Question

The probability that a certain hockey team will win any given game is 0.3723 based on their 13 year win history of 385 wins out of 1034 games played (as of a certain date). Their schedule for November contains 12 games. Let X = number of games won in November. What is the probability that the hockey team wins 6 games in November? (Round your answer to four decimal places.)

Answer 1

Answer:

0.1505 = 15.05% probability that the hockey team wins 6 games in November

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the team wins, or it does not. The probability of winning a game is independent of winning other games. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that a certain hockey team will win any given game is 0.3723

So $p = 0.3723$

12 games in November

So $n = 12$

What is the probability that the hockey team wins 6 games in November?

This is $P(X = 6)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{12,6}.(0.3723)^{6}.(1-0.3723)^{6} = 0.1505$

0.1505 = 15.05% probability that the hockey team wins 6 games in November