Question

An urn contains \(2 n\) balls, of which \(r\) are red. The balls are randomly removed in \(n\) successive pairs. Let \(X\) denote the number of pairs in which both balls are red. (a) Find \(E[X]\). (b) Find \(\operatorname{Var}(X)\).

Short answer

In summary, the expected value of \(X\) is \(E[X] = \frac{r(r-1)}{4}\), and the variance of \(X\) is \(\operatorname{Var}(X) = \sum_{i=1}^n \frac{(r-2i)^2(r-2i-1)^2}{(2n-2i)^2(2n-2i-1)^2} - \left(\frac{r(r-1)}{4}\right)^2\).

Step-by-step solution

Conditional Probabilities

Let's denote the probability that both balls in pair \(i\) are red by \(P(RR_i)\). The conditional probability of the first pair being two red balls, given that there are \(r\) red balls and \(2n-r\) other balls, is: \[P(RR_1)=\frac{r}{2n} \times \frac{r-1}{2n-1}\] Similarly, for the second pair, given that \((r-2)\) red balls and \((2n-4)\) other balls are left, the probability is: \[P(RR_2) = \frac{r-2}{2n-2} \times \frac{r-3}{2n-3}\] We can continue doing this until the last pair.

Expected Value

We can now find the expected value of \(X\), which is the sum of the probabilities for each pair. Using the linearity of expectation, we have: \[E[X] = \sum_{i=1}^n P(RR_i) = \frac{r(r-1)}{(2n)(2n-1)} + \frac{(r-2)(r-3)}{(2n-2)(2n-3)} + \dots\] To make this sum more manageable, we will rewrite it as a summation: \[E[X] = \sum_{i=1}^n \frac{(r-2(i-2))(r-2i-1)}{(2n-2(i-1))(2n-2i-1)}\]

Simplify the Expected Value Formula

The formula for \(E[X]\) can be simplified by using the fact that the sum of the first n natural numbers is given by \(n(n+1)/2\). Substituting \(i\) with \((i-1)\) in the summation formula, we get: \begin{align*} E[X] &= \sum_{i=0}^{n-1} \frac{(r-2i)(r-2i-1)}{(2n-2i)(2n-2i-1)} \\ &= \frac{1}{(2n)(2n-1)} \sum_{i=0}^{n-1}(r-2i)(r-2i-1)(2n-2i)(2n-2i-1) \\ \end{align*} Now, we know that the sum of the first n natural numbers is \(n(n+1)/2\). So, we can rewrite the summation part of the formula as follows: \[E[X] = \frac{1}{(2n)(2n-1)} \frac{r(r-1)(2n)(2n-1)}{2}\] Simplifying the expression, we find the expected value to be: \[E[X] = \frac{r(r-1)}{4}\]

Variance

Now we will find the variance of \(X\) using the formula: \[\operatorname{Var}(X) = E[X^2] - E[X]^2\] We already have the expected value, so now we can find the expected value of \(X^2\). Similar to the previous steps, we can first find the probability of having two red balls in the same pair for each pair \(i\). Then, we square the probabilities and sum them: \begin{align*} E[X^2] &= \sum_{i=1}^n P(RR_i)^2 \\ &= \frac{r^2(r-1)^2}{(2n)^2(2n-1)^2} +\frac{(r-2)^2(r-3)^2}{(2n-2)^2(2n-3)^2}+ \dots \end{align*} Now, we can substitute this value into the variance formula: \[\operatorname{Var}(X) = \sum_{i=1}^n \frac{(r-2i)^2(r-2i-1)^2}{(2n-2i)^2(2n-2i-1)^2} - \left(\frac{r(r-1)}{4}\right)^2\] Now the variance of \(X\) is given by the formula above. In summary, we have found the expected value of \(X\) to be \(E[X] = r(r-1)/4\), and the variance of \(X\) to be \(\operatorname{Var}(X) = \sum_{i=1}^n \frac{(r-2i)^2(r-2i-1)^2}{(2n-2i)^2(2n-2i-1)^2} - \left(\frac{r(r-1)}{4}\right)^2\).

Key concepts

Conditional Probabilities

Understanding conditional probabilities is crucial for solving many problems in probability theory, including those that involve a sequence of events, like drawing balls from an urn in pairs. In essence, conditional probabilities allow us to determine the likelihood of an event occurring given that another event has already taken place.
In the exercise given, the calculation of the probability that both balls in a pair are red hinges on the number of red balls and other balls remaining for each draw. This is why we calculate the probability conditionally, adjusting for the balls that have already been removed. As the number of red balls decreases with each pair that is drawn, the probability of the subsequent pair also being two red balls must account for this changed dynamic.

Example of Conditional Probability

Imagine an urn that initially contains 3 red balls and 2 blue balls. The conditional probability that the second ball is red, given that the first ball drawn was red, is calculated by considering the new composition of the urn: 2 red balls and 2 blue balls,resulting in a probability of \(\frac{2}{4}=0.5\).
If an exercise asks you to find the likelihood of multiple independent conditional events happening in sequence, you would calculate the individual probabilities at each step and multiply them, considering the change in conditions after each event. Visualizing the scenario or creating a clear probability tree might help in understanding the progression of events and the corresponding changes in probabilities.

Expected Value

The expected value is a fundamental concept in probability that reflects the long-term average outcome of a random variable if an experiment is repeated many times. It's a way of predicting what to expect on average from a random process.
To calculate the expected value, you multiply each possible outcome by the probability of that outcome, and then sum all these values. In the context of the problem at hand, the expected value of \(X\), the number of pairs in which both balls are red, is the sum of the probabilities of drawing a red pair for each draw. We use the linearity of expectation to add the probabilities together, making the calculation more straightforward.

Linearity of Expectation

The beauty of expected value is that it allows for the addition of probabilities of independent events, greatly simplifying the calculation process. This principle is called the linearity of expectation and does not require the events to be independent.
For instance, if one game has an expected payoff of \$10 and another independent game has an expected payoff of \$15, regardless of the first game's outcome, the total expected payoff from playing both games is \$(10+15)=\$25. This principle is applied in our probability model for the number of red pairs drawn from the urn.

Variance

Variance is a measure of how much the outcomes of a random variable differ from the expected value. It quantifies the spread or variability in the possible outcomes. A high variance implies that the outcomes are spread out over a wider range, while a low variance indicates that they are clustered closely around the expected value.
The variance is computed by taking the expected value of the squared deviation of each possible outcome from the expected value, and summing these. In other words, for each outcome, we find the difference between the outcome and the expected value, square that difference to make it positive, then weigh it by the probability of the outcome, and sum all these values.

Computing Variance

In the exercise presented, the variance of \(X\) is found using the formula \(\operatorname{Var}(X) = E[X^2] - E[X]^2\). The expectation of \(X\) squared is crucial as it considers the squared deviation from each possible outcome, thus reflecting the variability.
The calculated variance helps to understand the reliability of the expected value. If the variance is low, you can be more confident that the actual outcomes will usually be close to the expected value. Conversely, a high variance suggests that the actual outcomes could differ significantly from what is expected, indicating a less predictable scenario.
By understanding both expected value and variance, you get a fuller picture of the behavior of a random variable, which is why both are staples in the study of probability models.