Explanation:
the problem description is seemingly contradicting itself.
"without replacement" means that you don't put the picked item back into the bag or "pool" of items.
and that means that after Rebecka picked a puppy, Aaron CANNOT pick the same one at all.
but then it says that they both may select the same one. which means it is WITH replacement.
so, what is it now ?
I have to give you both answers, and you will have to clarify with your original question or directly with your teacher what is the true condition.
in any case, a probability is always
desired cases / total possible cases.
WITH replacement (they can both pick the same puppy) :
there are 10 puppies, from which 3 are poodles.
so, total possible cases = 10, desired cases = 3
the probability to pick a poodle at the 1st pick = 3/10.
the picked puppy goes back.
the probability to pick a poodle at the 2nd pick = 3/10.
the probability that both picks are poodles is
3/10 × 3/10 = 9/100 = 0.09
WITHOUT replacement :
the first picked puppy is not returned to the group of 10 puppies, so, it can't be picked again, and the number of available puppies gets less as each puppy is removed. this changes the probabilities.
at the beginning we have 10 puppies, from which 3 are poodles.
the probability to pick a poodle at the 1st pick = 3/10.
the picked puppy does NOT go back.
now we have 9 puppies for Aaron to pick from. and for our desired cases for the second pick we have to combine it with the outcome, that the first pick was a poodle.
that means we have only 2 poodles left in the group of 9 puppies.
the probability to pick a poodle at the 2nd pick = 2/9.
the probability that both picks are poodles is therefore
3/10 × 2/9 = 6/90 = 1/15 = 0.066666666...