Question

In an animal rescue facility are 21 animals.14 cats and 7 dogs. During the day 5 animals were adopted.what is the probabity that 3 cats and 2 dogs were adopted? Review ch 7 on probability

Answer 1

0.3756

Explanation:

I believe this is a combinatorics probability problem.

We first figure out in how many ways we can choose 3 cats out of 14 cats

Then we figure out how many ways we can choose 2 dogs out of 7 dogs

The total number of ways to select 3 cats and 2 dogs will be the product of these two

We then figure out in how many ways we can select 5 animals out of a total of 21 animals

The product we computed divided by the above number will give the probability

The number of ways we can choose k items out of n items is given by nCk i.e n Choose k

The formula is rather complicated but almost all calculators should be able to compute this given n and k

Let's deal with the selection of 3 cats from the population of 14 cats

This is given by 14C₃ which works out to 364

Number of ways for selecting 2 dogs from 7 dogs = 7C₂ = 21

Number of ways of selecting 5 animals from 21 = 21C₅ = 20349

The probability of selecting 3 cats and 2 dogs is therefore:

`(364 * 21)/(20349) = 0.3756`

That, I believe is the answer. Please let me know if it is correct and if you need more info.