A pair of dice is rolled. Find the probability of rolling
a) a sum not more than 1010,
b) a sum not less than 77,
c) a sum between 33 and 77 (exclusive).
Your question makes little sense.
The sum on the two dice is between 2 and 12
What's with all the weird numbers?
ok,
a) the prob of rolling a sum not more than 1010 = 1
b) prob of sum less than 77 = 1
so prob of sum not less than 77 = 0
c) prob of sum between 33 and 77 ??
there are 3 such numbers, namely 34, 35, and 36
so prob(of your event) = 3/36 = 1/12
To find the probability of rolling a certain sum with a pair of dice, we need to determine the number of ways that sum can be obtained and divide it by the total possible outcomes.
Each die has 6 faces numbered from 1 to 6. So the total number of outcomes when rolling two dice is 6 * 6 = 36.
a) To find the probability of rolling a sum not more than 10, we need to determine the number of ways to obtain a sum less than or equal to 10. We can calculate this by listing all the possible combinations:
2 (1+1), 3 (1+2 or 2+1), 4 (1+3, 2+2, 3+1), 5 (1+4, 2+3, 3+2, 4+1), 6 (1+5, 2+4, 3+3, 4+2, 5+1), 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), 8 (2+6, 3+5, 4+4, 5+3, 6+2), 9 (3+6, 4+5, 5+4, 6+3), 10 (4+6, 5+5, 6+4)
There are 21 possible combinations in which the sum is not more than 10. Therefore, the probability is 21/36.
b) To find the probability of rolling a sum not less than 7, we need to determine the number of ways to obtain a sum of 7 or greater. We can calculate this by listing all the possible combinations:
7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), 8 (2+6, 3+5, 4+4, 5+3, 6+2), 9 (3+6, 4+5, 5+4, 6+3), 10 (4+6, 5+5, 6+4), 11 (5+6, 6+5), 12 (6+6)
There are 15 possible combinations in which the sum is not less than 7. Therefore, the probability is 15/36.
c) To find the probability of rolling a sum between 33 and 77 (exclusive), we need to determine the number of ways to obtain a sum greater than 33 and less than 77. Since the largest possible sum is 12 (6+6) and the smallest possible sum is 2 (1+1), all sums between 33 and 77 are possible.
Therefore, the probability of rolling a sum between 33 and 77 (exclusive) is 36/36, which is equivalent to 1.
To find the probabilities of rolling different sums with a pair of dice, we first need to determine the total number of possible outcomes and the number of favorable outcomes for each situation.
The total number of outcomes when rolling a pair of dice is found by multiplying the number of outcomes on one die by the number of outcomes on the other die. Since each die has 6 sides numbered 1 through 6, the total number of outcomes would be 6 × 6 = 36.
Now let's consider each situation separately:
a) To find the probability of rolling a sum not more than 10, we need to count the favorable outcomes. In this case, we can list all the possible sums:
Sum 2: (1, 1)
Sum 3: (1, 2) (2, 1)
Sum 4: (1, 3) (2, 2) (3, 1)
Sum 5: (1, 4) (2, 3) (3, 2) (4, 1)
Sum 6: (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)
Sum 7: (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)
Sum 8: (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)
Sum 9: (3, 6) (4, 5) (5, 4) (6, 3)
Sum 10: (4, 6) (5, 5) (6, 4)
Sum 11: (5, 6) (6, 5)
Sum 12: (6, 6)
From the above list, we can see that there are 10 favorable outcomes.
Therefore, the probability of rolling a sum not more than 10 is 10/36, which simplifies to 5/18.
b) To find the probability of rolling a sum not less than 77, we need to count the favorable outcomes. Since the maximum sum we can get is 12, it's not possible to roll a sum less than 7. Therefore, all 36 outcomes are favorable.
Therefore, the probability of rolling a sum not less than 7 is 36/36, which simplifies to 1.
c) To find the probability of rolling a sum between 33 and 77 (exclusive), we need to count the favorable outcomes between 33 and 77 (exclusive). In this case, we can list all the possible sums again:
Sum 2: (1, 1)
Sum 3: (1, 2) (2, 1)
Sum 4: (1, 3) (2, 2) (3, 1)
Sum 5: (1, 4) (2, 3) (3, 2) (4, 1)
Sum 6: (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)
Sum 7: (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)
Sum 8: (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)
Sum 9: (3, 6) (4, 5) (5, 4) (6, 3)
Sum 10: (4, 6) (5, 5) (6, 4)
Sum 11: (5, 6) (6, 5)
Sum 12: (6, 6)
From the above list, we can see that there are 6 favorable outcomes (4, 5), (5, 4), (5, 5), (5, 6), (6, 5), (6, 6).
Therefore, the probability of rolling a sum between 33 and 77 (exclusive) is 6/36, which simplifies to 1/6.