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?
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?
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
$\displaystyle \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5}$
Hello, mdenham2!
Another approach . . .
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: .$\displaystyle P(\text{4 men}) \:=\:\frac{1}{70}$
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.