Question

a pet store has 11 puppies, including 4 poodles, 2 terriers, and 5 retrievers. if rebecca selects one puppy at random, the pet store replaces the puppy with a puppy of the same breed, then aaron chooses a puppy at random. find the probability that they both select a poodle
the probability is
(type an integer or decimal rounded to three decimal places as needed)

Explanation:

Step1: Probability of Rebecca selecting a poodle

The total number of puppies is \(n = 11\), and the number of poodles is \(m=4\). The probability of Rebecca selecting a poodle, \(P(R)\), is given by the formula \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So \(P(R)=\frac{4}{11}\).

Step2: Probability of Aaron selecting a poodle

Since the pet - store replaces the puppy (independent events), the total number of puppies is still \(n = 11\) and the number of poodles is still \(m = 4\). The probability of Aaron selecting a poodle, \(P(A)\), is \(P(A)=\frac{4}{11}\).

Step3: Probability of both events occurring

For independent events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\). Here, let \(A\) be Rebecca's selection and \(B\) be Aaron's selection. So \(P = P(R)\times P(A)=\frac{4}{11}\times\frac{4}{11}=\frac{16}{121}\approx0.132\)

Answer:

\(0.132\)