# 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?

Nov 13, 2015

The probability is $\frac{1}{9}$

#### 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

$P \left({R}_{p}\right) = \frac{4}{9}$

because among 9 dogs there are 4 poodles.

The second event is "Aaron selects a retriever".
This event has a probability of $P \left({A}_{r}\right) = \frac{2}{8}$, 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:

$P = P \left({R}_{p}\right) \cdot P \left({A}_{r}\right) = \frac{4}{9} \cdot \frac{2}{8} = \frac{4}{9} \cdot \frac{1}{4} = \frac{1}{9}$

Sep 5, 2016

$\frac{1}{4} \mathmr{and} \frac{1}{9}$ 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) = $\frac{2}{8} = \frac{1}{4}$

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) = $\frac{4}{9} \times \frac{2}{8}$

=$\frac{1}{9}$