Roll four dice one time. What is the probability that the sum will be odd?Can anyone explain/show in detail how to find the answer to this problem?
To find the probability of rolling four dice and getting an odd sum, we need to determine the total number of outcomes where the sum is odd and divide it by the total number of possible outcomes.
Step 1: Determine the total number of possible outcomes:
When rolling four dice, each die can have six possible outcomes (1, 2, 3, 4, 5, or 6). Since there are four dice, the total number of possible outcomes is 6 x 6 x 6 x 6 = 1296.
Step 2: Determine the number of outcomes where the sum is odd:
For the sum to be odd, we need an odd number of odd outcomes. There are three possibilities for the number of odd outcomes: 1 odd, 3 odds, or all 4 odds.
- Case 1: 1 odd outcome
There are four dice, so we choose one of them to be odd, which can be done in C(4,1) = 4 ways. For each odd outcome, there are three possibilities (1, 3, or 5) and three possibilities for each of the remaining three dice (even numbers: 2, 4, or 6). So, Case 1 has a total of 4 x 3^1 x 3^3 = 324 outcomes.
- Case 2: 3 odd outcomes
There are four dice, but this time we choose three of them to be odd, which can be done in C(4,3) = 4 ways. For each odd outcome, there are three possibilities, and for each even outcome, there are three possibilities. So, Case 2 has a total of 4 x 3^3 x 3^1 = 324 outcomes.
- Case 3: 4 odd outcomes
This case is straightforward. Since all the dice need to have odd outcomes, each die has three possibilities. So, Case 3 has a total of 3^4 = 81 outcomes.
Step 3: Calculate the probability:
Add up the number of outcomes for each case: 324 + 324 + 81 = 729
Now, divide the number of outcomes where the sum is odd (729) by the total number of possible outcomes (1296):
Probability = 729 / 1296
Simplifying the fraction, we get:
Probability = 9 / 16
Therefore, the probability of rolling four dice and getting an odd sum is 9/16.