Assume that there are 18 board
members: 12 females, and 6 males including Larry.
There are 3 tasks to be assigned. Note that assigning the same
people different tasks constitutes a different assignment.
Find the probability that both males and females are given a task.
Find the probability that Larry and at least one female
are given tasks.
To find the probability in both scenarios, we need to calculate the total number of possible assignments and the number of favorable assignments that meet the given conditions.
1. Probability that both males and females are given a task:
Step 1: Find the total number of possible assignments for the 3 tasks among the 18 board members.
The total number of possible assignments can be calculated using combinations. We select 3 members out of 18 without considering order. This can be denoted as C(18, 3), which equals 18! / (3! × (18-3)!) = 816.
Step 2: Find the number of favorable assignments where both males and females are given a task.
To calculate this, we need to consider that there are 6 males and 12 females. We select 1 male out of 6 and 2 females out of 12. This can be denoted as C(6, 1) × C(12, 2), which equals 6! / (1! × (6-1)!) × 12! / (2! × (12-2)!) = 6 × 66 = 396.
Step 3: Calculate the probability.
The probability is the ratio of favorable assignments to the total number of possible assignments. So, P(both males and females assigned) = 396 / 816 ≈ 0.485.
2. Probability that Larry and at least one female are given tasks:
Step 1: Find the total number of possible assignments for the 3 tasks among the 18 board members (same as before), which is 816.
Step 2: Find the number of favorable assignments where Larry and at least one female are given tasks.
To calculate this, we need to consider that there are 6 males (including Larry) and 12 females. We select Larry and at least 1 female for the tasks. This can be denoted as C(6, 1) × C(12, 1) + C(6, 1) × C(12, 2) + C(6, 1) × C(12, 3), which equals 6! / (1! × (6-1)!) × 12! / (1! × (12-1)!) + 6! / (1! × (6-1)!) × 12! / (2! × (12-2)!) + 6! / (1! × (6-1)!) × 12! / (3! × (12-3)!) = 6 × 12 + 6 × 66 + 6 × 220 = 432 + 396 + 1320 = 2148.
Step 3: Calculate the probability.
The probability is the ratio of favorable assignments to the total number of possible assignments. So, P(Larry and at least one female assigned) = 2148 / 816 ≈ 2.63.