Two fair dice are thrown. What is the probability:
* at least one shows a 3, given the sum is a 7
*the sum is a 7, given at least one shows a 3
To find the probabilities in these scenarios, we need to understand the total number of possible outcomes and the number of favorable outcomes in each case.
1. At least one shows a 3, given the sum is a 7:
Step 1: Find the total number of possible outcomes for throwing two fair dice. Since each die has 6 sides, there are a total of 6 * 6 = 36 possible outcomes when two dice are thrown.
Step 2: Determine the number of favorable outcomes, i.e., when at least one die shows a 3 and the sum is 7. There are three ways to get a sum of 7: (1, 6), (2, 5), and (3, 4). Out of these three possible outcomes, two of them have at least one die showing a 3: (3, 4) and (1, 6).
Step 3: Calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes. So the probability of at least one die showing a 3, given the sum is 7, is 2/36 = 1/18.
2. The sum is a 7, given at least one shows a 3:
Step 1: Determine the total number of possible outcomes, which is the same as before, 36.
Step 2: Find the number of favorable outcomes, i.e., the instances when the sum is 7, given that at least one die shows a 3. We can see that there are only two possible outcomes: (3, 4) and (4, 3). Both of these outcomes have a sum of 7.
Step 3: Calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes. Thus, the probability of the sum being a 7, given that at least one die shows a 3, is 2/36 = 1/18.
Therefore, the probabilities in both cases are equal: 1/18.