varun289 wrote:Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33
Here's one approach using counting techniques.
First recognize that P(at least one pair) = 1-P(no pairs)
We'll find P(no pairs)
First, the number of possible outcomes.
We have 12 cards and we select 4.
This is accomplished in 12C4 ways =
495
Aside: If anyone is interested, we have a free video on calculating combinations (like 12C4) in your head:
https://www.gmatprepnow.com/module/gmat-counting?id=789
Now we count the number of ways to select 4 different cards
with no pairs. In other words, we want 4 different card values.
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 =
15 ways.
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1-card we have
2 suits from which to choose.
For the 2-card we have
2 suits from which to choose.
For the 4-card we have
2 suits from which to choose.
For the 5-card we have
2 suits from which to choose.
So, for the selection 1,2,4,5 there are
2x
2x
2x
2=
16 possibilities.
So, the number of ways to select 4 cards such that there are no pairs is
15x
16=
240
So, the probability that there are no pairs =
240/
495 = 16/33
So, P(at least one pair) = 1- 16/33 = 17/33 =
C
Cheers,
Brent
