gmatter0913 wrote:
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) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)
This question can be solved using complementary probability or in other words 1-p. When translating the question into an equation, and translating the various probabilities for each card draw, we should use the 1-x technique which is a concept relating to pairs in probability.
So
We need to find the complementary probability of
not drawing a pair on each draw. There are four draws, so when we find the probability of not drawing a pair for each draw we will multiply those four probabilities and subtract from one.
First DrawThe probability of
not drawing a pair on the first draw is 12/12, or 1, because you cannot have a pair with two items of course
Second DrawHere is where we start to apply the 1-x technique
10 ( we subtract 2 from the numerator ) / 11 (we subtract 1 from the numerator
Third Draw 8 (again subtract 2 from the numerator)/ 10 (again subtract 1 from the numerator)
Fourth Draw 6 (again subtract 2 from the numerator)/9 (again subtract 1 from the numerator)
1- ( [1] X [10/11] x [8/10] x [6/9] )
1-(16/33)= 17/33
Possibly, the most fundamental example of using the complementary probability is a dice roll problem- on a dice numbered 1-6 what is the chance of rolling five
at least once. *note the words at least and exact are important in probability problems
1- (5^3/6^3)
Though a probability question that is more on par with this question is one regarding a pair of socks
https://gmatclub.com/forum/tony-owns-si ... 34352.html