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    4men & 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?
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    Quote Originally Posted by mdenham2 View Post
    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: \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

    \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5}
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    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: . _8C_4 \:=\:\frac{8!}{4!\,4!} \:=\:70 possible outcomes.


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

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

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


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

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    Quote Originally Posted by mdenham2 View Post
    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?
    Hi mdenham2.

    The difference between the two approaches is....

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

    The first person chosen being a man has probability \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 \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 = \frac{4(3)2}{8(7)6(5)}

    \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.
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