Hello, Skoz!

A committee of four members is selected randomly from a list of ten names.

The list has the names of five union members and five non-union workers.

a) What is the probability that the first person selected is a union member? 1/2

b) What is the probability that the first 2 people selected are union members? 5/10 x 4/9 = 2/9

c) What is the probability that all commitee members are union members? 1/42

All correct . . . good work!There are many ways to approach this one. .Here's one way . . .d) What is the probability that three of the four committee members are union members?

There are: . posslble committees.

We want 3 Union workers.

There are: . ways to choose them.

We want 1 non-union worker.

There are: . ways to choose him.

Hence, there are: . ways to have exactly 3 union members.

Therefore: .