aadikamagic wrote:
As a treat for her two crying children, a mother runs to the freezer in which she has two cherry ice pops, three orange ice pops, and four lemon-lime ice pops. If she chooses two at random to bring outside to the children, but realizes as she runs out the door that she cannot bring them different flavors without one invariably being jealous of the other and getting even more upset, what is the probability that she has to return to the freezer to make sure that they each receive the same flavor?
A. 1/9
B. 1/6
C. 5/18
D. 13/18
E. 5/6
Official solution from Veritas Prep.
The mother will have to return to the freezer if each child receives a different flavor, meaning that she does not have to return to the freezer if they each receive the same flavor. And in probability, it's often more convenient to mathematically account for "matches" (or pairs) than it is for non-matches, so in this case it may be helpful to use the "outcomes when the ice pops match" and subtract from one than it is to go the other route.
If the goal, then, is to match, then there are three ways to do it:
Both cherry: 2/9 * 1/8
Both orange: 3/9 * 2/8
Both lemon-lime: 4/9 * 3/8
And since the denominator is the same in each case (you start with 9, then for the second draw there are 8 left, so the denominator is always 9 * 8), it's easier if you don't reduce the fractions before adding:
(2 * 1) + (3 * 2) + (4 * 3) all over the common denominator leads to:
(2 + 6 + 12)/72
20/72 = 10/36 = 5/18
But remember - 5/18 is the probability that she got what she wanted on her random draw, and we want the probability that she did not. That's the other 13/18 of outcomes, so the answer is D.