Question

A group of 30 people, 12 females and 18 males, are electing 5 representatives.

a. How many ways can the 5 representatives be chosen from the group?


b. How many ways can the 5 representatives be 2 males and 3 females?

c. What is the probability that the representatives end up being 2 males and 3 females if they are randomly selected? Enter your answer as a decimal rounded to the nearest thousandth. (3 decimal places).

Answer 1

• Total number = 30
• Number of females = 12
• Number of males = 18

a) Combination of 5 out of 30:

• 30C5 = 30!/[5!*(30 - 5)!] = 142506

b) Combination of 2 males:

• 18C2 = 18!/[2!*(18 - 2)!] = 153

Combination of 3 females:

• 12C3 = 12!/[3!*(12 - 3)!] = 220

The number of ways with 2 males and 3 females:

• 153*220 = 33660

c) Required probability is:

• P(2 males and 3 females) = 33660 / 142506 = 0.236