# Probability of events, given independence and mutual exclusivity

I know $$P(A) = 0.5$$, $$P(B) = 0.2$$, and $$P(C) = 0.6$$

A and C are independent, and B and C are mutually exclusive.

I am interested in the following four quantities:

1. $$P(A \cap B \cap C)$$
2. $$P({A \cup B} | C)$$
3. $$P(A \setminus C)$$
4. $$P(A \cup B \cup C^{c}$$)

I know the values are respectively, 0, 0.5, 0.2, 0.7. But how do I find these given the information I know?

• Let's look at the first. $B$ and $C$ are mutually exclusive so... what doe sthat mean about $B\cap C$? What does that mean about $A\cap B\cap C = A\cap (B\cap C)$? – JMoravitz Oct 6 '19 at 16:26
• Since B, C are mutually exclusive, then $B \cap C$ is the empty set, so $P(A \cap \emptyset) = 0$. Thanks! – user581882 Oct 6 '19 at 16:38
• Good, now, $P(A\setminus C) = P(A)-P(A\cap C)$, what do you know about $P(A\cap C)$ since $A$ and $C$ are independent? – JMoravitz Oct 6 '19 at 16:43
• You seem to be doing fine then. What specific difficulties are having with the remaining two? Just confidence? – JMoravitz Oct 6 '19 at 16:48
• $P(A\cup B\mid C) = \dfrac{P((A\cup B)\cap C)}{P(C)}=\dfrac{P((A\cap C)\cup (B\cap C)}{P(C)}=\dots$ – JMoravitz Oct 6 '19 at 16:54

1. $$P(A\cap B\cap C)=P(A\cap (B\cap C)) = P(A\cap \phi)=P(\phi)=0$$
2. $$P(A\cup B | C) = P(A|C) + P(B|C)-P(A\cap B|C)=P(A|C)=P(A)=0.5$$ The first transition follows from the fact that $$P(\cdot|C)$$ is a probability function and regular rules hold, the second follows from the fact that $$B,C$$ and $$A\cap B,C$$ are mutually exclusive.
3. $$P(A \backslash C) = P(A)-P(A\cap C)=P(A)-P(A)P(C)=0.5-0.5\times0.6=0.2$$
4. $$P(A\cup B\cup \overline{C})=P(A)+P(B\cup \overline{C})- P(A\cap(B\cup \overline{C}))$$ $$C$$ and $$B$$ are mutually exculsive, that means that for every $$\omega\in B:\omega \notin C\rightarrow\omega\in \overline{C}$$, so $$B\subseteq \overline{C}$$.
$$P(A\cup B\cup\overline{C})=P(A)+P(\overline{C})-(A\cap \overline{C})=P(A)+P(\overline{C})-P(A)P(\overline{C})=0.5+0.4-0.5\times 0.4=0.7$$