Assume that there are 12 board members: 8 females, and 4 males including Carl. 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.
ind the probability that Carl and at least one female are given tasks.
2006-09-18 17:59:43 · 2 answers · asked by J R 2 in Science & Mathematics ➔ Mathematics
Calculate the probabilities that only males or only females are given the task; subtract their sum from one.
Total possibilities: 12P3 = 12 * 11 * 10 = 1320;
Only males: 4P3 = 4 * 3 * 2 = 24;
Only females: 8P3 = 8 * 7 * 6 = 336;
One gender: 24 + 336 = 360;
Mixed: 1320 - 360 = 960.
Probability: 960 / 1320 = 8 / 11 (72.73%)
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The number of ways to do this is
8 * 7 / 2 (two females) +
+ 8 * 3 (other male & female)
= 36 + 24 = 60.
Probability: 60 / 1320 = 1/22 (4.55%)
2006-09-18 18:16:38 · answer #1 · answered by dutch_prof 4 · 0⤊ 0⤋
2 females and 1 male would be in each group if you want it equally seperated between all groups.otherwise i dont know?
2006-09-19 01:03:37 · answer #2 · answered by dcjohnson5 2 · 0⤊ 0⤋