# 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 3 members of the same gender are selected?

30/165=6/33=0.bar(18)~=0.18=18%

#### Explanation:

Let's find the number of ways all boys will go, then find the number of ways all girls will go, then add them together. Then we'll find the probability.

We're working with combinations (we don't care in what order the kids are picked). That general formula is:

C_(n,k)=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

I'll use $\textcolor{b l u e}{\text{blue for boys}}$ and $\textcolor{red}{\text{red for girls}}$

All boys

From the population of 5 boys, 3 are picked. From the population of girls, none are picked:

color(blue)(C_(5,3))xxcolor(red)(C_(6,0))=color(blue)((5!)/((3!)(5-3)!))xxcolor(red)((6!)/((0!)(6-0)!))=color(blue)((5!)/(3!2!))xxcolor(red)((6!)/(0!6!))=color(blue)(120/12)xxcolor(red)(1)=10

All girls

From the population of 6 girls, 3 are picked. From the population of boys, none are picked:

color(blue)(C_(5,0))xxcolor(red)(C_(6,3))=color(blue)((5!)/((0!)(5-0)!))xxcolor(red)((6!)/((3!)(6-3)!))=color(blue)((5!)/(0!5!))xxcolor(red)((6!)/(3!3!))=color(blue)(1)xxcolor(red)(720/36)=20

Total ways

$10 + 20 = 30$

Probability

We now have the total number of ways to meet the condition of having 3 of the same sex go on the field trip. How many different ways can we select 11 children to go? There are 11 children in total and we're picking 3:

C_(11,3)=(11!)/((3!)(11-3)!)=(11!)/(3!8!)=(11xx10xx9xx8!)/(6xx8!)=990/6=165

And so that gives us:

30/165=6/33=0.bar(18)~=0.18=18%