Respuesta :
Answer:
[tex]P=0.4737[/tex]
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:
[tex]nCx=\frac{n!}{x!(n-x)!}[/tex]
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:
[tex]3C1*3C2 =\frac{3!}{1!(3-1)!}* \frac{3!}{2!(3-2)!}=9[/tex]
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:
[tex]3C2*3C1 =\frac{3!}{2!(3-2)!}* \frac{3!}{1!(3-1)!}=9[/tex]
3. Select 3 black goldfish and 0 yellow goldfish: There is 1 way. it is calculated as:
[tex]3C3*3C0 =\frac{3!}{3!(3-3)!}* \frac{3!}{0!(3-0)!}=1[/tex]
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:
[tex]P=\frac{9}{9+9+1} =0.4737[/tex]
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.
