$\left.\begin{array}{ll}{\left[\begin{array}{l}\text { Averages }\end{array}\right]} \\ {[\text { Pascal's Triangle and the Binomial Theorem }]}\end{array}\right]$

[ Products and Factorials $\quad]$

On board an airliner, there are $2 n$ passengers, and the airline has loaded $n$ portions of chicken and $n$ portions of fish for them. It is known that a passenger prefers chicken with a probability of 0.5 and fish with a probability of 0.5. We will call a passenger dissatisfied if they are left with what they do not prefer.

a) Find the most probable number of dissatisfied passengers.

b) Find the expected value of the number of dissatisfied passengers.

c) Find the variance of the number of dissatisfied passengers.
a) The number of dissatisfied passengers can be any from 0 to $n$. In the case of $n=1$, everything is obvious: there is either no dissatisfied passenger or one, and both cases are equally likely. We will further assume that $n>1$.

Let us introduce the random variable $\xi$ "Number of dissatisfied passengers". $\xi=0$ only if exactly $n$ passengers prefer chicken, and the other $n$ prefer fish. We will consider a success the event "a passenger wants chicken". Then

$$
\mathrm{P}(\xi=0)=\mathrm{P}(\{\text { there are $n$ successes in a series of $2n$ trials }\})=C_{2 n}^{n} \frac{1}{2^{2 n}} \text {. }
$$

Exactly one passenger will be dissatisfied if the number of passengers preferring chicken differs from $n$ by one, that is, the number of successes is

$n \pm 1$. Therefore, $\mathrm{P}(\xi=1)=\mathrm{P}(\{n+1$ success in a series of $2n$ trials $\})+\mathrm{P}(\{n-1$ success in a series of $2n$ trials $\})=2 C_{2 n}^{n-1} \frac{1}{2^{2 n}}$.

Reasoning similarly, we find that $\mathrm{P}(\xi=k)=2 C_{2 n}^{n-k} \frac{1}{2^{2 n}}$, where $k=1,2, \ldots, n$.

In the sequence of numbers $C_{2 n}^{m}$, there are $2n+1$ numbers (this is the $2n$-th row of Pascal's triangle). These numbers first increase for $m=0,1, \ldots, n$, and then decrease for $m=n, n+1, \ldots, 2n$. The middle number $C_{2 n}^{n}$ is larger than the others, but smaller than twice the previous one: $C_{2 n}^{n}1 \mathrm{P}(\xi=0)\mathrm{P}(\xi=2)>\ldots>\mathrm{P}(\xi=n)$.

The result is illustrated by the distribution diagram of the random variable $\xi$ for $n=5$. For comparison, the diagram of the binomial distribution $\frac{1}{2^{10}} C_{10}^{k}$ is given in a lighter color on the same graph.

![](https://cdn.mathpix.com/cropped/2024_05_06_696cdbe703e904f1264dg-21.jpg?height=680&width=1235&top_left_y=0&top_left_x=409)

Thus, the most probable number of dissatisfied passengers is one.

b) Let us find the expected value of the quantity $\xi+n$:

$$
\begin{aligned}
& \mathrm{E}(\xi+n)=n \mathrm{P}(\xi=0)+(n+1) \mathrm{P}(\xi=1)+(n+2) \mathrm{P}(\xi=2)+\ldots+2 n \mathrm{P}(\xi=n)= \\
& =n C_{2 n}^{n} \frac{1}{2^{2 n}}+\frac{1}{2^{2 n-1}}\left((n+1) C_{2 n}^{n-1}+(n+2) C_{2 n}^{n-2}+\ldots+(n+k) C_{2 n}^{n-k}+\ldots+2 n C_{2 n}^{0}\right)
\end{aligned}
$$

Notice that $(n+k) C_{2 n}^{n-k}=\frac{(n+k)(2 n)!}{(n-k)!(n+k)!}=2 n C_{2 n-1}^{n-k}$. Therefore, $\mathrm{E}(\xi+n)=$

$n C_{2 n}^{n} \frac{1}{2^{2 n}}+\frac{2 n}{2^{2 n-1}}\left(C_{2 n-1}^{n-1}+C_{2 n-1}^{n-2}+\ldots+C_{2 n-1}^{0}\right)$

In the parentheses, we have the sum of the first half of all numbers in the $(2n-1)$-th row of Pascal's triangle. The sum of all numbers in this row is $2^{2 n-1}$, so the sum in the parentheses is $2^{2 n-2}$. Thus, $\mathrm{E}(\xi+n)=n C_{2 n}^{n} \frac{1}{2^{2 n}}+n$, and $\mathrm{E} \xi=\mathrm{E}(\xi+n)-n=n C_{2 n}^{n} \frac{1}{2^{2 n}}$

Let us find this number approximately using Stirling's formula $\quad m!\approx \sqrt{2 \pi m} \cdot m^{m} e^{-m}$ :

$\mathrm{E} \xi \approx \frac{n}{2^{n}} \cdot \frac{\sqrt{4 \pi n} \cdot(2 n)^{2 n} e^{-2 n}}{\left(\sqrt{2 \pi n} \cdot n^{n} e^{-n}\right)^{2}}=\sqrt{\frac{n}{\pi}}$.

c) $\mathrm{E} \xi^{2}=0^{2} \cdot \mathrm{P}(\xi=0)+1^{2} \cdot \mathrm{P}(\xi=1)+\ldots+n^{2} \mathrm{P}(\xi=n)=1^{2} \cdot \frac{C_{2 n}^{n-1}}{2^{2 n-1}}+2^{2} \cdot \frac{C_{2 n}^{n-2}}{2^{2 n-1}}+\ldots+n^{2} \cdot \frac{C_{2 n}^{0}}{2^{2 n-1}}=$

$$
=(0-n)^{2} \cdot \frac{C_{2 n}^{0}}{2^{2 n}}+(1-n)^{2} \cdot \frac{C_{2 n}^{1}}{2^{2 n}}+(2-n)^{2} \cdot \frac{C_{2 n}^{2}}{2^{2 n}}+\ldots+(2 n-n)^{2} \cdot \frac{C_{2 n}^{2 n}}{2^{2 n}}
$$

This expression is exactly equal to the variance of the random variable $\eta$ "number of successes in a series of $2n$ Bernoulli trials with a success probability of $0.5$". As is known, it is equal to $2n \cdot 0.5 \cdot 0.5=n / 2$.

Therefore, $\mathrm{D} \xi=\mathrm{E} \zeta^{2}-(\mathrm{E} \xi)^{2}=\frac{n}{2}-\left(\frac{n C_{2 n}^{n}}{2^{2 n}}\right)^{2} \approx \frac{n}{2}-\frac{n}{\pi}=\frac{\pi-2}{2 \pi} n \approx 0.182 n$.

## Answer

a) If $n>1$, then the most probable is one dissatisfied; if $n=1$, then with equal probabilities either there is one dissatisfied, or there are no dissatisfied at all;

b) $\frac{n C_{2 n}^{n}}{2^{2 n}} \approx \sqrt{\frac{n}{\pi}} \approx 0.564 \sqrt{n}$; c) $\frac{n}{2}-\left(\frac{n C_{2 n}^{n}}{2^{2 n}}\right)^{2} \approx \frac{\pi-2}{2 \pi} n \approx 0.182 n$.