# Probability, conditional probability

m. Let $P(A) = 0.8$, $P(B) = 0.5$, and $P(A \cap B) = 0.4$.

Find $P(A^c \cap B^c)$ and $P(A^c \cup B^c)$

My answer: $P(A^c \cap B^c): 0.4$, $P(A^c \cup B^c): 0.8 + 0.5=1.3$

n. A box contains four $10$ dollar bills, six $5$ dollar bills, and two $1$ dollar bills. Two bills are taken at random from the box without replacement. What is the probability that both bills will be of the same denomination?

My answer: C$(12,2)$

o. A department has $7$ men and $5$ women on its faculty. What is the probability that women will outnumber men on a randomly selected ﬁve-member committee?

p. Of the $120$ students in a class, $30$ speak Chinese, $50$ speak Spanish, $75$ speak French, $12$ speak Spanish and Chinese, $30$ speak Spanish and French, and $15$ speak Chinese and French. Seven students speak all three languages. What is the probability that a randomly chosen student speaks none of these languages?

Any help? I want to know if I am headed in the right direction with these. How do I determine if women will outnumber men? It's a combination right?

For p, how do I do this one? Thank you

(m.) De Morgan's Laws are applied: $P(A^\complement\cup B^\complement) =1-P(A\cap B)\\P(A^\complement\cap B^\complement) = 1-P(A\cup B)$
(n.) $C(12,2)$ is not a probability measure.   It is the count of ways to select two items from twelve items.   That's just the size of the probability space.   You want the probability of selecting either two of four $\$10$, two of six$\$5$, or two of two $\$1\$ out of all the ways to select any two of twelve items. $$\frac{?}{^{12}C_2}$$
$$\lvert (C\cup F\cup S)^\complement\rvert = \lvert \Omega\rvert - \lvert C\rvert - \lvert F\rvert - \lvert S\rvert+ \lvert C\cap F\rvert+\lvert C\cap S\rvert+\lvert F\cap S\rvert - \lvert C\cap F\cap S\rvert$$