Question

An aquarium at a pet store contains six fish; three yellow goldfish and three black goldfish. On Sunday, a customer came to the store and randomly selected three fish to purchase. Suppose you know the customer purchased a black goldfish. What is the probability that two yellow goldfish and a black goldfish remain in the tank after the customer has left? Please simplify your answer to a decimal value and circle your answer. Ensure you show your work.

Answer 1

Answer:

$P=0.4737$

Step-by-step explanation:

First, we need to know that nCx give as the number of ways in which we can select x elements from a group of n. It is calculated as:

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

Then, to select 3 fish in which at least one a them is a black goldfish we can:

1. Select one black goldfish and 2 yellow goldfish: There are 9 different ways to do this. it is calculated as:

$3C1*3C2 =\frac{3!}{1!(3-1)!}* \frac{3!}{2!(3-2)!}=9$

Because we select 1 black goldfish from the 3 in aquarium and select 2 yellow goldfish from the 3 in the aquarium.

2. Select 2 black goldfish and 1 yellow goldfish: There are 9 different ways. it is calculated as:

$3C2*3C1 =\frac{3!}{2!(3-2)!}* \frac{3!}{1!(3-1)!}=9$

3.  Select 3 black goldfish and 0 yellow goldfish: There is 1 way. it is calculated as:

$3C3*3C0 =\frac{3!}{3!(3-3)!}* \frac{3!}{0!(3-0)!}=1$

Now, we identify that just in part 2 (Select 2 black goldfish and 1 yellow goldfish), two yellow goldfish and a black goldfish remain in the tank after the customer has left.

So, the probability that two yellow goldfish and a black goldfish remain in the tank after the customer has left given that the customer purchased a black goldfish is equal to:

$P=\frac{9}{9+9+1} =0.4737$

Because there are 19 ways in which the customer can select a black fish and from that 19 ways, there are 9 ways in which two yellow goldfish and a black goldfish remain in the tank.