Question

A class contains 5 boys and 6 girls. The teacher is told that 3 of the students can go on a school trip, and selects them at random. What is the probability that 2 boys and 1 girls are selected?

Answer 1

`36.36%`

Explanation:

so the total way to choose 3 students randomly from 11 results in
`C_k^n = n/(k!(n-k)!) = 11/(3!(11-3)!) =165`

now the ways to get 2 boys out of 5 is

`C_k^n = n/(k!(n-k)!) = 5/(2!(5-2)!) =10`

and 1 girl from 6
`C_k^n = n/(k!(n-k)!) = 6/(1!(6-1)!) = 6`

putting it all together

`(10*6)/165 = 36.36%`

one way to think about these problems is to consider smaller spaces. For instance lets consider just 3 people composed of 1 girl and 2 boys. If we wanted to see about 1 girl and 1 boy we could iterate through and see the following

`b_1, b_2`
`b_1, g`
`b_2, b_1`
`b_2, g`
`g, b_1`
`g, b_2`
we see there is a `4/6 = 2/3` chance. When we break down using the patterns of combinations as above we get
`1` way to choose a girl out of 1 and 2 ways to choose 2 guys from 2 possible for a numerator of 2.

For the denominator we have 3 choose 2 which is 3. Thus we can decompose our combinations and bring them together as needed. If you get stuck sometimes it helps to use a toy example that helps show the correct path.