Probability Q

Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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

We can solve this question using probability rules.

First, recognize that P(at least one pair) = 1 - P(no pairs)

P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33
= C

Cheers,
Brent

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

We can also solve the question 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 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 = 240/495 = 16/33

So, P(at least one pair) = 1- 16/33 = 17/33 = C

Cheers,
Brent

prachi18oct wrote:




Hi,

Can experts pls tell me where did I go wrong.

Probability of no pairs = fav outcomes/total outcomes

total outcomes = 12C4

fav outcome = 12 * 10 * 8 * 6

Prob of no pair = 12 * 10 * 8 * 6 * 4!/ 12 * 11 * 10 * 9

This is not giving me the answer. pls suggest

You almost have it.
However, your first calculation (total outcomes = 12C4 ) treats the cards as though order does NOT matter, but your second calculation (fav outcome = 12 * 10 * 8 * 6) treats the cards as though order DOES matter

Let's make order matter throughout and change total outcomes = 12C4 to total outcomes = 12 * 11 * 10 * 9

So, we get: P(no pairs) = (12 * 10 * 8 * 6)/(12 * 11 * 10 * 9 ) = 16/33

So, P(at least one pair) = 1 - 16/33 = [spoiler]17/33 = C[/spoiler]

Cheers,
Brent