# 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?

Apr 26, 2017

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 $\frac{4}{6} = \frac{2}{3}$ 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.