Question

four girls and six boys are in a spanish club. three of the people will be chosen at random to represent the group in a photograph.
what is the probability that one girl and two boys will be chosen?

Explanation:

Step1: Calculate total number of people

Total number of people \(n = 4 + 6=10\)
Number of ways to choose 3 people out of 10 is \(C(n,k)=\frac{n!}{k!(n - k)!}\), so \(C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120\)

Step2: Calculate number of ways to choose 1 girl and 2 boys

Number of ways to choose 1 girl out of 4: \(C(4,1)=\frac{4!}{1!(4 - 1)!}=4\)
Number of ways to choose 2 boys out of 6: \(C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15\)
By the multiplication principle, number of ways to choose 1 girl and 2 boys is \(C(4,1)\times C(6,2)=4\times15 = 60\)

Step3: Calculate the probability

Probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{60}{120}=0.5 = 50\%\)

Answer:

50%