Question

Let \(X\) denote the number of white balls selected when \(k\) balls are chosen at random from an urn containing \(n\) white and \(m\) black balls. (a) Compute \(P[X=i\\}\). (b) Let, for \(i=1,2, \ldots, k ; j=1,2, \ldots, n\) \(X_{i}=\left\\{\begin{array}{ll}1, & \text { if the } i \text { th ball selected is white } \\ 0, & \text { otherwise }\end{array}\right.\) \(Y_{j}=\left\\{\begin{array}{ll}1, & \text { if white ball } j \text { is selected } \\ 0, & \text { otherwise }\end{array}\right.\) Compute \(E[X]\) in two ways by expressing \(X\) first as a function of the \(X_{i} s\) and then of the \(Y_{j}\) s.

Short answer

The probability of selecting i white balls out of k total balls is given by: \[P[X=i] = \frac{\binom{n}{i} \binom{m}{k - i}}{\binom{n + m}{k}}\] The expected value of X can be calculated in two ways: 1. As a function of \(X_i\): \[E[X] = k \times \frac{n}{n + m}\] 2. As a function of \(Y_j\): \[E[X] = n \times \frac{k}{n + m}\] Both methods yield the same result for the expected value of X, demonstrating that the choice of representation does not change the final result for E[X].

Step-by-step solution

(a) Find the probability P[X=i]

To determine the probability of selecting i white balls out of k total balls, we can use the following formula: \[P[X=i] = \frac{\binom{n}{i} \binom{m}{k - i}}{\binom{n + m}{k}}\] Where: - \(\binom{n}{i}\) represents the number of ways to choose i white balls out of n white balls, - \(\binom{m}{k - i}\) represents the number of ways to choose (k - i) black balls out of m black balls, - \(\binom{n + m}{k}\) represents the total number of ways to choose k balls out of (n + m) balls.

(b.1) Calculate E[X] as a function of \(X_i\)

To find the expected value of X as a function of \(X_i\), we can first express X as the sum of \(X_i\)s: \[X = X_1 + X_2 + \cdots + X_k\] Now, we calculate the expected value E[X]: \[E[X] = E[X_1 + X_2 + \cdots + X_k] = E[X_1] + E[X_2] + \cdots + E[X_k]\] Since each \(X_i\) is an indicator variable representing whether the i-th ball selected is white or not, we can find the expected value E[\(X_i\)] as follows: \[E[X_i] = 1 \cdot P(X_i = 1) + 0 \cdot P(X_i = 0) = P(X_i = 1)\] The probability that the i-th ball selected is white is \(P(X_i = 1) = \frac{n}{n + m}\). Therefore, the expected value E[X] as a function of \(X_i\) is: \[E[X] = k \times \frac{n}{n + m}\]

(b.2) Calculate E[X] as a function of \(Y_j\)

To find the expected value of X as a function of \(Y_j\), we can first express X as the sum of \(Y_j\)s: \[X = Y_1 + Y_2 + \cdots + Y_n\] Now, we calculate the expected value E[X]: \[E[X] = E[Y_1 + Y_2 + \cdots + Y_n] = E[Y_1] + E[Y_2] + \cdots + E[Y_n]\] Since each \(Y_j\) is an indicator variable representing whether the j-th white ball is selected or not, we can find the expected value E[\(Y_j\)] as follows: \[E[Y_j] = 1 \cdot P(Y_j = 1) + 0 \cdot P(Y_j = 0) = P(Y_j = 1)\] The probability that the j-th white ball is selected is \(P(Y_j = 1) = \frac{k}{n + m}\). Therefore, the expected value E[X] as a function of \(Y_j\) is: \[E[X] = n \times \frac{k}{n + m}\] Both methods yield the same result for the expected value of X, which demonstrates that the choice of representation (i.e., as a function of \(X_i\) or \(Y_j\)) does not change the final result for E[X].