gmat_guy666 wrote:A hat contains 8 pieces of paper and each piece of paper has a different number on it. The numbers written on the paper are: 4,1,9,3,7,8,5,13
A piece of paper is picked at random from the hat and kept aside. After which another piece of paper is removed. What is the probability that the sum of the numbers written on both these pieces is 11 or 12?

(A) 1/7
(B) 3/28
(C) 5/28
(D) 1/14

As with many probability questions, we can also solve this using counting techniques.

IMPORTANT: This question can be treated as a combination question (where order does not matter), or one in which order matters. Many students will feel that the wording implies that order DOES matter (which is fine), so let's examine that solution first.

Total OUTCOMES: In how many ways can we remove 2 pieces of paper?
There are 8 ways to select the first piece.
There are 7 ways to select the second piece.
So, the number of way to select both pieces = (8)(7) = 56

Favorable OUTCOMES: In how many way can we get a sum of 11 or 12?
Select 3 then 8 (sum = 11)
Select 8 then 3 (sum = 11)
Select 4 then 7 (sum = 11)
Select 7 then 4 (sum = 11)
Select 4 then 8 (sum = 12)
Select 8 then 4 (sum = 12)
Select 3 then 9 (sum = 12)
Select 9 then 3 (sum = 12)
Select 7 then 5 (sum = 12)
Select 5 then 7 (sum = 12)
In total, there are 10 favorable outcomes.

So, P(sum is 11 or 12) = 10/56 = [spoiler]5/28 = C[/spoiler]

Cheers,
Brent

gmat_guy666 wrote:A hat contains 8 pieces of paper and each piece of paper has a different number on it. The numbers written on the paper are: 4,1,9,3,7,8,5,13
A piece of paper is picked at random from the hat and kept aside. After which another piece of paper is removed. What is the probability that the sum of the numbers written on both these pieces is 11 or 12?

(A) 1/7
(B) 3/28
(C) 5/28
(D) 1/14

We can also treat the paper selection as "order not mattering."
Notice that the probability question doesn't really change if we say that two pieces of paper are drawn at the same time (so order doesn't matter).

Total OUTCOMES: In how many ways can we remove 2 pieces of paper?
There are 8 pieces of paper. We want to select 2.
Since order doesn't matter, we can accomplish this task in 8C2 ways (= 28 ways)

If anyone is interested, we have a free video on calculating combinations (like 8C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Favorable OUTCOMES: In how many way can we get a sum of 11 or 12?
Select 3 and 8 (sum = 11)
Select 4 and 7 (sum = 11)
Select 4 and 8 (sum = 12)
Select 3 and 9 (sum = 12)
Select 7 and 5 (sum = 12)
In total, there are 5 favorable outcomes.

So, P(sum is 11 or 12) = 5/28 = C

Cheers,
Brent