How is the intersection of two independent sets such as $A$ and $B$ a multiplication process? That is, why is $P(A \cap B)=P(A)P(B)$?

  • $\begingroup$ Problem is quite unclear. Total answers of 2-D matrix product? Independent events? What is it all about? And did you want to understand these in K.G? $\endgroup$
    – Ganesh
    Oct 20, 2012 at 22:44

2 Answers 2


Why does $P(A)\cap P(B)=P(A)\times P(B) $ for independent events $A$ and $B$?

Your original question noted that multiplication usually computes the number of combinations, and you can model probabilities in a similar manner. Suppose you have two events, $A$ and $B$, such that $A$ occurs 40% of the time, and $B$ occurs 60% of the time. We can make a matrix of possible outcomes by multiplying these probabilities by some number of occurrences. Then we look to see which combinations are the event $(A,B)$.

Probability Matrix

We see that 6 of the 25 occurrences are $(A,B)$; 24% of all occurrences. We could have computed this by multiplying the number of $A$ by the number of $B$ and dividing by the total number of occurrences: $$\frac{\#A\times \#B}{(\#A+\#\neg A)\times(\#B+\#\neg B)}=\frac{\#A}{(\#A+\#\neg A)}\frac{\#B}{(\#B+\#\neg B)}=P(A)P(B)$$


There is a universe $\Omega$ of possible outcomes (or histories) $\omega$. Any reasonable subset $A\subset\Omega$ is called an event and is assigned a certain probability $P(A)\in[0,1]$. Two events $A$ and $B$ are called independent if $$P(A\cap B)=P(A)\,P(B)\ .\qquad(*)$$ So what you call a "multiplication process" is actually the definition of independence.

Of course this definition is derived from our intuition about "independence": Knowing that $B$ has indeed happened should not alter the prospects of the occurrence of $A$. This means that the fraction of outcomes $\omega\in A$ among all outcomes $\omega\in B$ should be the same as the fraction of outcomes $\omega\in A$ among all outcomes $\omega\in\Omega$. Written as a formula this is $${P(A\cap B)\over P(B)}={P(A)\over P(\Omega)}=P(A)\ ,$$ which coincides with $(*)$ if $P(B)\ne0$.


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