`=>` Consider that there are two bags I and II. Bag I contains 2 white and 3 red balls and Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the bags. We can find the probability of selecting any of the bags (i.e.`1/2` ) or probability of drawing a ball of a particular colour (say white) from a particular bag (say Bag I).

`=>` A set of events `E_1, E_2, ..., E_n` is said to represent a partition of the sample space S if

(a) `E_ i ∩ E_j = φ, i ≠ j, i, j = 1, 2, 3, ..., n`

(b) `E_1 ∪ Ε_2 ∪ ... ∪ E_n= S` and

(c) `P(E_i) > 0` for all i = 1, 2, ..., n.

`=>` In other words, the events `E_1, E_2, ..., E_n` represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have nonzero probabilities.

`=>` As an example, we see that any nonempty event E and its complement E′ form a partition of the sample space S since they satisfy` E ∩ E′ = φ` and` E ∪ E′ = S`.

`=>` From the Venn diagram in Fig 13.3, one can easily observe that if E and F are any two events associated with a sample space S, then the set` {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′}` is a partition of the sample space S. It may be mentioned that the partition of a sample space is not unique. There can be several partitions of the same sample space.

`=>` Let `{E_1, E_2,...,E_n}` be a partition of the sample space S, and suppose that each of the events `E_1, E_2,..., E_n` has nonzero probability of occurrence. Let A be any event associated with S, then


`color{blue}{P(A) = P(E_1) P(A|E_1) + P(E_2) P(A|E_2) + ... + P(E_n) P(A|E_n)}`

`color{blue}{= sum_(j=1)^n P(E_j) P(A|E_j)}`


`color{red} "Proof"` Given that `E_1, E_2,..., E_n` is a partition of the sample space S (Fig). Therefore,

`color{orange} {S = E_1 ∪ E_2 ∪ ... ∪ E_n}` ... (1)

and `E_i ∩ E_j = φ, i ≠ j, i, j = 1, 2, ..., n`

Now, we know that for any event A,

`A = A ∩ S`

`= A ∩ (E_1 ∪ E_2 ∪ ... ∪ E_n)`

`= (A ∩ E_1) ∪ (A ∩ E_2) ∪ ...∪ (A ∩ E_n)`

Also `A ∩ E_i` and `A ∩ E_j` are respectively the subsets of `E_i` and `E_j` . We know that
`E_i` and `E_j` are disjoint, for i ≠ j , therefore, `A ∩ E_i` and `A ∩ E_j` are also disjoint for all

`i ≠ j, i, j = 1, 2, ..., n`.

Thus, `P(A) = P [(A ∩ E_1) ∪ (A ∩ E_2)∪ .....∪ (A ∩ E_n)]`

`= P (A ∩ E_1) + P (A ∩ E_2) + ... + P (A ∩ E_n)`

Now, by multiplication rule of probability, we have

`P(A ∩ E_i) = P(E_i) P(A|E_i)` as `P (E_i) ≠ 0 ∀ i = 1,2,..., n`

Therefore, `P (A) = P (E_1) P (A|E_1) + P (E_2) P (A|E_2) + ... + P (E_n)P(A|E_n)`


or `color{green}{P(A) = sum_(j=1)^n P(E_j) P(A |E_j)}`

`=>` Bayes’ Theorem If `E_1, E_2 ,..., E_n` are n non empty events which constitute a partitionof sample space S, i.e. `E_1, E_2 ,..., E_n` are pairwise disjoint and `E_1∪ E_2∪ ... ∪ E_n = S` and A is any event of nonzero probability, then

`"Proof : "` `color{blue}{P(E_i | A) = (P(E_i) P(A|E_i))/(Sigma_(j =1)^n P(E_j ) P (A|E_j))}` for any i = 1, 2, 3, ..., n

`color {red} "Proof"` By formula of conditional probability, we know that

`P(E_r|A) = (P(AnnE_i))/(P(A))`

`= (P(E_i)P(A|E_i))/(P(A) )` (by multiplication rule of probability)

`= (P(E_i ) (PA | E_i))/(Sigma_(j=1)^n P(E_j) P(A| E_j))` (by the result of theorem of total probability)

`color{blue} ul"Remark"` :

The following terminology is generally used when Bayes' theorem is applied.
The events `E_1, E_2, ..., E_n` are called hypotheses.

`=>` The probability `P(E_i)` is called the priori probability of the hypothesis `E_i`

`=>` The conditional probability `P(E_i |A) `is called a posteriori probability of the hypothesis `E_i`.

`=>` Bayes' theorem is also called the formula for the probability of "causes". Since the `E_i's` are a partition of the sample space S, one and only one of the events `E_i` occurs (i.e. one of the events `E_i` must occur and only one can occur). Hence, the above formula gives us the probability of a particular `E_i` (i.e. a "Cause"), given that the event A has occurred.
The Bayes' theorem has its applications in variety of situations, few of which are illustrated in following examples.