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?

1 Answer
Will
Apr 26, 2017

36.36%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)!) =165Cnk=nk!(n−k)!=113!(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)!) =10Cnk=nk!(n−k)!=52!(5−2)!=10

and 1 girl from 6
C_k^n = n/(k!(n-k)!) = 6/(1!(6-1)!) = 6Cnk=nk!(n−k)!=61!(6−1)!=6

putting it all together

(10*6)/165 = 36.36%10⋅6165=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_2b1,b2
b_1, gb1,g
b_2, b_1b2,b1
b_2, gb2,g
g, b_1g,b1
g, b_2g,b2
we see there is a 4/6 = 2/346=23 chance. When we break down using the patterns of combinations as above we get
11 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.

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