`=>` 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)}`
`=>` 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)}`