Question

A fish bowl contains 12 gray and 3 red fish, all the same size. If the fish are selected
at random from the bowl, one at a time without replacement, what is the probability
that the third red fish is the seventh fish selected?

Answer 1

Using the hypergeometric distribution, it is found that there is a 0.033 = 3.3% probability  that the third red fish is the seventh fish selected.

The fishes are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:  

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

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

The parameters are:

• x is the number of successes.
• N is the size of the population.
• n is the size of the sample.
• k is the total number of desired outcomes.

In this problem:

• During the first 6 fish, we want two red.
• The seventh being red, with a 1/9 probability, as considering 2 of the first 6 are red, one of the 9 remaining will be red.
• 15 fish, thus $N = 15$.
• 3 are red, thus $k = 3$.
• 6 fish, thus $n = 6$.

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

$P(X = 2) = h(2,15,6,3) = \frac{C_{3,2}*C_{12,4}}{C_{15,6}} = 0.2967$

$p = 0.2967 \times \frac{1}{9} = 0.033$

0.033 = 3.3% probability  that the third red fish is the seventh fish selected.

A similar problem is given at https://brainly.com/question/24008577