Ashishkapoor7 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, for each value from 1 to 6, there are two cards in the deck with that 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?
8/33
62/165
17/33
103/165
25/33
I tackled this question by recognizing that P(at least one pair) = 1-P(no pairs)
First, the number of possible outcomes:
We have 12 cards and we select 4.
This is accomplished in 12C4 ways = 495
Now we count the number of ways to select 4 different values with no pairs. In other words, we wand 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 2x2x2x2-16 possibilities.
So, the number of ways to select 4 cards such that there are no pairs is 15x16=240
So, the probability that there are no pairs = 240/(11x5x9) = 48/99 = 16/33
So, P(at least one pair) = 1- 16/33 =17/33
Cheers,
Brent
