Probability Question

Dec 2009
61
0
A group of eight is to be chosen at random from eight men and six women.
What is the probability that
i. the group contains the same number of men and women?
ii. the group contains at most one woman?
iii. the group contains the oldest person but not the youngest person?
 
Dec 2009
61
0
Well heres what I did
(8 choose 4) x (6 choose 4) = 70 x 15 = 1050
(14 choose 8) = 3003

so the probability = 1050/3003.
I dont know if thats right though
 
Dec 2009
3,120
1,342
Well heres what I did
(8 choose 4) x (6 choose 4) = 70 x 15 = 1050
(14 choose 8) = 3003

so the probability = 1050/3003.
I dont know if thats right though
Yes, that's fine since any group of 4 men can go with any group of 4 women,
which is why the selections are multiplied.

(ii)

At most 1 woman is 1 woman or no women,
which means 1 woman and 7 men or all 8 men.

\(\displaystyle P=\frac{\binom{8}{7}\binom{6}{1}+\binom{8}{8}\binom{6}{0}}{\binom{14}{8}}\)

(iii)

If the group contains the oldest person and excludes the youngest person,
then the oldest person can go with 7 of the remaining 12.

The number of ways this can happen is \(\displaystyle \binom{12}{7}\)

hence \(\displaystyle P=\frac{\binom{12}{7}}{\binom{14}{8}}\)