# probability of winning $k$ out of $n$ games?

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.

1. What is the probability Team $A$ wins the series in $4$ games?

2. 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?

• Well, its easiest to imagine that all $7$ games are played and compute the probability that $A$ wins at least $4$. This differs from what you said in that we also consider scenarios in which $A$ wins more than $4$ games.
– lulu
Mar 2, 2018 at 17:55

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 $k=4$

• Why are subtracting 1 from both n and k. I'm kinda confused on that Mar 2, 2018 at 18:19
• For a team to win the series in $n=5$ games, they must win $3$ of the first $4$
– Remy
Mar 2, 2018 at 18:22