E(X) = 1.147 - average number of red balls
E(Y) = 1.853 - average number of blue balls
Var(X) = 0.425 - variance of X around its expected value
Cov(X,Y) = -0.425 - negative correlation between X and Y
Here's how to compute the expected values, variances, and covariance in the given scenario:
a) Expected Value of X (E(X))
This represents the average number of red balls selected.
Since there are 3 draws without replacement, we can calculate the probability of each outcome and its corresponding number of red balls:
3 red balls: (5/13 * 4/12 * 3/11) * 3 = 5/143
2 red balls: (5/13 * 4/12 * 8/11) * 2 = 40/143
1 red ball: (5/13 * 8/12 * 3/11) * 1 = 120/143
0 red balls: (8/13 * 4/12 * 3/11) * 0 = 0
Therefore, E(X) = (5/143 * 3) + (40/143 * 2) + (120/143 * 1) + (0) = 1.147.
b) Expected Value of Y (E(Y))
This represents the average number of blue balls selected. Since Y = 3 - X, we can simply subtract E(X) from 3:
E(Y) = 3 - E(X) = 3 - 1.147 = 1.853.
c) Variance of X (Var(X))
This measures the spread of X around its expected value. We can calculate it using the formula:
Var(X) = Σ[P(X=x) * (x - E(X))^2]
Following the same calculations as for E(X), we get:
Var(X) = (5/143 * (3 - 1.147)^2) + (40/143 * (2 - 1.147)^2) + (120/143 * (1 - 1.147)^2) = 0.425.
d) Covariance of X and Y (Cov(X,Y))
This measures the dependence between X and Y.
Since X and Y are negatively correlated (more red balls implies fewer blue and vice versa), the covariance will be negative.
We can calculate it using the formula:
Cov(X,Y) = Σ[P(X=x,Y=y) * (x - E(X)) * (y - E(Y))]
However, calculating all the joint probabilities for each outcome is quite tedious.
We can leverage the fact that Cov(X,Y) = -Var(X) for negatively correlated binary variables (like X and Y in this case).
Therefore, Cov(X,Y) = -0.425.