Question

A school debate team has 4 girls and 6 boys. A total of 3 of the team members will be chosen to participate in the district debate. What is the probability that 1 girl and 2 boys will be selected?

Answer 1

There are 4+6=10 people in the team.
First calculate in how many ways you can choose 3 people from a group of 10 people, using combination:
$(^{10} _3)=\frac{10!}{3!(10-3)!}=\frac{10!}{3! \times 7!}=\frac{7! \times 8 \times 9 \times 10}{6 \times 7!}=\frac{8 \times 9 \times 10}{6}=\frac{720}{6}=120$
You can choose 3 people in 120 ways.

You can select 1 girl from a group of 4 girls in 4 ways.
Calculate in how many ways you can choose 2 boys from a group of 6 boys:
$(^6 _2)=\frac{6!}{2! (6-2)!}=\frac{6!}{2! \times 4!}=\frac{4! \times 5 \times 6}{2 \times 4!}=\frac{5 \times 6}{2}=\frac{30}{2}=15$
You can choose 2 boys in 15 ways.
Using the rule of product, you can calculate that you can choose 1 girl and 2 boys in 4×15=60 ways.

The probability of choosing 1 girl and 2 boys is the number of ways you can choose 1 girl and 2 boys divided by the number of ways you can choose 3 people from the group.
 $P=\frac{60}{120}=\frac{1}{2}=50\%$

The probability is 1/2, or 50%.