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Numerical Ability
Probability
In how many ways a team of 11 must be selected from 5 men and 11 women such that the team must comprise of not more than 3 men?
Read Solution (Total 10)
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- (11c11)+(5c1*11c10)+(5c2*11c9)+(5c3*11c8)
- 10 years agoHelpfull: Yes(18) No(2)
- (11c11)+(5c1*11c10)+(5c2*11c9)+(5c3*11c8)=2256 ways
- 10 years agoHelpfull: Yes(11) No(1)
- 5C1*11C8+5C2*11C9+5C1*11C10+5C0*11C11
=2256 - 10 years agoHelpfull: Yes(6) No(1)
- let w and m denotes the probability of selecting women and men.
if only women are selected then P(w)=11C11/16C11
if 1 man and 10 women are selected then P(w)*P(m)=(5C1*11C10)/16C11
if 2 men and 9 women are selected then P(w)*P(m)=(5C2*11C9)/16C11
if 3 men and 8 women are selected then p(w)*P(m)=(5C3*11C8))/16C11
total probability=(11C11 + 11C10*5C1 + 11C9*5C2 + 11C8*5C3)/16C11
=47/91.
- 10 years agoHelpfull: Yes(5) No(0)
- 47/91
(11C11 + 11C10*5C1 + 11C9*5C2 + 11C8*5C3)/16C11 - 10 years agoHelpfull: Yes(3) No(0)
- 5c3*11c8 + 5c2*11c9 + 5c1*11c10
- 10 years agoHelpfull: Yes(1) No(1)
- 11C11 + 11C10*5C1+11C9*5C2 + 11C8*5C3
- 10 years agoHelpfull: Yes(1) No(1)
- (11c11)+(5c1*11c10)+(5c2*11c9)+(5c3*11c8)
1+55+825+150
1030 ans - 10 years agoHelpfull: Yes(0) No(4)
- (11c11)+(5c1*11c10)+(5c2+11c9)+(5c3+11c8)
- 10 years agoHelpfull: Yes(0) No(0)
- (11c11)+(5c1*11c10)+(5c2+11c9)+(5c3+11c8)
- 10 years agoHelpfull: Yes(0) No(1)
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