SOLUTION: Discipline | Male | Female | Total ----------- ----------- ----------- --� Mathematics | 6 | 4 | 10 | ----------- ----------- ----------- --� Science | 5 | 2 | 7

SOLUTION: Discipline | Male | Female | Total ----------- ----------- ----------- --� Mathematics | 6 | 4 | 10 | ----------- ----------- ----------- --� Science | 5 | 2 | 7 | -----------

Question 358426: Discipline | Male | Female | Total
--------------------------------------�
Mathematics | 6 | 4 | 10 |
--------------------------------------�
Science | 5 | 2 | 7 |
--------------------------------------�
Engineering | 1 | 0 | 1 |
--------------------------------------�
Total | 12 | 6 | 18 |
--------------------------------------�
If four faculty members are randomly chosen, what is the probability that you choose four men or four mathematics instructors?
Found 2 solutions by Jk22, jrfrunner:
Answer by Jk22(389) (Show Source): You can put this solution on YOUR website!
We pick up the number of people in either cases :

p=(math(male)+math(female)+sci(male)+eng(male))/36=(4+6+5+1)/36=16/36=8/18=4/9

or we take the sum of people in math and male, less people both, since this would be counted 2 times :

p=(math+male-math&male)/36=(10+12-6)/36=16/36
Answer by jrfrunner(365) (Show Source): You can put this solution on YOUR website!
If four faculty members are randomly chosen, what is the probability that you choose four men or four mathematics instructors?
==
Let A=Event of selecting four men
Let B=Event of selecting four math instructors
==
basics of probabilities
P(AUB)=P(A)+P(B)-P(AnB)
--
selecting 4 men: P(A)=12C4*6C0/18C4
explanation: There are 18C4 of selecting 4 individuals from the 18
of these, there are 12C4 of selecting the 4 men and 6C0
of selecting 0 women
P(A)=495*1/3060=0.162
--
selecting 4 math instructors P(B)=10C4*8C0/18C4
explanation: There are 18C4 of selecting 4 individuals from the 18
of these, there are 10C4 of selecting the 4 math instructors
and 8C0 of selecting 0 non-math instructors
P(B)=210*1/3060=0.069
--
selecting 4 who are both male and math instructors
P(AnB)=6C4*12C0/18C4
explanation: There are 18C4 ways of selecting 4 individuals from the 18
There are 6C4 of selecting 4 of the 6 who are both male/math
and 12C0 of selecting 0 who are the others.
P(AnB)=15/3060=0.005
--
answer P(AUB)=P(A)+P(B)-P(AnB)=0.162+0.069-0.005=0.226

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