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 > 1/2?
(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10
OA:
E
All of the solutions that I've found used combinations. Can somebody help me with this problem using only probability. Here's my solution, and why I think it's B. Please let me know where I went wrong.
Statement 1, I know why it's wrong.
For statement 2,
The only way that the probability of both men is: 2/10 * 1/9 = 1/45 < 1/10
That means that there will be 8 women, so 8/10 * 7/9 = 56/90 > 1/2. Sufficient.
(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10
OA:
E
All of the solutions that I've found used combinations. Can somebody help me with this problem using only probability. Here's my solution, and why I think it's B. Please let me know where I went wrong.
Statement 1, I know why it's wrong.
For statement 2,
The only way that the probability of both men is: 2/10 * 1/9 = 1/45 < 1/10
That means that there will be 8 women, so 8/10 * 7/9 = 56/90 > 1/2. Sufficient.
