# Theorem of Compound Probability (Multiplication Theorem)

## Statement

If two events A and B are independent, then the probability of their simultaneous occurrence is equal to the product of their individual probabilities.

## Proof

Let *n1 *and *n2 * the total number of possible cases for events A and B respectively, have *m1* and *m2* as their respective favorable cases. Then,

P(A) = Probability of occurrence of an event A = m1/n1

P(B) = Probability of occurrence of an event B = m2/n2

Since the total number of possible cases for the events A and B are n1 and n2 respectively and since we can combine each of n1 possible cases of the event A with each of the n2 possible cases of the event B, so the total number of possible cases for their simultaneous occurrence is n1.n2. In the same way, the total number of favorable cases for the simultaneous occurrence of the events A and B is m1.m2.

Now,

P(A∩B) = Probability of simultaneous occurrence of the events A and B

i.e. P(A∩B) = m1.m2 / n1.n2 = m1/n1 . m2/n2 = P(A).P(B)

### Case I:

If A, B and C are three independent events, then the probability of their simultaneous occurrence P(A∩B∩C) is given by;

P(A∩B∩C) = P(A).P(B).P(C)

### Case II:

The probability that the event A will not happen is ** 1 – P(A)**.

Therefore, the probability that none of the two independent events A and B will happen is [1 – P(A)].[1 – P(B)].

∴the probability that at least one event will happen= 1 – [1 – P(A)].[1 – P(B)]

#### Note:

The probability of the simultaneous occurrence of two independent events A and B can also be written as P(AB) or P(A and B).

*This question has been asked for 4 marks in the Grade 12 final examination Mathematics.*