Bunuel
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4
Kudos for a correct solution.
MANHATTAN GMAT OFFICIAL SOLUTION:Consider one specific case: blue first, then green, then red. By the domino-effect rule, the probability of this case is:
\(\frac{7 \ blue}{16 \ total}*\frac{5 \ green}{15 \ total}*\frac{4 red}{14 \ total}=\frac{1}{24}\).
Now consider another case: green first, then red, then blue. The probability of this case is \(\frac{5 \ green}{16 \ total}*\frac{4 red}{15 \ total}*\frac{7 \ blue}{14 \ total}=\frac{1}{24}\). Notice that all we have done is swap around the numerators. We get the same final probability! This is no accident; the order in which the balls come out does not matter.
Because the three desired gum balls can come out in any order, there are 3! = 6 different cases. All of these cases must have the same probability. Therefore, the overall probability is 6*1/24 = 1/4.
In general, when you have a symmetrical problem with multiple equivalent cases, calculate the probability of one case (often by using the domino-effect rule). Then multiply by the number of cases. Use combinatorics to calculate the number of cases, if necessary.
Remember that when you apply a symmetry argument, the situation must truly be symmetrical. In the case above, if you swapped the order of "red" and "green" emerging from the gumball machine, nothing would change about the problem. As a result, we can use symmetry to simplify the computation.