Question

Three fair dice are cast. In 10 independent casts, let \(X\) be the number of times all three faces are alike and let \(Y\) be the number of times only two faces are alike. Find the joint \(\mathrm{pmf}\) of \(X\) and \(Y\) and compute \(E(6 X Y)\).

Short answer

The joint function \(P (X = x, Y = y) = \binom{10}{x} (p_1)^x (1 - p_1)^{10 - x} \binom{10}{y} (p_2)^y (1 - p_2)^{10 - y}\) and the expected value \(E(6XY) = 250/36\).

Step-by-step solution

Define the probability for \(X\) and \(Y\)

Denote by \(p_1\) the probability that when three dice are thrown, all faces are the same, and by \(p_2\) the probability that exactly two faces are the same. There are \(6\) ways the three faces can be the same and total of \(6^3 = 216\) possible throws, therefore \(p_1 = 6/216 = 1/36\). Counting the cases when two faces are the same, the first two are the same in 6 ways and the third is different in 5 ways so \(p_2 = 6*5/216 = 5/36\).

Compute the Joint pmf

Let \(Y_i = 1\) if in the ith throw there are exactly two faces alike and \(0\) otherwise, and define \(X_i\) similarly for three faces the same. We find the joint distribution of \(X\) and \(Y\) by noting that \(X = \sum_{i=1}^{10} X_i\) and \(Y = \sum_{i=1}^{10} Y_i\), and both \(X\) and \(Y\) follow binomial distributions, where \(X\) ~ \(B(10, p_1)\) and \(Y\) ~ \(B(10, p_2)\). Therefore, \(P (X = x, Y = y) = P(X = x) P(Y = y) = \binom{10}{x} (p_1)^x (1 - p_1)^{10 - x} \binom{10}{y} (p_2)^y (1 - p_2)^{10 - y}\) because 10 independent casts were made.

Compute the Expected value \(E(6XY)\)

The expected value is given by \(E(6XY) = 6E(XY) = 6E(X)E(Y)\) because \(X\) and \(Y\) are independent. Since for any binomial distribution \(B(n, p)\) the expected value is \(np\), we find \(E(X) = 10 p_1 = 10 / 36\) and \(E(Y) = 10 p_2 = 50 / 36\). Therefore, \(E(6XY) = 6 * 10/36 * 50/36 = 250/36\).