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- Thread starter Toonzaka
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- Feb 13, 2012

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Wellcome in MHB Toonzaka!... if we suppose that each fellow has equal probability $\displaystyle p=\frac{1}{x}$ to be in a car and any car contain all N people, the the probability to have n peple in a car is...Quick question for you all...

If I had X number of train cars and I wanted to know the probability of having Y or more people in a car when I have N total of people. How would I go about solving this?

Thanks!

$\displaystyle p_{n} = \binom {N}{n} p^{n}\ (1-p)^{N-n}\ (1)$

... so that the requested probability is...

$\displaystyle P= \sum_{n=y}^{N} p_{n}\ (2)$

Kind regards

$\chi$ $\sigma$

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I just really want to learn how to solve this type of problem and why we would solve it that way.

Also, thanks for the reply and welcome .

There is some ambiguity in this question. Do you mean having Y or more people in one particular car (for example, the first car), or do you mean having Y or more people in the most crowded car?Quick question for you all...

If I had X number of train cars and I wanted to know the probability of having Y or more people in a car when I have N total of people. How would I go about solving this?

Thanks!

If you mean having Y or more people in one particular car, ChiSigma has given the answer-- it's a binomial distribution.

But if you are interested in the probability of having Y or more people in the most crowded car, that's more complicated. In that case, it's more convenient to look at the complementary event-- having at most Y-1 people in the most crowded car. If you know that probability, you can just subtract it from 1 to get the probability of having Y or more people in the most crowded car. The only way I know to solve the problem involves generating functions, so I will just quote the result. Let $k = Y-1$. The first step is to expand the polynomial

$$\left(1 + \frac{z}{1!} + \frac{z^2}{2!} + \dots + \frac{z^k}{k!} \right) ^X$$

Just multiply it out or use the multinomial theorem. (If you have access to a computer algebra system or know how to use Wolfram Alpha, this would be a good place to use it.) Let $a_N$ be the coefficient of $z^N$ after the expansion. Then the probability that the most crowded car will have at most Y-1 people in it is $\frac{N! \; a_N}{X^N}$, and the probability that the most crowded car will have Y or more people in it is $1 - \frac{N! \; a_N}{X^N}$.

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