If 2 different representatives are to be selected at random from a group of 10 employees
and if p is the probability that both representatives selected will be women, is p >0.5
(1) More than 50% of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 10%
.
probability
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- thephoenix
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- thephoenix
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Ststement 1. More than 50% of the 10 employees are women.
Possible options for the 10 members are
M W
4 6
3 7
2 8
1 9
for 4 men and 6 women prob that both are women is 6/10 * 5/9 = 5/30. this is <0.5
for 2 men and 8 women prob ...... is 8/10 * 7/9 = 28/45 this is >0.5
St.1 insuff.
Statement 2. The probability that both representatives selected will be men is less than 10%
take same options ( you will have to try with men greater than women too)
for 3 men and 7 women prob that both are men is 3/10 * 2/9 = 1/15 this is < 10%
and prob that both are women is 7/10 * 6/9 = 7/15 , this is < 0.5
for 2 men and 8 women prob that both are men is 2/10 * 1/9 = 1/45 this is < 10%
but prob that both are women is 8/10 * 7/9 = 28/45 this is >0.5
so St 2 also insufficient.
Combining both cases will result in same set of sols as case 2 as we have only taken options where the women are greater in no than men.
I hope you understand my explanations.
Possible options for the 10 members are
M W
4 6
3 7
2 8
1 9
for 4 men and 6 women prob that both are women is 6/10 * 5/9 = 5/30. this is <0.5
for 2 men and 8 women prob ...... is 8/10 * 7/9 = 28/45 this is >0.5
St.1 insuff.
Statement 2. The probability that both representatives selected will be men is less than 10%
take same options ( you will have to try with men greater than women too)
for 3 men and 7 women prob that both are men is 3/10 * 2/9 = 1/15 this is < 10%
and prob that both are women is 7/10 * 6/9 = 7/15 , this is < 0.5
for 2 men and 8 women prob that both are men is 2/10 * 1/9 = 1/45 this is < 10%
but prob that both are women is 8/10 * 7/9 = 28/45 this is >0.5
so St 2 also insufficient.
Combining both cases will result in same set of sols as case 2 as we have only taken options where the women are greater in no than men.
I hope you understand my explanations.
- thephoenix
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thsnks !!!!trindadesn wrote:Ststement 1. More than 50% of the 10 employees are women.
Possible options for the 10 members are
M W
4 6
3 7
2 8
1 9
for 4 men and 6 women prob that both are women is 6/10 * 5/9 = 5/30. this is <0.5
for 2 men and 8 women prob ...... is 8/10 * 7/9 = 28/45 this is >0.5
St.1 insuff.
Statement 2. The probability that both representatives selected will be men is less than 10%
take same options ( you will have to try with men greater than women too)
for 3 men and 7 women prob that both are men is 3/10 * 2/9 = 1/15 this is < 10%
and prob that both are women is 7/10 * 6/9 = 7/15 , this is < 0.5
for 2 men and 8 women prob that both are men is 2/10 * 1/9 = 1/45 this is < 10%
but prob that both are women is 8/10 * 7/9 = 28/45 this is >0.5
so St 2 also insufficient.
Combining both cases will result in same set of sols as case 2 as we have only taken options where the women are greater in no than men.
I hope you understand my explanations.
is this the only way to solve ??