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I am completely a beginner in Probability and I really struggle to understand why the probability of two independent events is the product of their probabilities, that is, I cannot understand how we can prove the following formula:

$$P(A \text{ and }B) = P(A \cap B) = P(A).P(B)$$

If I take the example of rolling two dices, by using the figure below, I can see that by multiplying the probabilities we can indeed obtain the result. Let's say that our purpose is to find the probability that by rolling the first dice we obtain two and by rolling the second dice we obtain 5 (independent events as rolling one dice doesn't affect the rolling of the other). This can be shown with the following graphic as a series of ordered outcomes $(x,y)$ where $x$ is the number obtained after rolling the first dice and $y$ is the number obtained after rolling the second dice.

enter image description here

Now just by looking to the picture, if one asks me what is the probability of obtaining $(2,5)$, first of all, I can see that there are 36 possible outcomes in total. And then,

  • If $P(A)$ = obtaining 2 after rolling the first dice, we can see that this equals to $$\dfrac{6}{36} = \dfrac{\text{Number of outcomes where 2 is obtained after rolling the first dice (2nd column in the picture)}}{\text{Number of all possible outcomes}} = \dfrac{1}{6}$$

  • If $P(B)$ = obtaining 5 after rolling the 2nd dice, we can see that this equals to $$\dfrac{6}{36} = \dfrac{\text{Number of outcomes where 5 is obtained after rolling the 2nd dice (5th row in the picture)}}{\text{Number of all possible outcomes}} = \dfrac{1}{6}$$

Now just by looking to the picture, I can see that at the intersection of the 2nd column and the 5th row, there is only one entry, among all 36 possible outcomes that is equal to $(2, 5)$ (I have shown in the picture with a small red rectangle). Consequently, the probability of having $(2, 5) = \dfrac{1}{36}$

Now if I multiply $P(A)$ and $P(B)$ I can see that indeed I obtain the same value:

$$P(A).P(B) = \dfrac{1}{6} . \dfrac{1}{6} = \dfrac{1}{36}$$

Yet, still I cannot grasp the meaning and logic behind this and I desperately try to find somewhat a clear algebraic proof of the fact that the product of the probabilities of two independent events is the proability that both occur.

So far, I have been googling and everytime I saw a page presenting the subject, it was just the definition and the formula without explaining clearly the reason that we have to multiply them. I would appreciate if you could kindly make some clarification.

Odyssey2001
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  • Think of the probability of an event as the percentage of times it occurs when the experiment is repeated. – David Mitra Feb 19 '23 at 14:45
  • That is indeed the definition of independence; to obtain a proof of it, you'd have to pick some other definition, and then tell us what it is. – Adam Feb 19 '23 at 14:46
  • Events A and B are said to be independent if knowing whether A occurred does not change the probability of B i.e. $\Pr(B|A) = \Pr(B) \implies \Pr(B \cap A) = \Pr(B)\Pr(A)$ – LemNon Feb 19 '23 at 14:51
  • Are you just asking for intuition? You can't prove a definition, after all. But for intuition, consider a discrete case. Like the joint outcome of flipping a fair coin and tossing a fair die. Each of the $12$ outcomes are equally probable, so... – lulu Feb 19 '23 at 15:11
  • Two events $A$ and $B$ are *defined* as independent iff $P(A\cap B)=P(A)P(B),$ and mutually exclusive iff $P(A\cap B)=0.\quad$ Conditional probability: intuition and visualisation. – ryang Feb 19 '23 at 15:33
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    You need not prove definition. You need to convince yourself that definition makes sense and leads to the same answers as your intuition. – kludg Feb 19 '23 at 15:58

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Working in some sample space $\Omega$ having equiprobable outcomes we can find the probability of some event $A\subseteq\Omega$ by application of the rule: that $$\text{probability }=\text{ number of favourable outcomes divided by number of possible outcomes}$$ In mathematical notation:$$P(A)=\frac{n(A)}{n(\Omega)}$$ If we are only interested in outcomes that appear in $B\subseteq\Omega$ then we get on a similar way:$$P(A|B)=\frac{n(A\cap B)}{n(B)}$$ We speak of independence in the special case where restriction to outcomes in $B$ does not affect the original probability, i.e. the case in which $P(A|B)=P(A)$.

We are dealing with it if and only if we have the equality:$$P(A\cap B)=\frac{n(A\cap B)}{n(\Omega)}=\frac{n(A)}{n(\Omega)}\frac{n(B)}{n(\Omega)}=P(A)P(B)$$

drhab
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