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).

a) Combination of 5 out of 30:

b) Combination of 2 males:

Combination of 3 females:

The number of ways with 2 males and 3 females:

c) Required probability is:

50

step-by-step explanation:

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)