There are 9 candidates total, 5 male and 4 female, who run for 4 committee positions. Suppose that each person is equally likely
to be elected. What is the probability that 2 males and 2 females are elected?
1 answer:
Use the hypergeometric distribution.
M=number of Men=5
F=number of women=4
m=number of men elected=2
f=number ow women elected=2.
Assuming equal chance to get elected, then
P(2M,2F)=C(M,m)*C(F,f)/C(M+F,m+f)
=C(5,2)*C(4,2)/C(9,4)
=10*6/126
=10/21
Reference: Hypergeometric distribution.
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$15 thanks hope this helps
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j would be 130 cuz sides that are congruent have congruent angles so you add those two angles up and subtract from 180
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432cm^3
Step-by-step explanation:
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B
Step-by-step explanation:
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