Assume that there are 14 board members: 9 females, and 5 males including Carl. There are 4 tasks to be assigned. Note that assigning the same people different tasks constitutes a different assignment. Find the probability that Carl and at least one female are given tasks
To find the probability that Carl and at least one female are given tasks, we need to calculate the number of favorable outcomes and the total number of possible outcomes.
First, let's find the number of favorable outcomes. Since Carl is already assigned one of the tasks, we have 3 remaining tasks to assign to the rest of the board members. There are 13 people to choose from (excluding Carl), out of which 9 are females. So, the number of ways to assign the remaining tasks to the females is 9C3.
Next, let's find the total number of possible outcomes. For each of the remaining 3 tasks, we have 13 people to choose from (excluding Carl). So, the total number of ways to assign these tasks is 13C3.
Therefore, the probability that Carl and at least one female are given tasks is:
P = Number of favorable outcomes / Total number of possible outcomes
P = (9C3) / (13C3)
Now, we can calculate this probability:
P = (9! / (3! * 6!)) / (13! / (3! * 10!))
P = (9 * 8 * 7) / (13 * 12 * 11)
P = 504 / 1716
P ≈ 0.2937
Therefore, the probability that Carl and at least one female are given tasks is approximately 0.2937.