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

`P=0.4737`

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=(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 =(3!)/(1!(3-1)!)* (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 =(3!)/(2!(3-2)!)* (3!)/(1!(3-1)!)=9`

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

`3C3*3C0 =(3!)/(3!(3-3)!)* (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=(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.