I am working on a probability problem involving three events $A$, $B$ and $C.$ I am given $Pr(A),$ $Pr(B),$ $Pr(C),$ $Pr(A \cup B),$ Pr$(A \cup C),$ $Pr(B \cup C)$ and we are asked to find:
$Pr(A^c \cap B^c \cap C)$
I am not sure if this is correct, but this is what I have done thus far:
$((A^c \cap B^c) \cap C) >> ((A^c \cap B^c)^c \cap C) >> ((A \cup B) \cap C)$
Then by distributing we get $(A \cap C) \cup (A \cap B)$
I am not sure if I am doing this right or not. Any thoughts?
2 Answers
$A^c \cap B^c \cap C = (A \cup B \cup C^c)^c$. 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.
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.
