Question

A TV studio has brought in 8 boy kittens and 10 girl kittens for a cat food commercial.

Answer 1

Answer:

0.323 = 32.3% probability that the director chooses 3 boy kittens and 5 girl kittens.

Step-by-step explanation:

The kittens are chosen without replacement, which means that the hypergeometric 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.

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

A TV studio has brought in 8 boy kittens and 10 girl kittens for a cat food commercial.

This means that $N = 8 + 10 = 18$

We want 3 boys, so $k = 8$

The director is going to choose 8 of these kittens at random to be in the commercial.

This means that $n = 8$

What is the probability that the director chooses 3 boy kittens and 5 girl kittens?

This is P(X = 3).

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

$P(X = 3) = h(3,18,8,8) = \frac{C_{8,3}*C_{10,5}}{C_{18,8}} = 0.323$

0.323 = 32.3% probability that the director chooses 3 boy kittens and 5 girl kittens.