The waiting room for jury selection has 40 women and 30 men to choose from. A jury of 12 is selected from these 70 people, what is the probability that exactly 5 of the jury members will be women ?
I tried to use binomial probability to work it out. but it doesn't work because of the probability of success in one trial is not the same in all those 5 women.
How can I work it out?
Thanks!
Probability question?
2009-05-19 2:57 pm
回答 (3)
2009-05-19 3:26 pm
✔ 最佳答案
5 women out of 40 may be chosen in 40C5 ways. The other 7 men of the jury may be chosen from 30 men in 30C7 ways. Therefore, 5 women (and 7 men) may be chosen in (40C5)(30C7) ways.The total number of ways of choosing 12 people out of 70 is 70C12.
P(5 women) = P(5 women, 7 men) = (40C5)(30C7) / 70C12 ----(1)
40C5 =658,008
30C7 =2,035,800
70C12 =10,638,894,058,520
You can simplify it if you want. the answer is ----(1)
This is an example of Hypergeometric distribution (not Binomial)
2009-05-19 3:10 pm
12/70 = jury
40/70 = women
probability all women = 480/4900
probability all men = 360/4900
suspect the answer is 380/4900 = 7.755%
40/70 = women
probability all women = 480/4900
probability all men = 360/4900
suspect the answer is 380/4900 = 7.755%
2009-05-19 3:10 pm
You are right that binomial is not appropriate.
Total number of all selections without restriction is 70C12.
Number of selections 5W + 7M = 40C5*30C7
Probability of getting 5W, 7M = 40C5*30C7/70C12
= (40!/5!*35!)*(30!/7!*23!)/(70!/12!*58!) = 0.126 (3 dec. pl.)
Total number of all selections without restriction is 70C12.
Number of selections 5W + 7M = 40C5*30C7
Probability of getting 5W, 7M = 40C5*30C7/70C12
= (40!/5!*35!)*(30!/7!*23!)/(70!/12!*58!) = 0.126 (3 dec. pl.)
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