probs: help!

[demoskopico]

I know the answer intuitively but I am not able to follow the right formula, please help me!

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

[kevin]

Suppose that there are n women (and thus 10 - n men).

How can we write (1) in terms of n?

[demoskopico]

I know the answer intuitively but I am not able to follow the right formula, please help me!

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

nC2/10C2>1/2?

1. n>5. let's try with 6: 6C2/10C2=1/3, thus p>1/3 insuff
2. (10-n)C2/10C2<1/10, thus (10-n)C2/45<1/10, thus (10-n)C2<4.5, so 10-n can be 3 or less, which means that there are less than 3 men and more than 7 women.
7C2/10C2 = 21/45 = 7/15 which is < 1/2


E doesn't work as well....I would have liked to solve it with the formula

N=NCM*P^m * q^n-m. can it work? how?

[kevin]

nC2/10C2>1/2?

1. n>5. let's try with 6: 6C2/10C2=1/3, thus p>1/3 insuff
2. (10-n)C2/10C2<1/10, thus (10-n)C2/45<1/10, thus (10-n)C2<4.5, so 10-n can be 3 or less, which means that there are less than 3 men and more than 7 women.
7C2/10C2 = 21/45 = 7/15 which is < 1/2


E doesn't work as well....I would have liked to solve it with the formula

N=NCM*P^m * q^n-m. can it work? how?

Isn't this formula for independent trials?

[demoskopico]

Isn't this formula for independent trials?



mm, yes .... I have to work hard to improve probabilities, perms and combs. In 4 months I have improved only a little bit