Probability -- Deck of Cards -- Help Needed !!!!

hemanthkumarmn 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?

A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33

P(at least 1 pair of the SAME value) = 1 - P(selecting 4 DIFFERENT values).

P(selecting 4 different values):
The first card selected can be ANY VALUE.
P(2nd card is a different value) = 10/11. (Of the 11 cards remaining, any but the mate of the 1st card, leaving 11-1 = 10 good options.)
P(3rd card is a different value) = 8/10. (Of 10 cards remaining, any but the mates of the first 2 cards, leaving 10-2 = 8 good options.)
P(4th card is a different value) = 6/9. (Of the 9 cards remaining, any but the mates of the first 3 cards, leaving 9-3 = 6 good options.)
Since we want all of these events to happen, we multiply the probabilities:
10/11 * 8/10 * 6/9 = 16/33.

Thus:
P(at least 1 pair of the same value) = 1 - 16/33 = 17/33.

The correct answer is C.

GMATGuruNY thanks for your help.. Is there a method to solve this problem without using P(at least 1 pair of the SAME value) = 1 - P(selecting 4 DIFFERENT values). I mean directly finding the at least value..


Hemanth

hemanthkumarmn wrote:GMATGuruNY thanks for your help.. Is there a method to solve this problem without using P(at least 1 pair of the SAME value) = 1 - P(selecting 4 DIFFERENT values). I mean directly finding the at least value..

Hemanth

Favorable outcome 1: Selecting exactly ONE matching pair
Case 1: The first two cards are of the same value, the 3rd and 4th cards are of different values
The 1st card selected can be any of the 12 cards.
P(2nd card is of the same value) = 1/11. (Of the 11 remaining cards, 1 is of the same value as the 1st.)
The next card selected can any of the 10 remaining cards.
P(4th card is of a different value from the 3rd) = 8/9. (Of the 9 remaining cards, 8 are of a different value from the 3rd.)
Since we want both events to happen, we multiply the probabilities:
1/11 * 8/9 = 8/99.

Remaining cases:
Of the 4 cards selected, any combination of 2 could be the matching pair.
Number of combinations of 2 that can be formed from 4 options = 4C2 = (4*3)/(2*1) = 6.
To account for the remaining cases, we multiply by 6:
6 * 8/99 = 16/33.

Favorable outcome 2: Selecting TWO matching pairs
Case 1: The 1st and the 2nd cards match AND the 3rd and the 4th cards match
The first card selected can be any of the 12 cards.
P(2nd card is of the same value) = 1/11. (Of the 11 remaining cards, 1 is of the same value as the 1st.)
The next card selected can any of the 10 remaining cards.
P(4th card is of the same value as the 3rd) = 1/9. (Of the 9 remaining cards, 1 is of the same value as the 3rd.)
Since we want both events to happen, we multiply the probabilities:
1/11 * 1/9 = 1/99.

Remaining cases:
The following ways will yield two matching pairs:
XXYY
XYXY
XYYX
Total ways = 3.
To account for the remaining cases, we multiply by 3:
3 * 1/99 = 1/33.

Since either outcome is favorable, we add the probabilities:
16/33 + 1/33 = 17/33.

The approach suggested in my initial post seems much easier.

GMATGuruNY : True that the earlier solution was lot more simpler, yet out of curiosity wanted to know the alternate.. Anyways thanks a ton..

Hemanth