Assume that there are 15 board members: 9 females, and 6 males including Carl. There are 3 tasks to be assigned. Note that assigning the same people different tasks constitutes a different assignment.
(2) Find the probability that Carl and at least one female are given tasks.
p(Carl) = 1/15
assuming Carl is male, that still leaves 9 females out of the 14 people left.
So, P(Carl,female) = 1/15 * 9/14
And there 3P2 ways that she and Carl can be assigned tasks.
To find the probability that Carl and at least one female are given tasks, we need to find the total number of possible assignments where Carl and at least one female are assigned tasks, and then divide it by the total number of possible assignments.
Let's first find the total number of possible assignments where Carl and at least one female are assigned tasks.
To assign tasks, we will use the concept of combinations.
For the first task, Carl can be assigned, and there are 14 remaining board members to choose from (9 females and 5 males).
For the second task, we can choose any two people from the remaining 14 board members (including females and males).
So, the total number of possible assignments where Carl and at least one female are assigned tasks is:
1 * C(14, 2) = 1 * (14! / (2! * (14-2)!)) = 1 * (14! / (2! * 12!)) = (14 * 13) / (2 * 1) = 91
Now, let's find the total number of possible assignments.
For the first task, there are 15 board members to choose from.
For the second task, there are 14 remaining board members to choose from.
For the third task, there are 13 remaining board members to choose from.
So, the total number of possible assignments is:
15 * 14 * 13 = 2,730
Finally, the probability that Carl and at least one female are given tasks is:
91 / 2,730 = 0.0334 (rounded to four decimal places)
Therefore, the probability is approximately 0.0334 or 3.34%.
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 number of total outcomes.
First, let's calculate the number of favorable outcomes, which is the number of ways Carl and at least one female can be assigned tasks.
Since Carl is one of the board members, he must be assigned one of the tasks. Therefore, we can assign the remaining two tasks among the remaining 14 board members (8 females and 6 males excluding Carl). The number of ways to do this is given by:
C(14, 2) = 14! / (2!(14-2)!) = 91
Next, let's calculate the number of total outcomes, which is the total number of ways the 3 tasks can be assigned among the 15 board members. This can be calculated as:
C(15, 3) = 15! / (3!(15-3)!) = 455
Now, we can calculate the probability by dividing the number of favorable outcomes by the number of total outcomes:
P(Carl and at least one female) = Favorable outcomes / Total outcomes
= 91 / 455
= 1 / 5
Therefore, the probability that Carl and at least one female are given tasks is 1/5.