91Ó°ÊÓ

The correlation coefficient, often denoted by \( \rho \), is a measure of the strength and direction of the linear relationship between two random variables. In this exercise, we are looking at \( Y_1 \) (number of white balls) and \( Y_2 \) (number of black balls) drawn from a box without replacement. The correlation coefficient lies between -1 and 1. A value close to 1 implies a strong positive relationship, -1 indicates a strong negative relationship, and 0 suggests no linear relationship.

To compute \( \rho(Y_1, Y_2) \), we use the formula:

• \( \rho(Y_1, Y_2) = \frac{Cov(Y_1, Y_2)}{\sqrt{Var[Y_1] \cdot Var[Y_2]}} \)

Substituting the values from our steps, the correlation coefficient involves simplifying the expression for covariance and variances. Keep in mind that because we're using balls from a finite pool (like in this box example), the hypergeometric distribution accounts for the lack of replacement. So, this specific scenario often results in a negative correlation, as obtaining more of one type decreases the chance of having more of the other type.

The expected value, or mean, is a fundamental concept in probability that provides an average outcome if an experiment is repeated many times. With the hypergeometric distribution, the expected value of the random variables (such as the number of white balls, \( Y_1 \), drawn from the box) is calculated using:

• \( E[Y_1] = n \cdot \frac{N_1}{N} = n p_1 \)
• \( E[Y_2] = n \cdot \frac{N_2}{N} = n p_2 \)

where \( n \) is the sample size, and \( p_1 \) and \( p_2 \) are the probabilities of selecting a white or black ball, respectively.

These formulas provide a quick estimate for how many white or black balls we expect in a sample of size \( n \). They are derived from the proportions of the different colored balls in the entire population. Expected value helps us anticipate outcomes and make informed decisions in probabilistic contexts like this one.

Variance measures the spread of the random variable values around the expected value. For the hypergeometric distribution, this can be slightly more complex due to the finite nature of the sampling pool. In this scenario, the variance tells us how much \( Y_1 \) (number of white balls) and \( Y_2 \) (number of black balls) are likely to differ from their expected values when balls are drawn from the box without replacement.

Variance formulas for hypergeometric distribution are:

• \( Var[Y_1] = n p_1 (1-p_1) \frac{N-N_1}{N-1} \)
• \( Var[Y_2] = n p_2 (1-p_2) \frac{N-N_2}{N-1} \)

Here, the terms \( (1-p_1) \) and \( (1-p_2) \) represent the complement probabilities, while \( \frac{N-N_1}{N-1} \) and \( \frac{N-N_2}{N-1} \) adjust the variance because of the limited total number of balls. Understanding variance helps predict the reliability and variability of your results.

Covariance is a measure of how two random variables change together. In this problem, we are concerned with the covariance between \( Y_1 \) (white balls) and \( Y_2 \) (black balls). The covariance for the hypergeometric distribution differs due to sampling without replacement, and here it affects \( Y_1 \) and \( Y_2 \) negatively.

The formula for covariance given in the solution is:

• \( Cov(Y_1, Y_2) = -n \frac{N_1}{N} \frac{N_2}{N} \frac{N-N_3}{N-1} = -n p_1 p_2 \frac{N-N_3}{N-1} \)

The negative sign here indicates that as the count of one type of ball increases, the count of the other is more likely to decrease in the sample. This is a direct result of the concept of removal, which makes the sampling dependent and reduces the availability of one type as more are picked. Understanding covariance is essential to identifying how two variables interrelate in scenarios like sampling without replacement.