How do i solve P(A\B∪C)

Question

I just started set theory is math and with lack of understanding I've come here for some much-needed assistance. How would I go about determining what the probability is if its A given B or C?

I've had a hard time finding any useful information elsewhere of the internet, most resources don't mention how to do a P with a relative complement with more than $2$ things in the p

does it mean the P() of a and not the number of b or c, I'm confused with that

Edit: $\text{A B C}$ is set up in a three-circle Venn diagram.

$\text{A} = 4$
$\text{B} = 5$
$\text{C} = 8$

$\text{A}$ and $\text{B}$ and $\text{C}$ is $2$, $\text{A}$ and $\text{B}$ is $6$, $\text{A}$ and $\text{C}$ is $7$, $\text{B}$ and $\text{C}$ is $1$

Answer 1

Assume the question is asking to find $p=P(A|B\cup C)$.

$p=\frac{\text{number of favorable events}} {\text{total number of events}} = \frac{7+2+6}{7+2+6+8+1+5}=\frac{15}{29} $

Explanation: number of favorable events are those elements in A which also belong to a bigger event $(B\cup C)$. The total number of events are those that are "given" to the right of the vertical bar, that is, $(B\cup C)$.

Note that the numbers in Venn diagram mean the number of elements in each segment. For example, number $4$ in $A$ means that there are four elements there and nowhere else. It does not mean that there is "element $4$" there. Same for all other segments.

Also, note that what you wrote in your question is not correct: "A and B and C is 2, A and B is 6, A and C is 7, B and C is 1"

For example $A\cap B = 8$, not $6$.

Answer 2

Given the verbal description ($A$ given $B$ or $C$), the most plausible (to me anyway) interpretation of the question is that there is a slight typo in the notation: $P(A\backslash B\cup C)$ should be $P(A|B\cup C)$.

You can just compute $P(A|B\cup C)$ by counting. First, $B\cup C$ entails $6+5+2+1+7+8=29$ outcomes. Of these 29 outcomes, $6+2+7=15$ are in $A$. So the desired probability is $\frac{15}{29}$.

The implicit assumption of the question is that all the elements each carries the same weight.