Introduction:
In this problem, we are given independent trials consisting of rolling a fair die. We need to find the probability that the number 2 appears before the numbers 3 or 5. Let's solve this step by step.

Step 1: Determining the possible outcomes:
When we roll a fair die, the possible outcomes are the numbers 1, 2, 3, 4, 5, and 6.

Step 2: Analyzing the desired outcomes:
We need to determine the probability that the number 2 appears before the numbers 3 or 5. Let's consider the two cases separately:

Case 1: 2 appears before 3:
The possible outcomes in this case are:
2, 3, 4, 5, 6
2, 4, 3, 5, 6
2, 4, 6, 3, 5
...

Case 2: 2 appears before 5:
The possible outcomes in this case are:
2, 5, 3, 4, 6
2, 4, 5, 3, 6
2, 4, 6, 5, 3
...

Step 3: Calculating probabilities:
To calculate the probability of the two cases, we need to determine the number of favorable outcomes for each case and divide it by the total number of possible outcomes.

Case 1: 2 appears before 3:
In this case, we can see that the favorable outcomes are those where 2 appears before 3. The possible outcomes in each trial are 2, 3, 4, 5, and 6. Out of these, only one outcome favors our condition (2 appears before 3). Therefore, the probability of this case is 1/5.

Case 2: 2 appears before 5:
Similarly, in this case, the favorable outcomes are those where 2 appears before 5. The possible outcomes in each trial are 2, 3, 4, 5, and 6. Out of these, two outcomes favor our condition (2 appears before 5): (2, 5, 3, 4, 6) and (2, 4, 5, 3, 6). Therefore, the probability of this case is 2/5.

Step 4: Combining the probabilities:
To find the overall probability, we need to consider both cases. Since the trials are independent, we can multiply the probabilities of each case together.

Probability of Case 1: 1/5
Probability of Case 2: 2/5

Overall probability = (1/5) * (2/5) = 2/25 ≈ 0.08

Conclusion:
The probability that the number 2 appears before the numbers 3 or 5 is approximately 0.08.