Probability - playing cards

[paridhi]

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?

(A) 8/33

(B) 62/165

(C) 17/33

(D) 103/165

(E) 25/33

Can someone please tell me the approach to do this question?

[4GMAT_Mumbai]

Hi,

Interesting problem ... Perhaps amongst the tough ones in GMAT.

Atleast 1 card repeating = total - No cards repeating.

Total card possibilities = 12 C 4 = 495

No card repeatings = 5 ways of achieving this

4 cards from deck1 (or)
3 cards from d1 and 1 of the other cards from d2 (or)
2 cards from d1 and 2 of the other cards from d2 (or)
1 card from d1 and 3 of the other cards from d2 (or)
4 cards from d2

=
6C4 +

6C3 times 3C1 +

6C2 times 4C2 +

6C1 times 5C3 +

6C4

= 240 ways

Atleast 1 card repeating = total - No cards repeating.

Atleast 1 card repeating = 495 - 240

P(the above event) = (495 - 240) / 495. Choice C

Look forward to a better method or pls highlight if there are some issues here.

Thanks.

[Rahul@gurome]

Let p be the probability that Bill finds no pair of cards that have the same value. Then probability that Bill finds at least one pair of cards that have the same value = 1- p
Total no. of ways in which 4 cards can be drawn from 12 cards = 12C4 = 495

Number of ways in which Bill finds no pair of cards with the same value:
There are 6 cards within a suit and we want to select 4 cards. So, this can be done in 6C4 = 15 ways.
For 15 values each card can be of either suit.
1st card can be chosen from 2 suits, similarly 2nd, 3rd & 4th card can be chosen from 2 suits. We can do this 2^4 = 16 ways
So, number of ways to choose 4 cards so that no pairs exist = 15×16 = 240

So, p = 240/495 = 16/ 33
Hence 1- p = 1-16/33 =17/33

[spoiler]The correct answer is (C).[/spoiler]

[meshtrap]

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?

(A) 8/33

(B) 62/165

(C) 17/33

(D) 103/165

(E) 25/33

Can someone please tell me the approach to do this question?

Since the problem asks us to find atleast one pair of cards that have the same value, a simpler approach would be to find the probability if none of the cards have the same value and subtract that number from 1.
P(A) = 1 - P(~A)

Thus, P(finding one pair) = 1 - P(not finding a single pair)

P(~Single Pair):

Calculate the denominator as follows:

4 cards can be chosen out of 12 cards in 12C4 = (12*11*10*9)/4! ways

Calculate the numerator as follows:

Since nCr = (nPr)/r!, calculate an ordered pair(permutation) and divide it by r!

1st card can be chosen in 12 ways
2nd------------------------ in 10 ways(if 2 is chosen, the other 2 in the deck cannot be chosen)
3rd------------------------- in 8 ways
4th--------------------------in 6 ways

Thus, numerator = (12*10*8*6)/4!

Thus, P(~Single pair) = num/den = (12*10*8*6)/(12*11*10*9) = 16/33

Thus P(Atleast one pair) = 1-16/33 = 17/33

Hope it helps,
meshtrap

[paridhi]

Thanks everyone for such good explanations.

OA Is C