Bunuel wrote:
Official Solution:
If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?
A. 36
B. 60
C. 72
D. 80
E. 100
Although
eliminating one extreme is the fastest way to solve the problem, in this case by subtracting the number of ways in which only men could form the committee,
per the official solution, it may be useful to take the time to understand in practice how you
could solve this or a similar probability question by using another application of combinations. Namely, consider that
a probability is nothing more than a desired subset within a total, or, in mathematical language,
\(P = \frac{desired}{total}\)
In this case, we can work up to the answer by looking at the committees that could contain 1, 2, or 3 women, respectively. In the first case, 4C1 works nicely--from four total women, we could choose one. But that would mean the other two slots would have to be occupied by men, so out of six men, you would need to choose two: 6C2. Since these are
dependent actions--the three slots would have to be filled by 1 woman
and 2 men in such a scenario--we would need to multiply them: 4C1 * 6C2 (or 4 * 15, which is 60).
Next, we could repeat the process with 2 women instead: 4C2. We need one man to fill the third slot: 6C1. Together, 4C2 * 6C1 = 6 * 6 = 36.
Finally, we could repeat the process with 3 women filling the three slots, bypassing the men altogether: 4C3 = 4.
Altogether, then, we would have 60 or 36 or 4, or 60 + 36 + 4 different ways of choosing the women to fill the slots of the committee under the given conditions. 60 + 36 + 4 = 100.
As also noted in the official solution, there are 10C3 ways in which to choose 3 people from a pool of 10. This becomes our total from the earlier fraction: 10C3 = 120. Written out in full, then, we get
\(P = \frac{(4C1 * 6C2 + 4C2 * 6C1 + 4C3)}{10C3}\)
The probability boils down to
\(\frac{100}{120}\)
I will reiterate:
this would not be the most efficient way to solve the question at hand. However, if you understand how such a solution operates, then you might be able to approach another type of question that is also based on combinations or probability with more confidence, and that is the goal.
Good luck with your studies.
- Andrew