5.1.
Contracts for two construction jobs are randomly assigned to one or more of three firms, A, B, and C. Let \(Y_1\) denote the number of contracts assigned to firm A and \(Y_2\) the number of contracts assigned to firm B. Recall that each firm can receive 0, 1, or 2 contracts.
(a)
Find the joint probability function for \(Y_1\) and \(Y_2\text{.}\)
Solution.
Treating the two construction jobs as distinguishable, each job has 3 possible firms it can be assigned to, for a total of \(3^2 = 9\) possible assignment of the two jobs. Let the ordered triple \((j_1 j_2, , )\) represent the assignment of jobs 1 and 2 both to firm A, with other assignment ordered triples defined similarly. Then the 9 possible assignments are \(\{(j_1 j_2, , ), (j_1, j_2, ), (j_1, , j_2), (j_2, j_1, ), ( , j_1 j_2, ), ( , j_1, j_2), (j_2, , j_1), ( , j_2, j_1), ( , , j_1 j_2)\}\text{.}\) Each probability given by the joint probability function for \(Y_1\) and \(Y_2\text{,}\) \(P(Y_1 = y_1, Y_2 = y_2)\text{,}\) will be the number of assignments where firm A is assigned \(y_1\) jobs and firm B is assigned \(y_2\) jobs, all divided by the total number of possible assignments, 9. The table below gives all such probabilities \(p(y_1, y_2) = P(Y_1 = y_1, Y_2 = y_2)\text{:}\)
| \(y_1\) | |||
| \(y_2\) | 0 | 1 | 2 |
| 0 | 1/9 | 2/9 | 1/9 |
| 1 | 2/9 | 2/9 | 0 |
| 2 | 1/9 | 0 | 0 |
(b)
Find \(F(1, 0)\text{.}\)
Solution.
\begin{equation*}
F(1, 0) = \sum_{t_1 \leq 1} \sum_{t_2 \leq 0} p(t_1, t_2) = p(0, 0) + p(1, 0) = \frac{1}{9} + \frac{2}{9} = \frac{3}{9} = \frac{1}{3}
\end{equation*}
