sanaa.rizwan wrote:If 2 different representatives are to be selected at random from group of 10 employees and if p is the probability that both representatives selected will be women, is p > ½

1. more than ½ of the 10 employees are women
2. the probability that both representatives selected will be men is less than 1/10

Possible number of way to select 2 out of 10 = 10C2 = 45

Statement 1: More than 1/2 of the 10 employees are women.
Number of can be 6, 7, 8, 9 or 10.

• For number of women = 6, Possible number of selecting two women out of 6 = 6C2 = 15. Thus probability = 15/45 = 1/3 < 1/2
  
  For number of women = 9, Possible number of selecting two women out of 9 = 9C2 = 36. Thus probability = 36/45 = 4/5 > 1/2

Not sufficient.

Statement 2: The probability that both representatives selected will be men is less than 0.1. This means (Probability that both representative will be women + Probability that one of the representative will be man and other woman) > (1 - 0.1) = 0.9

But we don't have any idea about "Probability that one of the representative will be man and other woman".

Not sufficient.

1 & 2 Together: No relevant new information.

Not sufficient.

The correct answer is E.