Question

A chef is choosing from 13 different entrees for an important banquet, 3 of which contain spinach. The guests will have a choice of 4 entrees. The chef has decided to choose the 4 at random. Write your answers as a decimals rounded to the nearest thousandth. What is the probability that none of the chosen entrees contain spinach

Answer 1

Answer:

0.294 = 29.4% probability that none of the chosen entrees contain spinach.

Step-by-step explanation:

The choices of entrees are chosen without replacement, which means that the hypergeomtric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

13 different entrees for an important banquet

This means that $N = 13$

3 of which contain spinach.

This means that $k = 3$

The guests will have a choice of 4 entrees.

This means that $n = 4$

What is the probability that none of the chosen entrees contain spinach?

This is P(X = 0). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,13,4,3) = \frac{C_{3,0}*C_{10,4}}{C_{13,4}} = 0.294$

0.294 = 29.4% probability that none of the chosen entrees contain spinach.