paridhi 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
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
