## Law of Total Probability

For two events A and B associated with a sample space S, the sample space can be divided into a set A ∩ B′, A ∩ B, A′ ∩ B, A′ ∩ B′. This set is said to be mutually disjoint or pairwise disjoint because any pair of sets in it is disjoint. Elements of this set are better known as a partition of sample space.

This can be represented by the Venn diagram as shown in fig. 1. In cases where the probability of occurrence of one event depends on the occurrence of other events, we use total probability theorem.

**Derivation of total probability theorem:**

Definition:

Let events C_{1}, C_{2} . . . C_{n} form partitions of the sample space S, where all the events have a non-zero probability of occurrence. For any event, A associated with S, according to the total probability theorem, P(A)

P(A):

\( \sum\limits_{k=0}^{n} P(C_k)P(A| C_k)\)

**Proof:** From the figure 2, {C_{1}, C_{2}, . . . . , C_{n}} is the partitions of the sample space S such that, C_{i} ∩ C_{k} = φ, where i ≠ k and i, k = 1, 2,…,n also all the events C_{1},C_{2} . . . . C_{n} have non zero probability. Sample space S can be given as,

S = C_{1} ∪ C_{2} ∪ . . . . . ∪ C_{n}

For any event A,

A = A ∩ S

= A ∩ (C_{1} ∪ C_{2}∪ . . . . ∪ C_{n})

= (A ∩ C_{1}) ∪ (A ∩ C_{2}) ∪ … ∪ (A ∩ C_{n}) . . . . . (1)

We know that A ∩ C_{i} and A ∩ C_{k} are the subsets of C_{i} and C_{k}. Here, C_{i} and C_{k} are disjoint for i ≠ k. since they are mutually independent events which imply that A ∩ C_{i} and A ∩ C_{k} are also disjoint for all i ≠ k. Thus,

P(A) = P [(A ∩ C_{1}) ∪ (A ∩ C_{2}) ∪ ….. ∪ (A ∩ C_{n})]

= P (A ∩ C_{1}) + P (A ∩ C_{2}) + … + P (A ∩ C_{n}) . . . . . . . (2)

We know that,

P(A ∩ C_{i}) = P(C_{i}) P(A|C_{i})(By multiplication rule of probability) . . . . (3)

Using (2) and (3), (1) can be rewritten as,

P(A) = P(C_{1})P(A| C_{1}) + P(C_{2})P(A|C_{2}) + P(C_{3})P(A| C_{3}) + . . . . . + P(C_{n})P(A| C_{n})

Hence, the theorem can be stated in form of equation as,

\(P(A) = \sum\limits_{k=0}^{n} P(C_k)P(A| C_k)\)

### Solved Examples on the Law of Total Probability

**Illustration:** **A person has undertaken a mining job. The probabilities of completion of job on time with and without rain are 0.42 and 0.90 respectively. If the probability that it will rain is 0.45, then determine the probability that the mining job will be completed on time.**

** Solution:** Let A be the event that the mining job will be completed on time and B be the event that it rains. We have,

P(B) = 0.45,

P(no rain) = P(B′) = 1 − P(B) = 1 − 0.45 = 0.55

By multiplication law of probability,

P(A|B) = 0.42

P(A|B′) = 0.90

Since, events B and B′ form partitions of the sample space S, by total probability theorem, we have

P(A) = P(B) P(A|B) + P(B′) P(A|B′)

=0.45 × 0.42 + 0.55 × 0.9

= 0.189 + 0.495 = 0.684

So, the probability that the job will be completed on time is 0.684.