From: Javier, a secondary school student.

Hello. Can you please help me with the following? I don't know how to set it up.

A committee consisting of 6 people is to be selected from 8 parents and 4 teachers. Find the probability of selecting 3 parents and 3 teachers.

Thanks so very much.


Hi Javier.

You can use a simple ratio to solve your problem:

(The number of committees that have exactly three parents and  three teachers)

divided by:
(the number of committees that have six people)

Let’s look at the numerator first:

First, you choose three parents to fill the three parent seats from the group of eight. How many such choices are there? For the first seat, you have eight choices. For the second seat you have just seven (because you already took one parent), for the third seat you have six choices.

This is correct if the order of the three chosen parents is important, but we are just forming a committee, so the order doesn't matter. That means we need to divide by the number of ways you can re-arrange the three parents. How many ways can you re-arrange three things? The answer is 3! (three factorial). That's 3x2x1=6.

We have 8x7x6/6 = 56 different parent sub-committees. Let's compose the teacher sub-committees in a similar way now.

Choose three teachers to fill the teacher seats from the group of four.  That’s four choices for the first seat, three for the second and two for the third. Let's divide again by the number of ways we can re-arrange the three teachers we've chosen (again, this is 3!). That makes the number of different teacher sub-committees 4x3x2/6 = 4.

Now at last we can multiply the two subcommittee counts together to find the total number of 6-person committees that have three teachers and three parents: 4x56 = 224.

The denominator is simpler: you have six seats and a group of twelve to choose from. Use the same principle I described for the numerator. This time, you'll need to account for re-arrangement of the six people (by dividing by 6!).

Hope this helps!