Question

Teams A and B are playing a series of games. The series is over as soon as a team wins a total of three games. Team A is two times more likely to win a game than Team B. What is the probability that Team A wins the series?

Answer 1

The probability that A wins the series is:

`P(x=3) = (n!)/((n-3)!3!)(2p)^(3)(1-2p)^(n-3)`

Explanation:

Since team A and B are playing a series of games, let n be the total number of games played. Since there is a probability that either A or B will win if any of the two teams wins three games, then this experiment follows a binomial distribution. Binomial, because the experiment(i.e games played) is carried out n number of times.

Now, if the probability that team B wins the series is p, then the probability that team A wins the series is 2p(from the question).

Since our aim is to know the probability of A winning the series, then the probability of success is 2p and probability of failure is 1-2p

The probability density function for binomial distribution is given as:

`P(x) = (n!)/((n-x)!x!)p^(x)q^(n-x)`

Where:

n = total number of trials

x = total number of successes

p = probability of success

q = probability of failure

Therefore, we have, after substituting:

`P(x=3) = (n!)/((n-3)!3!)(2p)^(3)(1-2p)^(n-3)`

Which is the required answer.