How to prove the following results?

Question

I was researching about the Bayes theorem, I would like to prove it in order to understand more I found that the prove uses the following equations:

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

I would like to see if some one could help me to prove that equations using Venn diagrams in order to be more graphic and also providing me the theorems that I would need to prove that result,

Thanks any how, I appreciate the support.

Answer 1

It seems you may be looking for an intuitive justification of the definition $$P(A|B) = P(A \cap B)/P(B).$$ Here are three examples that might help match the definition with intuition. I hope that, somewhere among the three, there is something that persuades you that the definition of conditional probability is intuitively sound.

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At an imaginary small college enrollments in schools of Arts, Business, and Science according to gender are as follows:

College     A       B       S         Total
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Male       100     500     400        1000
Female     200     400     600        1200
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Total      300     900    1000        2200

Suppose a student is chosen at random, given that a Science major was chosen, what is the probability a woman was chosen?

Method 1: Logic. Knowing that a Science major was chosen, we are interested only in the column headed 'S'. Ignore the rest of the table. There are 1000 Science majors of whom 600 are women, so $P(W|S) = 600/1000 = 0.6.$

Method 2: Use definition.
$$P(W|S) = \frac{P(W \cap S)}{P(S)} = \frac{600/2200}{1000/2200} = \frac{600}{1000} = 0.6.$$

The definition gets the same answer as we got by logic. It may look trivial on account of the highly rounded numbers, but this shows that the definition is in accord with real applications.

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An urn contains three Red balls and two Green balls. What is the probability of getting two Red balls if two balls are selected at random without replacement?

Method 1. Combinaations. $$P(R_1 \cap R_2) = \frac{{3 \choose 2}{2 \choose 0}}{{5 \choose 2}} = \frac 3 {10} = 0.3.$$

Method 2. The definition of conditional probability implies the 'general multiplication rule': $P(R_1 \cap R_2) = P(R_1)\cdot P(R_2|R_1).$

Obviously, $P(R_1) = 3/5.$ Then on the second draw, it is intuitive to say $P(R_2|R_1) = 2/4$ because the remaining contents of the urn are four balls of which two are red. Then multiplying, we have $$P(R_1 \cap R_2) = \frac{3}{5}\cdot\frac{2}{4} = \frac{6}{20} = 0.3.$$

Again here an intuitive view of a conditional probability $P(R_2|R_1)$ leads to an answer obtained by another method (combinatorics).

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A fair die is rolled. We are told that the result is a 'big' number, where $B = \{4,5,6\}.$ Given this conditional information, what is the probability that the result is 'even', where $E=\{2,4,6\}?$ Because we know the result is Big, it seems reasonable to assign probability $1/3$ to each of the outcomes 4, 5, and 6. Of these, two are even. So intuitively, $P(E|B) = 2/3.$

But the definition of conditional probability gives $$P(E|B) = \frac{P(E \cap B)}{P(B)} = \frac{P\{4,6\}}{1/2} = \frac{2/6}{1/2} = 2/3.$$