What is the probability of one person each from six groups being on the same committee?

"A committee of six people is formed to investigate connections between business and the education system. It consists of people chosen from 5 students, 8 teachers, 5 principals, 10 parents, 4 professors, and 6 businessmen. If the members are chosen at random, what is the probability that there is one person from each group on the committee?"

So far I understand the committee has to be 6 people out of 34 people total, so I assume the denominator has to be something like 34C6. I'm also assuming the numerators go 5C1 (for students), 8C1 (for teachers), 5C1 (for principals), 10C1 (for parents), 4C1 (for professors), and 6C1 (for businessmen). After that I get stuck on how to calculate them all together.

Solution 1:

As N.F.Taussig's comment indicated, the response below does not pertain to the (edited) problem as it currently exists. Initially, there was no mention of a group of $(4)$ professors, so $6$ people had to be drawn from $5$ groups. Analysis left in, re original problem. Certainly the comment of J Moravitz pertains to the current problem.

Shooting gallery blues: tough to hit a moving target.

Your analysis is good, and your denominator is spot on. Simply multiply the (partial) factors that you have identified together, to form the numerator. So, $\binom{5}{1} \times \binom{8}{1} \times \cdots $.

The one complication is that you have $5$ separate cases to add together, when forming the numerator, since the $6$-th person might come from any of the groups. So, Case 1 (for example) would have $\binom{5}{2}$ as the 1st factor, rather than $\binom{5}{1}.$

In general, for problems like this

• The numerator and denominator must be computed in a consistent manner

• You must avoid over-counting (the same situation more than once)

• You must strive to identify mutually exclusive cases. For example, the case of $2$ people from the students is mutually exclusive with the case of $2$ people from the teachers.

Final answer:

$$\frac{S_1 + S_2 + S_3 + S_4 + S_5}{\binom{34}{6}}.$$

$\displaystyle S_1 = \binom{5}{2}\times \binom{8}{1} \times \binom{5}{1} \times \binom{10}{1} \times \binom{6}{1}.$

$\displaystyle S_2 = \binom{5}{1}\times \binom{8}{2} \times \binom{5}{1} \times \binom{10}{1} \times \binom{6}{1}.$

$\displaystyle S_3 = \binom{5}{1}\times \binom{8}{1} \times \binom{5}{2} \times \binom{10}{1} \times \binom{6}{1}.$

$\displaystyle S_4 = \binom{5}{1}\times \binom{8}{1} \times \binom{5}{1} \times \binom{10}{2} \times \binom{6}{1}.$

$\displaystyle S_5 = \binom{5}{1}\times \binom{8}{1} \times \binom{5}{1} \times \binom{10}{1} \times \binom{6}{2}.$

Actually, the above approach represents the direct approach. Problems like this will also succumb to Inclusion-Exclusion. In fact, there is a moderate conflict as to which approach should be used. Inclusion-Exclusion generalizes better, but the direct approach is often easier, if the problem is simple enough, as in this case.

For what it's worth, the Inclusion-Exclusion approach would have $A_k$ represent the subset of distributions where group $k$ was omitted $~: ~k \in \{1,2,3,4,5\}.$

Then, the numerator would become

$$\binom{34}{6} - |A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5|.$$