Question

The probability that a certain hockey team will win any given game is 0.3773 based on their 13 year win history of 389 wins out of 1031 games played (as of a certain date). Their schedule for November contains 12 games. Let X = number of games won in November.
Find the probability that the hockey team wins at least 3 games in November. (Round your answer to four decimal places.)

Answer 1

Answer:

0.8895 = 88.95% probability that the hockey team wins at least 3 games in November.

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the teams wins, or they do not win. The probability of the team winning a game is independent of any other game, which means that the binomial probability distribution is used 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 0.3773.

This means that $p = 0.3773$

Their schedule for November contains 12 games.

This means that $n = 12$

Find the probability that the hockey team wins at least 3 games in November.

This is:

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

So

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

$P(X = 0) = C_{12,0}.(0.3773)^{0}.(0.6227)^{12} = 0.0034$

$P(X = 1) = C_{12,1}.(0.3773)^{1}.(0.6227)^{11} = 0.0247$

$P(X = 2) = C_{12,2}.(0.3773)^{2}.(0.6227)^{10} = 0.0824$

Then

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0034 + 0.0247 + 0.0824 = 0.1105$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1105 = 0.8895$

0.8895 = 88.95% probability that the hockey team wins at least 3 games in November.