My professor gave our class a study guide for our midterm. On that study guide, there were 10 prompts he wanted us to know. Out of the 10, there were 3 prompts that I wasn't confident in. For the midterm, he picked 5 prompts out of the 10. Out of those 5 prompts, 3 of the prompts I wasn't confident managed to be on there. I want to know the probability of that happening.
If you could show the steps to solving it, that would be great.
I got 30%
To find the probability of the three prompts you weren't confident in being chosen for the midterm, you can use the concept of combinations.
Step 1: Determine the total number of ways the professor can choose 5 prompts out of the 10. This can be calculated using the combination formula, denoted as C(n, r), which represents the number of ways to choose r items from a set of n items without regard to their order. In this case, n = 10 (total number of prompts) and r = 5 (number of prompts the professor chooses). So, the total number of ways the prompts can be chosen is C(10, 5).
Step 2: Calculate the number of favorable outcomes, which means the number of ways the three prompts you weren't confident in can be chosen out of the 10 prompts. Since we know that the professor included those 3 prompts in the midterm, we need to calculate C(3, 3).
Step 3: Calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes. So, the probability is C(3, 3) / C(10, 5).
To simplify the calculations, you can use the principles of combinations:
C(10, 5) = 10! / (5! * (10-5)!) = (10 * 9 * 8 * 7 * 6) / (5 * 4 * 3 * 2 * 1) = 252
C(3, 3) = 3! / (3! * (3-3)!) = 1
Therefore, the probability is 1/252 or approximately 0.00397, which means there is a 0.397% chance of the three prompts you weren't confident in being chosen for the midterm.