I'm stuck on a probability question in my math class. Please explain how you got your answer so I can learn it. Thank you!

There are 8 people (4 men & 4 women). Four people are chosen at random. What is the probability that all were men?

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- Jan 26th 2010, 07:18 AMmdenham24men & 4women. 4 chosen randomly. Probability all men?
I'm stuck on a probability question in my math class. Please explain how you got your answer so I can learn it. Thank you!

There are 8 people (4 men & 4 women). Four people are chosen at random. What is the probability that all were men? - Jan 26th 2010, 07:26 AMe^(i*pi)
The Probability of an event happening is basically: $\displaystyle \frac{\text {amount of desired outcomes}}{\text {total amount of outcomes}}$ The key here is knowing that the person picked out is

**not**returned to the group

$\displaystyle \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5}$ - Jan 26th 2010, 09:33 AMSoroban
Hello, mdenham2!

Another approach . . .

Quote:

There are 4 men & 4 women.

Four people are chosen at random.

What is the probability that all were men?

There are: .$\displaystyle _8C_4 \:=\:\frac{8!}{4!\,4!} \:=\:70$ possible outcomes.

We want 4 men from the available 4 men: .$\displaystyle _4C_4 \:=\:1$ way.

. . and no women from the available 4 women: .$\displaystyle _4C_0 \:=\:1$ way.

Hence, there are: .$\displaystyle 1\cdot1\:=\:1$ way to choose 4 men.

Therefore: .$\displaystyle P(\text{4 men}) \:=\:\frac{1}{70}$

- Jan 26th 2010, 10:08 AMArchie Meade
Hi mdenham2.

The difference between the two approaches is....

$\displaystyle \color{blue}e^{\wedge}(i*pi)$

The first person chosen being a man has probability $\displaystyle \frac{4}{8}$

Now there are 7 people left, of whom 3 are men,

therefore the probability that the second person chosen is a man is $\displaystyle \frac{3}{7}$

There are now 6 people left, of whom 2 are men,

so there are 2 chances out of 6 of chosing a 3rd successive man.

Finally a 1 out of 5 chance the 4th will be a man if the first 3 chosen were men.

These calculations follow if you imagine choosing the people 1 by 1,

one after another.

Multiply the probabilities at each stage for the overall probability.

Or, there are 4 ways the first man can be chosen, 3 ways for the second,

2 for the third and only 1 for the 4th.

Hence there are 4(3)2 ways to pick 4 men in a row,

while there are 8(7}6(5) ways to choose 4 people in a row from 8.

Probability = $\displaystyle \frac{4(3)2}{8(7)6(5)}$

$\displaystyle \color{blue}Soroban$

You may also consider taking 4 people out of the group of 8__at once__.

Soroban's calculations show how to do this.