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Two teams play $7$ games and a team wins a series if they win $4$ out of $7$ games. Team $A$ has a probability of $.75$ winning a single game.
What is the probability Team $A$ wins the series in $4$ games?
What is the probability Team $A$ wins the series in any $n$ valid games?
What I tried:
$P(X = k) =\binom{n}{k}\ p^k (1-p)^{n-k}$
$=\binom{7}{4}\ 0.75^4 (1-.75)^{7-4}$
$= .01730$
B) Add up probability of Team $A$ winning the series in $4$, $5$, $6$, $7$ games.
Am I right?
Not quite. In the formula, $n$ is the number of games played in the series, which is $4$ in this case. That's a given. We want the probability that they win all $4$ of those games. This comes out to be $${4 \choose 4}\left(0.75\right)^4\cdot\left(0.25\right)^0=0.75^4$$
This makes sense because for them to win the series in $4$ games, they have to win each of the first $4$ games, each with probability $0.75$.
For (b), note that this is a negative binomial since the winning team need not just win $4$ games, but they also must win the final game of the series.
The formula for $n$ trials given $k$ successes is
$$P(X=n)={n-1 \choose k-1}p^k(1-p)^{n-k}$$
which can be interpreted as team A winning $3$ of the first $n-1$ games and then winning the next.
You'll need to sum for all valid values of $n$ which are $n\in\{4,5,6,7\}$ and $k$ stays fixed at $k=4$
