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
To solve this question, you need to determine the number of possible outcomes that either have an even sum or a sum less than eight, and then divide it by the total number of possible outcomes when two fair dice are rolled.
Here are the steps to solve the problem:
Step 1: Determine the number of possible outcomes where the sum is either even or less than eight:
- To have an even sum, we need either both dice to be even or both dice to be odd.
- Number of ways to get an even sum: 3 (2 even numbers: 2 and 4, and 1 odd number: 6)
- Number of ways to get an odd sum: 3 (3 odd numbers: 1, 3, and 5)
- Total number of ways to get an even or odd sum: 3 + 3 = 6
- To have a sum less than eight, we need to list all the possibilities and count them:
- 1+1, 1+2, 1+3, 1+4, 1+5, 1+6 = 6 possibilities
- 2+1, 2+2, 2+3, 2+4, 2+5 = 5 possibilities
- 3+1, 3+2, 3+3, 3+4 = 4 possibilities
- 4+1, 4+2, 4+3 = 3 possibilities
- 5+1, 5+2 = 2 possibilities
- 6+1 = 1 possibility
- Total number of ways to get a sum less than eight: 6 + 5 + 4 + 3 + 2 + 1 = 21
Step 2: Determine the total number of possible outcomes when two fair dice are rolled:
- For a standard pair of dice, each die has six faces, so there are 6 × 6 = 36 possible outcomes.
Step 3: Calculate the probability:
- Probability = Number of favorable outcomes / Total number of possible outcomes
- Probability = (Number of outcomes with an even or odd sum + Number of outcomes with a sum less than eight) / Total number of possible outcomes
- Probability = (6 + 21) / 36
- Probability = 27 / 36
- Probability = 3 / 4
Therefore, the probability of getting a total showing that is either even or less than eight when rolling two fair dice is 3/4 or 75%.