# combinations and permutations

• Jun 1st 2010, 03:32 PM
knicksfan426
combinations and permutations
There are 5 women and 7 men who are reporters for a magazine. There are 6 articles that need to be written. If 6 reporters are chosen at random to write the articles, what is the probability that 3 are men and 3 are women?

I got 40,000/117,649. I'm doubting my answer because its such a huge fraction. Please let me know what you guys get and explain!!
• Jun 1st 2010, 03:34 PM
knicksfan426
Also, how do u do this?

A family with 5 children is selected at random. What is the probability that the family has atleast 3 boys?
• Jun 1st 2010, 03:44 PM
pickslides
Are you sure this is trigonometry?

Quote:

Originally Posted by knicksfan426

I got 40,000/117,649. I'm doubting my answer because its such a huge fraction. Please let me know what you guys get and explain!!

Quote:

Originally Posted by knicksfan426
There are 5 women and 7 men who are reporters for a magazine. There are 6 articles that need to be written. If 6 reporters are chosen at random to write the articles, what is the probability that 3 are men and 3 are women?

I was thinking $\frac{^7C_3+^5C_3}{^{12}C_6}\approx 0.049$
• Jun 1st 2010, 03:49 PM
pickslides
Quote:

Originally Posted by knicksfan426
Also, how do u do this?

A family with 5 children is selected at random. What is the probability that the family has atleast 3 boys?

It is assumed the probability of having a boy is equal to the probability of having a girl i.e. 0.5?

If so use X~ $\text{Bi(5,0.5)}$ and find $P(X\geq 3) = P(X=3)+P(X=4)+P(X=5)$