Answer :
Men = 6
Women = 5
Total number = 6+5 = 11
First question:
This is a question of combination.
Total number of selecting 3 people from the group = 11C3 = 11!/[3!(11-3)!]= 165 ways
Second question:
This is question of combination where only men are considered.
Total number of collecting 3 men from 6 men = 6C3 = 6!/[3!*(6-3)!]= 20 ways
Third question:
The probability that the 3 people selected will all be men is given by:
1st selection: 6/11
2nd selection: 5/10
3rd selection: 4/9
The probability = 6/11*5/10*4/9 = 4/33
Women = 5
Total number = 6+5 = 11
First question:
This is a question of combination.
Total number of selecting 3 people from the group = 11C3 = 11!/[3!(11-3)!]= 165 ways
Second question:
This is question of combination where only men are considered.
Total number of collecting 3 men from 6 men = 6C3 = 6!/[3!*(6-3)!]= 20 ways
Third question:
The probability that the 3 people selected will all be men is given by:
1st selection: 6/11
2nd selection: 5/10
3rd selection: 4/9
The probability = 6/11*5/10*4/9 = 4/33
1.
[tex] \displaystyle
\binom{11}{3}=\dfrac{11!}{3!8!}=\dfrac{9\cdot10\cdot11}{2\cdot3}=165 [/tex]
2.
[tex] \displaystyle\binom{6}{3}=\dfrac{6!}{3!3!}=\dfrac{4\cdot5\cdot6}{2\cdot3}=20 [/tex]
3.
[tex] |\Omega|=165\\
|A|=20\\\\
P(A)=\dfrac{20}{165}=\dfrac{4}{33}\approx12\% [/tex]