Question

Rework problem 26 from the chapter 3 review section of your texting involving the rolling of two fair dice. Answer the following questions.

1) What is the probability that the sum is 4, given that at least one of the numbers rolled is even?
2) What is the probability that at least one number is even, given that the sum is 4?
3) What is the probability that at least one number is even, given that the sum is 3?

Answer 1

The probability that the sum is 4, given that at least one of the numbers rolled is even, is 1/12.

The probability that at least one number is even, given that the sum is 4, is 1/3.

The probability that at least one number is even, given that the sum is 3, is 0.

Step-by-step explanation:

To solve this problem, we need to first determine the sample space of throwing two fair dice. Since each die has 12 sides, the sample space consists of 12 x 12 = 144 possible outcomes.

1) To find the probability that the sum is 4, given that at least one of the numbers rolled is even, we need to determine the favorable outcomes.

The only way to get a sum of 4 is by rolling a 1 and a 3. Given that at least one number is even, we know that the first die must be an odd number (since an even number would guarantee that the sum is even).

There are 6 odd numbers on a 12-sided die. So, the favorable outcomes are (1, 3), (3, 1), (1, 5), (1, 7), (1, 9), and (1, 11), a total of 6 outcomes.

Therefore, the probability of the sum being 4, given that at least one of the numbers rolled is even, is 6/72 = 1/12.

2) To find the probability that at least one number is even, given that the sum is 4, we need to determine the favorable outcomes.

From the previous calculation, we know that the favorable outcomes are (1, 3), (3, 1), (1, 5), (1, 7), (1, 9), and (1, 11), a total of 6 outcomes.

Out of these 6 outcomes, only (1, 5) and (1, 9) have at least one even number.

Therefore, the probability that at least one number is even, given that the sum is 4, is 2/6 = 1/3.

3) To find the probability that at least one number is even, given that the sum is 3, we need to determine the favorable outcomes. The only way to get a sum of 3 is by rolling a 1 and a 2.

Given that at least one number is even, we know that the first die must be an odd number. There are 6 odd numbers on a 12-sided die.

So, the favorable outcomes are (1, 2), (1, 4), (1, 6), (1, 8), (1, 10), and (1, 12), a total of 6 outcomes. Out of these 6 outcomes, none of them have an even number.

Therefore, the probability that at least one number is even, given that the sum is 3, is 0/72 = 0.