Find the probability that the second dog removed is a terrier.?

Question

There are 7 puppies in a pet store window: 5 poodles and 2 terriers. Two dogs are removed randomly, one after the other without replacement.

a. 11/42
b. 2/7
c. 12/49
d. 5/12
e. none of the others

Answer 1

The answer to this question is (b) 2/7.

When removing two dogs there are four possible scenarios:

(1) Remove poodle, then remove terrier.
(2) Remove poodle, then remove another poodle.
(3) Remove terrier, then remove another terrier.
(4) Remove terrier, then remove poodle.

In each case, take the probability of the first even happening and multiply it by the probability of the second even happening. For instance:

Probability that a poodle is first selected = 5(# of poodles originally) / 7(total # of dogs).
Probability that a terrier is then selected = 2(# of terriers left) / 6(total # of dogs after a poodle is removed).
Multiply 5/7 and 2/6 to get 10/42.

Repeat this with each of the remaining three cases to get
(1) 10/42
(2) 20/42
(3) 2/42
(4) 10/42

(Note that these add up to 1, which means that we have all of the possible cases)

Now, just add the probabilities of the two cases with a terrier being selected second to get:

10/42(first case) + 2/42(third case) = 12/42

Reduce 12/42 down to 2/7 to get your answer.

I hope this helps, good luck with future probability problems!

Answer 2

The first dog to be picked can be anything, so the probability of it being a poodle is 5/7, and the probabiity of it being a terrier is 2/7.

2 Cases is possible:
Poodle Terrier: 5/7 x 2/6
Terrier Terrier: 2/7 x 1/6

Since they are BOTH liable individual cases, you add them up, so 5/21 + 1/21 = 2/7.

Hope this helps=)

Answer 3

In the first round of picking, the probabilities are:

probability of picking a poodle: 5/7
probability of picking a terrier: 2/7

since there is no replacement after the first dog is picked, there leaves 6 dogs for the second round of picking. Also, there is now a 2 different cases, depending on the type of dog that was picked in the first round.

If a poodle was picked in round 1:
the probability of picking a poodle in this round is 4/6 = 2/3.
the probability of picking a terrier in this round is 2/6 = 1/3.

If a terrier was picked in round 1:
the probability of picking a poodle in this round is 5/6
the probability of picking a terrier in this round is 1/6

So, there is 4 cases in total:
PT, PP, TP, TT

The question only asks for 'second dog removed is terrier' so only consider PT and TT.

now, PT = prob. of poodle (1st round) x prob of terrier (2nd round)
PT = (5/7) x (1/3) = 5/21

also, TT = prob of terrier (1st round) x prob of terrier (2nd round)
TT = (2/7) x (1/6) = 2/42

Adding these two probabilities:
PT + TT = 5/21 + 2/42=> 12/42 => 6/21 => 2/7

so your answer is b.

Answer 4

5/7x2/6+2/7x1/6=2/7

b. 2/7

Answer 5

Its sounds tricky....