Question

The probability that the san jose sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). an upcoming monthly schedule contains 12 games. what is the probability that the san jose sharks win at least 6 games in that upcoming month? let x = number of games won in that upcoming month. (round your answer to four decimal places.)

Answer 1

Answer:

0.2572 = 25.72% probability that the san jose sharks win at least 6 games in that upcoming month

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the Sharks win, or they do not. The probability of the Sharks winning each game is independent of 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 the san jose sharks will win any given game is 0.3694

This means that $p = 0.3694$

An upcoming monthly schedule contains 12 games

This means that $n = 12$

What is the probability that the san jose sharks win at least 6 games in that upcoming month?

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 6) = C_{12,6}.(0.3694)^{6}.(0.6306)^{6} = 0.1476$

$P(X = 7) = C_{12,7}.(0.3694)^{7}.(0.6306)^{5} = 0.0741$

$P(X = 8) = C_{12,8}.(0.3694)^{8}.(0.6306)^{4} = 0.0271$

$P(X = 9) = C_{12,9}.(0.3694)^{9}.(0.6306)^{3} =0.0071$

$P(X = 10) = C_{12,10}.(0.3694)^{10}.(0.6306)^{2} = 0.0012$

$P(X = 11) = C_{12,11}.(0.3694)^{11}.(0.6306)^{1} = 0.0001$

$P(X = 12) = C_{12,12}.(0.3694)^{12}.(0.6306)^{0} \cong 0$

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1476 + 0.0741 + 0.0271 + 0.0071 + 0.0012 + 0.0001 + 0.0000 = 0.2572$

0.2572 = 25.72% probability that the san jose sharks win at least 6 games in that upcoming month

Answer 2

$X$ follows a binomial distribution $\mathcal B(12,0.3694)$. We get

$\mathbb P(X\ge6)=\displaystyle\sum_{x=6}^{12}p_X(x)$

where $p_X(x)$ is the PMF of the distribution given by

$p_X(x)=\begin{cases}\dbinom{12}x0.3694^x(1-0.3694)^{12-x}&\text{for }0\le x\le12\\\\0&\text{otherwise}\end{cases}$

Using a calculator, you'd find

$\mathbb P(X\ge6)\approx0.2573$