Intersection of three events

Question

I am working on a probability problem involving three events, A B and C, I am given Pr(A), Pr(B), Pr(C), Pr(A), Pr(A $\bigcup$ B), Pr(A $\bigcup$ C),Pr(B $\bigcup$ C) and we are asked to find

Pr($A^c$ $\bigcap$ $B^c$ $\bigcap$ C)

I am not sure if this is correct, but this is what I am doing thus far...

(($A^c$ $\bigcap$ $B^c$) $\bigcap$ C) >> (($A^c$ $\bigcap$ $B^c$)$^c$ $\bigcap$ C) >> ((A $\bigcup$ B) $\bigcap$ C)

Then by distributing we get (A $\bigcap$ C) $\bigcup$ (A $\bigcap$ B)

I am not sure if I am doing this right or not, any thoughts?

Answer 1

$A^c \cap B^c \cap C = (A \cup B \cup C^c)^c$. So $\mathbb P(A^c \cap B^c \cap C) = 1 - \mathbb P(A \cup B \cup C^c)$.

Then expand $\mathbb P(A \cup B \cup C^c) = \mathbb P(A) + \mathbb P(B) + \mathbb P(C^c) - \mathbb P(A \cap B) - \mathbb P(A \cap C^c) - \mathbb P(B \cap C^c) + \mathbb P(A \cap B \cap C^c)$.

From this, you should be able to finish.

Answer 2

Your idea was good, but you have something wrong.

Just need to remember that:

• $\mathbb{P}((something)^{c}) = 1 - \mathbb{P}(something)$.
• Probability of union of $A$,$B$ and $C$ is the same as sum of probabilities for individual $A$,$B$ and $C$.
  But this is only truth if $A$,$B$,$C$ do not have elements in common (because if they had, you'd be counting those elements twice). So you can say $P(A \cup B \cup C) = P(A) + P(B) + P(C)$ for any $A,B,C$ if you subtract the intersections between every combination of $A$,$B$ and $C$

Explanation of the second point? Two roads.

First, if A,B and C are disjoint (no elements in common), then

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) - P(A\cap B \cap C)$ would reduce to

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - 0 -0 -0 -0 -0$ (probability of empty set is 0)

Second, if A,B,C are not disjoint (some elements in common), then, when we say $P(A) + P(B)$, we are talking about some elements that are in $A$ but also in $B$, so taking the sum means we count them twice, so we subtract $P(A\cap B)$ (which are the repeated elements) to 'make it even'. You need to do this for every two different sets in $A \cup B \cup C$.

Using those two rules, you can get the answer to your homework.