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.
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