+1\right)$$
$$=p\left(1+d(2d-1)\right)<p$$
where the last inequality is because $2d-1<0$ for $0<d<\frac12$. Indeed, the inequality is reversed compared to what is printed: conversely, $P[T]>p$ when $p<\frac12$.
A: I'll add an answer to point out that you can approach the analysis somewhat differently by extending the series to three games even when the same team wins the first two.  This is permissible because the third game doesn't change the outcome in those cases; that is, after all, why they then don't usually play the third game.  We'll have them play it just to simplify things a bit.
Then the team with home advantage wins its two home games with probability $p^3$ (the home team wins all three games) plus $2p(1-p)^2$ (it splits its home games and wins away).  It wins all three games with probability $p^2(1-p)$, for a total of
$$
P(\text{home team wins series}) = 2p-3p^2+2p^3
$$
To see that this is actually less than $p$, we write
\begin{align}
2p-3p^2+2p^3-p & = p(2p^3-3p+1) \\
               & = p(2p-1)(p-1)
\end{align}
This shows that for the interval $0 \leq p \leq 1$, the left-hand side is positive when $0 < p < 1/2$, and negative when $1/2 < p < 1$.  The latter is our situation, and since $(2p-3p^2+2p^3)-p > 0$, it must be the case that $P(\text{home team wins the series}) = 2p-3p^2+2p^3 > p$.
