## Rings in soup

### Task number: 3534

A cook has mistakenly dropped two distinct rings into some soup. The soup has been divided among 25 guests, including 8 women.

#### Variant

What is the probability that a single person will get both rings?

#### Hint

First propose a probability space \((\Omega,\mathcal F, P)\).

#### Solution

The set of elementary events \(\Omega\) is formed from all possible ways to distribute two distinct rings among 25 people, or the set of all possible mappings from a two-element set of rings into a set of 25 people. And so \(|\Omega|=25^2=625\).

The exercise assignment does not specify whether some ways to distribute the rings should be more probable than others, so we will assume a uniform probability distribution. Or for every \(\omega\in\Omega: P(\omega)=\frac{1}{|\Omega|}\).

The event where both rings are given to the same person corresponds to the set \(A\in \mathcal F\) (or \(A\subseteq\Omega\)) containing mappings which have the same value for both components. There are the same number of these as there are elements in the destination set, i.e. 25.

#### Answer

The resulting probability is \(\frac{1}{25}\).

#### Variant

What is the probability that no man receives a ring?

#### Solution

There are \(8^2\) mappings into the eight-element set of women.

#### Answer

The resulting probability is \(\frac{64}{625}\).

#### Variant

What is the probability that two men will have rings in their soup?

#### Solution

The positive outcomes correspond to injective mappings into a seventeen-element set of men, of which there are \(17{\cdot} 16\).

#### Answer

The resulting probability is \(\frac{272}{625}\).

#### Variant

What is the probability that one man and one woman will have rings in their soup?

#### Solution

We first choose the people who will receive the rings. We can choose a man in 17 ways and a woman in eight ways. We can assign the rings to each such pair in two different ways. So the number of positive outcomes is \(17{\cdot} 8\cdot 2=272\).

#### Answer

The resulting probability is \(\frac{272}{625}\).

#### Variant

How do the probabilities change if the rings are identical?

#### Solution

The set of events \(\Omega\) contains a unordered selection of at most two elements from 25, and so \(|\Omega|=\binom{25}{2}+\binom{25}{1}=\binom{26}{2}=325\).

The probabilities of elementary events are not identically distributed, because an event where both rings are in the same bowl is half as likely as an event where they are in two separate bowls (similarly as in a throw of two identical dice). So the elementary events where both rings belong to a single individual have probability \(P(\omega)=\frac{1}{625}\) and the events where distinct people receive the rings have probability \(P(\omega)=\frac{2}{625}\).

Here are the probabilities of the events under investigation:

- \(A\) … the same person receives both rings, \(P(A)=25\cdot\frac{1}{625}=\frac{1}{25}\)
- \(B\) … no man receives a ring, \(P(B)=8\cdot\frac{1}{625}+\binom{8}{2}\cdot\frac{2}{625}=\frac{8+56}{625}=\frac{64}{625}\)
- \(C\) … two men receive rings, \(P(C)=\binom{17}{2}\cdot\frac{2}{625}=\frac{272}{625}\)
- \(D\) … one man and one woman receive rings, \(P(D)=(17{\cdot} 8)\cdot\frac{2}{625}=\frac{272}{625}\)

In this case the probabilities have not changed. We can in fact further justify the validity of this model, which may appear unnatural at first glance, by noting that the probabilities of events that do not depend on differences between the rings should remain unchanged even when we take such differences into account.

#### Comment

By the way, if we decided that each of these elementary events had identical probability \(\frac{1}{325}\), then the probabilities of the investigated events would turn out as follows:

- \(|A|=25\), \(P(A)=\frac{25}{325}=\frac{1}{13}\)
- \(|B|=\binom{9}{2}\), \(P(B)=\frac{36}{325}\)
- \(|C|=\binom{17}{2}\), \(P(C)=\frac{136}{325}\)
- \(|D|=17{\cdot} 8\), \(P(D)=\frac{136}{325}\)

But of course this proposed probability space would not agree with reality.