What's $P(A_1\cap A_2\cap A_3\cap A_4) $?

Question

If $P(A_1\cap A_2\cap...\cap A_{n-1}) > 0$ so :

$$P(A_1\cap A_2\cap A_3\cap A_4) = P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_2\cap A_1)\cdot P(A_4|A_3\cap A_2\cap A_1)$$

I don't understand how to calculate $P(A_1\cap A_2\cap...A_{n-1})$ if I don't know whether they're independent or not.

Answer 1

The formula you have given $$P(A_1\cap A_2\cap A_3\cap A_4) = P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_2\cap A_1)\cdot P(A_4|A_3\cap A_2\cap A_1)$$ is a general formula. Independency is one of the special cases because if $A_1$, $A_2$, $A_3$, $A_4$ are independent, then

• $P(A_2|A_1) = P(A_2)$,

• $P(A_3|A_2\cap A_1) = P(A_3)$,

• $P(A_4|A_3\cap A_2\cap A_1) = P(A_4)$

so $P(A_1\cap A_2\cap A_3\cap A_4) = P(A_1)\cdot P(A_2) \cdot P(A_3) \cdot P(A_4)$ as required according to the definition of independence. So you can generalize the formula you have given to $n$ events with the same logic you used for $4$ events.

Answer 2

Just proceed in the same iterative pattern.   You do not need to know if the events are independent if you know all the conditional probabilities.

$$\begin{align}\mathsf P(\bigcap_{k=1}^n A_k) &=\mathsf P(A_1)\prod_{k=2}^n\mathsf P(A_k\mid \bigcap_{j=1}^{k-1} A_j) \\[3ex] \mathsf P(A_1\cap \ldots\cap A_n) & = \mathsf P(A_1)\cdots\mathsf P(A_k\mid A_1\cap\ldots\cap A_{k-1})\cdots\mathsf P(A_n\mid A_1\cap\ldots\cap A_{n-1}) \end{align}$$

Answer 3

Note that $P(A\mid B)$ is defined by $\frac{P(A\cap B)}{P(B)}$ which yields $P(A\cap B)=P(A\mid B)P(B)$. So \begin{align} P(A_1\cap A_2 \cap A_3 \cap A_4)&=P(A_4\cap (A_3 \cap A_2 \cap A_1))\\&=P(A_4\mid A_3 \cap A_2 \cap A_1)P(A_3 \cap (A_2 \cap A_1))\\&=P(A_4\mid A_3 \cap A_2 \cap A_1)P(A_3 \mid A_2 \cap A_1)P(A_2\cap A_1)\\&=P(A_4\mid A_3 \cap A_2 \cap A_1)P(A_3 \mid A_2 \cap A_1)P(A_2\mid A_1)P(A_1). \end{align}