Help! There are 5 black males and 6 black females. There are 8 white males and 4 white females. Suppose you choose 3 randomly. What is the probability that you have atleast a black and atleast a female within the 3?

Question

anonymous · Answer

@mathmate @jim_thompson5910 @completeidiot

anonymous · Answer

n(S)=C(23,3)=1771
$$E=B \cap F \ E'=B' \cup F'$$ $$n(B)=C(11,3)=165 \ n(F)=C(10,3)=120$$ $$n(B \cap F)=C(6,3)=20 \ n(E')=C(11,3)+C(10,3)-C(6,3)=265 \ n(E)=n(S)-n(E')=1771-265=1506$$

That's what I have, but it doesn't seem right!

mathmate · Answer

Have you done Binomial distribution?

anonymous · Answer

I don't believe so unless they have a different name, but if they would help solve this problem, then I'm up for learning about them.

mathmate · Answer

I would separate the problem into two, one for M/F, and the other one B/W.
Calculate P(F) for 3 picks, and P(B) for 3 picks.
Then we can find $ P(F) \cup P(B) $.

anonymous · Answer

Wouldn't it be the intersection because you need one of each and not just one which you would have in a union?

mathmate · Answer

You're right, my bad!

mathmate · Answer

In a binomial experiment with 3 trials, 
N=3 (trials)
p=10/23 (female)
q=1-p=13/23 (male)
P(all male)=C(3,3)p^0q^3=1*(10/23)^0*(13/23)^3=.1806
P(at least one female)=1-.1806=.8194

Similarly for B/W.

anonymous · Answer

Thanks!

mathmate · Answer

You're welcome!  :)