Mathematics Asked by AndroidV11 on September 27, 2020

Yes I am not aware of how to edit in LaTeX or make elegant edits, so apologies for not being able to do such editing on my own.

I am aware that $P(A cap B)$ for dependent events has $2$ variants based on the the $2!$ number of permutations in which $A$ and $B$ can be written.

$$P(A cap B) = P(A | B) P(B)$$

$$P(B cap A) = P(B | A) P(A)$$

My questions are,

- What is the equivalent of this for three events given $A, B$ and $C$?
- If I am right to assume, can we make $3!$ or $6$ versions of such a formula? What would they look like?
- This one does not need me to see the formula of permuted variants, but what would the intersection of $n$ sets look like in some summary notation like summation?

I am particularly interested in wanting to see a sample problem and solution involving the intersection of three dependent events, but I don’t think this site would find it nice if someone just goes around asking for x type of problem and solution even if it was simple just for me to get an illustration. But if it is allowed within the rules, I would appreciate it.

If $mathbb P(Acap B)ne 0$ then $$ mathbb P(Acap Bcap C)=mathbb P(A) mathbb P(Bmid A) mathbb P(Cmid Acap B) $$ And there are $3!=6$ such versions. For $n$ events $A_1,ldots, A_n$, if $mathbb P(A_1capldotscap A_{n-1})neq 0$, $$ mathbb P(A_1capldotscap A_n)=prod_{i=1}^n mathbb Pleft(A_iBiggm | bigcap_{j=0}^{i-1} A_{j}right) $$ where it is assumed that $A_0=Omega$.

Correct answer by NCh on September 27, 2020

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