# Thread: Looking for probability formula

1. ## 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.

2. 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.

3. Originally Posted by ANDS!
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?

4. 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.