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Math Help - Looking for probability formula

  1. #1
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    Looking for probability formula

    Been awhile since I've been in school, so forgotten a lot, and I came across a problem that I having trouble figuring out.

    Looking for the formulas so I can understand it.

    Given 100 people randomly choosing a number from 1 - 100, what is the probability that exactly 3 people will have picked the same number.

    Same as above but instead what is the probability that at least 3 people will have picked the same number.

    I can do it with small numbers doing probability trees but obviously with 100 numbers and 100 people the number of combinations would be large. Would prefer a formula that would be able to calculate the answer(s).

    Thanks.
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  2. #2
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    Assuming you are drawing a number with replacement, each person choosing a number is a single independent experiment where the odds of success are 1/100, and the odds of failure are 99/100. So your distribution would be:

    X~Bin(100,0.01)=

    P(X=x)=\left(\begin{array}{c}100\\x\end{array}\rig  ht)\left(\frac{1}{100}\right)^{x}\left(\frac{99}{1  00}\right)^{100-x}

    And you are trying to find P(X=3). It's really small. As it should be. For your second part, you are finding P(X\geq3). This would be insane to do for x=3, 4, 5. . .100 - so simply calculate the compliment and subtract it from 1:

    1-P(X<3);

    Here you are only solving three, so it shouldn't be that bad.
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  3. #3
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    Quote Originally Posted by ANDS! View Post
    Assuming you are drawing a number with replacement, each person choosing a number is a single independent experiment where the odds of success are 1/100, and the odds of failure are 99/100. So your distribution would be:

    X~Bin(100,0.01)=

    P(X=x)=\left(\begin{array}{c}100\\x\end{array}\rig  ht)\left(\frac{1}{100}\right)^{x}\left(\frac{99}{1  00}\right)^{100-x}

    And you are trying to find P(X=3). It's really small. As it should be. For your second part, you are finding P(X\geq3). This would be insane to do for x=3, 4, 5. . .100 - so simply calculate the compliment and subtract it from 1:

    1-P(X<3);

    Here you are only solving three, so it shouldn't be that bad.
    Thanks for the quick reply! I don't ever remember seeing a formula like that when I was in school, so I guess this stuff is way beyond what I ever learned.

    What numbers are for which variables? For instance if I wanted to change it to chance of x people picking same number given 50 people choosing numbers from 1 to 75.

    Would it be (50/x) (1/75)^x (74/75)^(50-x)? Would that be right?
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  4. #4
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    Yes. That would be right. You no doubt encountered it, but depending on the class level might not have been explicitly this.

    Or you just forgot.
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