Question

An ordinary die is rolled twice. What is the probability of each of the following events?

Event A: The sum is greater than 7.
Event B: The sum is divisible by 5.
a. P(A) = ____, P(B) = ____
b. P(A) = ____, P(B) = ____
c. P(A) = ____, P(B) = ____
d. P(A) = ____, P(B) = ____

Answer 1

P(A) = 5/12, P(B) = 7/36

Step-by-step explanation:

P(A): To figure this out you need to use casework:

Case 1: The first die rolls a 1. If the first die rolls a 1, then there are no possibilities that the sum of the 2 die will be greater than 7 since the largest possible number you can get on the second die is 6 making the total 7, which isn't greater than 7.

Case 2: The first die rolls a 2. If the first die rolls a 2, then the only possible number is 6 since, rolling a 5 would make it 7, which isn't more than 7.

Case 3: The first die rolls a 3. If the first die rolls a 3, then there are 2 different possibilities. The first one is the second die rolls a 6, making the total 9. The second possibility is the second die rolls a 5, making the total 8.

You might be seeing a pattern right now which is each time there is one more possibility added. This logic makes sense since each time the first die adds by 1 the second die then can lower the original lowest number by 1. I will still add the other cases but keep this in mind.

Case 4: The first die rolls a 4. If the first die rolls a 4, then there are 3 possibilities. The first is, the second die rolls a 6 making the total 10. The second is, that the second die rolls a 5 making the total 9. The third is, the second die rolls a 4 making the total 8.

Case 5: The first die rolls a 5. If the first die rolls a 5, then there are 4 possibilities. The first is, the second die rolls a 6 making a total of 11. The second is, the second die rolls a 5 making a total of 10. The third is, the second die rolls a 4 making a total of 9. The last is, the second die rolls a 3 making a total of 8.

Case 6, the Final Case: The first die rolls a 6. If the first die rolls a 6, then there are 5 possibilities. The first is, the second die rolls a 6 making a total of 12. The second is, the second die rolls a 5 making a total of 11. The third is, the second die rolls a 4 making a total of 10. The fourth is, the second die rolls a 3 making a total of 9. The last is, the second die rolls a 2 making a total of 8.

Since the 2 die makes 6x6=36 total combinations and out of those 0+1+2+3+4+5=15 combinations are greater than 7. So the answer is 15/36 or 5/12

P(B): This question is simpler than the first since there are only 2 cases. The way to think about this is since the largest combination you can create is 12 and the lowest is 2 so between these 2, there are only 2 numbers that are divisible by 5. The first one is 5, and the second one is 10. You could create these cases in your mind but writing them out makes sure you got everything correct.

Case 1: The total is 5. Since 0 isn't a number that can appear on a die, you cancel out the possibilities 0+5 and 5+0. That means the only possibilities left are 1+4, 2+3, 3+2, and 4+1.

Case 2: The total is 10. The largest number on a die is 6 so we can immediately cancel out the possibilities 10+0, 9+1, 8+2, 7+3, 3+7, 2+8, 1+9, and 0+10. This leaves only 4+6, 5+5, and 6+4.

Since the 2 die makes 6x6=36 total combinations and out of those 4+3=7 combinations are divisible by 5. So the answer is 7/36.

Don't think of these questions as very complicated things and before writing anything down you should think about it first and then start calculating. Sometimes you don't even need to calculate much to figure out the answers.

Answer 2

To calculate the probability of two events when rolling a six-sided die twice, you must analyze the number of outcomes that satisfy the conditions for each event compared to the total number of possible outcomes, which is 36.

Step-by-step explanation:

Probability of Rolling a Die Twice

When an ordinary six-sided die is rolled twice, there are several events that we could be interested in, such as the sum of the numbers rolled or the divisibility of that sum. To find the probability of a specific event, you count the number of favorable outcomes and divide by the total possible outcomes.

Event A: The sum is greater than 7

For Event A, the sum is greater than 7. The possible outcomes that fulfill this condition are when the sum of the two rolls is either 8, 9, 10, 11, or 12. To find P(A), calculate the number of ways you can roll two dice and get these sums versus the total number of possible outcomes, which is 36 (6 sides per die, squared).

Event B: The sum is divisible by 5

For Event B, the sum of the two dice is divisible by 5. The sums that fulfill this condition are 5, 10, and 15. Since 15 cannot be rolled with two six-sided dice, we only consider the sums of 5 and 10. Calculate P(B) similarly to P(A), by counting the outcomes that result in sums of 5 or 10.