Learn Unit VI: Probability with Free Lessons & Tips

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.

To solve this problem, let's first determine the total number of ways two numbers can be selected from the integers 1 through 9.

To find the probability that the target will be hit if both person A and person B shoot at it, we need to consider the complementary events: the event that the target is not hit by either person A or person B.

To solve this problem, let's first list all possible outcomes when a coin is tossed twice:

1. HH (both heads)

2. HT (head, then tail)

3. TH (tail, then head)

4. TT (both tails)

Now, let's consider the condition that the outcome is at most one tail. This means either zero tails or one tail.

The outcomes with at most one tail are: HH, HT, TH. Among these outcomes, only HT and TH have both head and tail.

So, the favorable outcomes are HT and TH.

Now, let's calculate the probability of getting at most one tail:

To find the probability of getting a 4, 5, or 6 on the first toss and a 1, 2, 3, or 4 on the second toss when an unbiased die is tossed twice, we need to calculate the probability of each individual outcome and then sum them up.