MHB Calculate Probability of Y or More People in N Train Cars

[Toonzaka1]

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!

 

[chisigma]

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!

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... $\displaystyle p_{n} = \binom {N}{n} p^{n}\ (1-p)^{N-n}\ (1)$ ... so that the requested probability is...$\displaystyle P= \sum_{n=y}^{N} p_{n}\ (2)$ Kind regards $\chi$ $\sigma$

 

[Toonzaka1]

I believe I understand why we chose that equation to solve this. Could I say that if each person choosing a car is random and equal that the probability of person n1 selecting car c1 is the same as them selecting c2, c3,...,cx? Meaning in the binomial formula we can say that p = 1/cx where cx is the number of cars?

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 :D .

 

[awkward]

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!

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?

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 $1 - \frac{N! \; a_N}{X^N}$.