If two fair dice are rolled, what is the probability that a total showing is either even or less than eight?
How would I solve this question?
What are the steps?
Thanks
If you are doing more of these dice-rolling problems it might be a good idea to have a matrix for 2 dice showing rows and columns from 1 to 6, and sums as the entries of the matrix
you will see that there are 18 even sums and 18 odd sums.
For a sum of less than 8 we have:
1,6 1,5, 1,4 1,3 1,2 1,1 ---- 6 of them
2,5 2,4 2,3 2,2 2,1 ----5 of them
3,4 3,3 3,2 3,1 ------4 of them
4,3 4,2 4,1 -------- 3 of them
5,2 5,1 -------- 2 of them
6,1 -------- 1 more
but in this total of 21 , 12 are already counted in the odds
so P(odd or <8) = (18 + 21 - 12)/36
= 3/4
To solve this question, you need to understand the concept of probability and work through the steps involved. Here's how you can approach this problem:
Step 1: Determine the total number of outcomes when rolling two fair dice. Each die has six sides numbered 1 to 6, so the total number of outcomes is 6 multiplied by 6, which equals 36.
Step 2: Identify the favorable outcomes. We want to find the total showing that is either even or less than eight. Let's break it down:
a) Even outcomes: When rolling a fair die, the number can be even if it is either 2, 4, or 6. There are three possible even outcomes.
b) Outcomes less than eight: In this case, we need to find the outcomes that sum up to less than eight. We can do this by listing all the possible combinations:
- For (1, 1), the sum is 2
- For (1, 2), the sum is 3
- For (1, 3), the sum is 4
- For (1, 4), the sum is 5
- For (1, 5), the sum is 6
- For (1, 6), the sum is 7
- For (2, 1), the sum is 3
- For (2, 2), the sum is 4
- For (2, 3), the sum is 5
- For (2, 4), the sum is 6
- For (2, 5), the sum is 7
- For (3, 1), the sum is 4
- For (3, 2), the sum is 5
- For (3, 3), the sum is 6
- For (3, 4), the sum is 7
- For (4, 1), the sum is 5
- For (4, 2), the sum is 6
- For (4, 3), the sum is 7
- For (5, 1), the sum is 6
- For (5, 2), the sum is 7
- For (6, 1), the sum is 7
In total, there are 21 outcomes that sum up to less than eight.
Step 3: Calculate the probability. Divide the number of favorable outcomes by the total number of outcomes.
a) Even outcomes: There are three favorable outcomes, so the probability of getting an even number on either die is 3/36, which simplifies to 1/12.
b) Outcomes less than eight: There are 21 favorable outcomes, so the probability of getting a sum less than eight is 21/36, which simplifies to 7/12.
Step 4: Find the probability of the given condition. Since we want to find the probability that the total showing is either even or less than eight, we need to add the probabilities from step 3a and 3b together:
Probability = Probability of even outcomes + Probability of outcomes less than eight
Probability = 1/12 + 7/12
Probability = 8/12
Therefore, the probability that a total showing is either even or less than eight when two fair dice are rolled is 8/12, which simplifies to 2/3 or approximately 0.67.
I hope this explanation helps! Let me know if you have any further questions.
To solve this question, you can follow these steps:
Step 1: Determine all the possible outcomes when rolling two fair dice. Each die has six sides with numbers from 1 to 6, so the total number of outcomes is 6 * 6 = 36.
Step 2: Identify the outcomes that would result in a total either being even or less than eight. We can list all the favorable outcomes: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1), so there are 15 favorable outcomes.
Step 3: Determine the probability by dividing the number of favorable outcomes by the total number of outcomes. Therefore, the probability is 15/36.
Simplifying the fraction, the probability that a total showing is either even or less than eight when rolling two fair dice is 5/12 or approximately 0.4167.
Let me know if you need help with anything else!