To find the conditional probability that the number 4 has arrived at least once given that the sum of the numbers is 6 when a die is thrown twice, we need to first determine all the possible outcomes where the sum of the numbers is 6 and then calculate the probability that at least one of the numbers is 4 among those outcomes.

Let's list all the outcomes when two dice are thrown:

1. (1, 1)
2. (1, 2)
3. (1, 3)
4. (1, 4)
5. (1, 5)
6. (1, 6)
7. (2, 1)
8. (2, 2)
9. (2, 3)
10. (2, 4)
11. (2, 5)
12. (2, 6)
13. (3, 1)
14. (3, 2)
15. (3, 3)
16. (3, 4)
17. (3, 5)
18. (3, 6)
19. (4, 1)
20. (4, 2)
21. (4, 3)
22. (4, 4)
23. (4, 5)
24. (4, 6)
25. (5, 1)
26. (5, 2)
27. (5, 3)
28. (5, 4)
29. (5, 5)
30. (5, 6)
31. (6, 1)
32. (6, 2)
33. (6, 3)
34. (6, 4)
35. (6, 5)
36. (6, 6)

Now, let's identify the outcomes where the sum of the numbers is 6:

1. (1, 5)
2. (2, 4)
3. (3, 3)
4. (4, 2)
5. (5, 1)

Among these outcomes, we can see that the number 4 appears at least once in the outcomes (2, 4) and (4, 2).

So, the conditional probability that the number 4 has arrived at least once given that the sum of the numbers is 6 is the probability of these outcomes occurring, which is \( \frac{2}{5} \).

Therefore, the conditional probability is \( \frac{2}{5} \).

Let's solve the problem step by step:

(i) If events A and B are mutually exclusive, it means that they cannot occur simultaneously. In other words, if event A occurs, then event B cannot occur, and vice versa.

To find the probability that the problem is solved when both persons A and B try to solve it independently, we can use the complement rule. The probability that the problem is solved is equal to 1 minus the probability that both A and B fail to solve it.

To solve this problem, we need to understand that drawing cards with replacement means that each time a card is drawn, it is returned to the deck before the next draw. This implies that each draw is independent of the others.

To determine whether events A and B are independent, we need to check if the occurrence of one event affects the probability of the other event.