A pet store has 4 poodles, 3 terriers, and 2 retrievers. If Rebecca and Aaron (in that order) each select a puppy at random, without replacement, what is the probability that Aaron selects a retriever given that Rebecca selects a poodle?

2 Answers
Nov 13, 2015

The probability is 1/919

Explanation:

In this task you have 2 events and you are looking for a conditional probability. The first event is "Rebecca chooses a poodle". The probability of this event is

P(R_p)=4/9P(Rp)=49

because among 9 dogs there are 4 poodles.

The second event is "Aaron selects a retriever".
This event has a probability of P(A_r)=2/8P(Ar)=28, because after Rebecca's choice there are 8 pets in total and among them there are 2 retrievers.

To calculate probability of both events ("Rebeca selects a poodle and Aaron selects a retriever") you have to multiply both calculated probabilities:

P=P(R_p)*P(A_r)=4/9*2/8=4/9*1/4=1/9P=P(Rp)P(Ar)=4928=4914=19

Sep 5, 2016

1/4 or 1/9 14or19 depending on the interpretation of the question.

Explanation:

There are 2 ways of interpreting this question...

I the understand the question as asking just about Aaron's choice. The fact that Rebecca has already chosen a poodle means that there are 8 puppies left.

So, just for Aaron, the probability is:

P(retriever) = 2/8 = 1/428=14

However, if the question is to be read as;

"What is the probability that Rebecca chooses a poodle and Aaron chooses a retriever in this order", then we need to include Rebecca's choice.

P(poodle,retriever) = 4/9 xx2/849×28

=1/919