I've been trying to solve this question, but my answer doesn't match the correct one. So I am trying to see where I am going wrong.
Agent 1 has 2 tasks: Task 1 and Task 2. The time it takes Agent 1 to complete task 1 and 2 follows an exp. distribution with mean 6 and mean 4. Agent 2 also has 2 tasks: Task 1 and Task 2. The time it takes Agent 2 to complete task 1 and 2 follows an exp. distribution with mean 3 and mean 1. Task 2 can only be started is Task 1 is complete for both agents. Assuming all the times are independent, what is the probability that Agent 1 completes task 2 first.
The way I understand the problem is that there are 2 possible outcomes. Agent 1 completes Task 1 and 2 before Agent 2 completes task 1 and 2 or Agent 1 completes Task1 after Agent 2 but completes task 2 before Agent 2.
1st Possibility: $$ P(T_{A1Task1} < T_{A2Task1}) \times P(T_{A1Task2} < T_{A2Task2}) $$
$$ P(T_{A1Task1} < T_{A2Task1}) = \frac{\frac{1}{6}}{\frac{1}{6} + \frac{1}{3}} = \frac{1}{3} $$
$$ P(T_{A1Task2} < T_{A2Task2}) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{1}} = \frac{1}{5} $$
$$ P(T_{A1Task1} < T_{A2Task1}) \times P(T_{A1Task2} < T_{A1Task2}) = \frac{1}{15} $$
2nd Possibility:
$$ P(T_{A2Task1} < T_{A1Task1} ) = \frac{\frac{1}{3}}{\frac{1}{6} + \frac{1}{3}} = \frac{2}{3} $$
$$ P(T_{A2Task1} < T_{A1Task1} ) \times P(T_{A1Task2} < T_{A1Task2}) = \frac{2}{3} \times \frac{1}{5} = \frac{2}{15} $$
So the total probability should be $\frac{3}{15} = \frac{1}{5} $. But this is incorrect. Where am I going wrong?
The challenge with trying to compute this in terms of the summary events of which agent finished each task first is that these are not independent.
For example, in your first case: if Agent 1 finishes Task 1 before Agent 2, then this increases their probability that they will also finish Task 2 first, as they get a head start.
As it stand calculating the product assumes that they first wait for Agent 2 to complete Task 1, and only then do they start Task 2.
To correct this approach you may wish to try first conditioning on the specific times at which Agent 1 and Agent 2 finish the first task, which reduces the calculation to comparing the times for Task 2 only. You then need to integrate out the conditioning over all possible times for Task 1.
Since the exponential distribution does not play particularly nicely under addition/subtraction this calculation will require some patience!