Math Help - Probability question

1. Probability question

A health safety comittee is to be selected from all the people who work at at a local factory. The committee is to consist of four members selected randomly from a list of ten names submitted by the shop leader. The list has the names of five union members and five workers who are not union members.

a) What is the probability that the first person selected from the list is a union member. 1/2

b)What is the probability that the first 2 people selected from the list are union members. 5/10 x 4/9 (tree diagram)=2/9

c)What is the probability that all commitee members are union members. 1/42 (used a tree diagram).

d)What is the probability that three of the four committee members are union members.

I dont know how to get this answer...

2. 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!
d) What is the probability that three of the four committee members are union members?
There are many ways to approach this one. .Here's one way . . .

There are: . ${10\choose4} \:=\:{\color{blue}210}$ posslble committees.

We want 3 Union workers.
There are: . ${5\choose3} \:=\:10$ ways to choose them.

We want 1 non-union worker.
There are: . ${5\choose1} \:=\:5$ ways to choose him.

Hence, there are: . $10\times5 \:=\:{\color{blue}50}$ ways to have exactly 3 union members.

Therefore: . $P(\text{3 union, 1 non-union}) \;=\;\frac{50}{210} \;=\;\frac{5}{21}$

3. thanks but the answer is 5/84 according to the back of the book.

4. Hello again, Skoz!

Thanks, but the answer is 5/84 according to the back of the book.
The book is wrong . . .
The author must have done something like this . . .

What is the probability that the first three are union and the last is non-union?

. . $\begin{array}{cccc}\text{First is Union:} & P(\text{1st U}) &=& \frac{5}{10}\\ \\[-4mm]
\text{Second is Union:} & P(\text{2nd U}) &=& \frac{4}{9} \\ \\[-4mm]
\text{Third is Union:} & P(\text{3rd U}) &=& \frac{3}{8} \\ \\[-4mm]
\text{Fourth is Non-union:} & P(\text{4th N}) &=& \frac{5}{7}
\end{array}$

Therefore: . $P(UUUN) \:=\:\frac{5}{10} \times\frac{4}{9} \times\frac{3}{8} \times\frac{5}{7} \;=\;\frac{5}{84}$

But the non-union member could be in any of FOUR places in the selection.

And that's why the correct answer is four times larger: . $4 \times\frac{5}{84} \:=\:\frac{5}{21}$