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

Answer:

The probability that A wins the series is:

$P(x=3) = \frac{n!}{(n-3)!3!}(2p)^{3}(1-2p)^{n-3}$

Step-by-step 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:

$P(x) = \frac{n!}{(n-x)!x!}p^{x}q^{n-x}$

Where:

n = total number of trials

x = total number of successes

p = probability of success

q = probability of failure

Therefore, we have, after substituting:

$P(x=3) = \frac{n!}{(n-3)!3!}(2p)^{3}(1-2p)^{n-3}$

Which is the required answer.