Multiplication Theorem of Probability (Theorem of Compound Probability),Extension of Multiplication Theorem of Probability to n Events

Multiplication Theorem of Probability (Theorem Of Compound Probability)

We can state a law that" the probability of simultaneous occurrence of two events A and B is equal to

(i) the product of the probability of the event A and the conditional probability of B assuming that A has already occurred .

(ii) the product of the probability of the event B and the conditional probability of A given that B has already occurred .

This law is also called the multiplicative law of probability.

Theorem 11.

Let A and B be two dependent events in a sample space S, the probability simultaneous occurrence of A and B is given by

$$ P(A \cap B)=P(A).P(\frac {B}{A})\; \; ;P (A)> 0$$

$$=P(B).P(\frac {A}{B})\; \; ;P (B)> 0$$

Proof:

Let A and B be two dependent events in a sample space S associated with a random experiment. According to the definition of probability

$$ P(A)= \frac {n(A)}{n(S)}, \; \; \; \; \; P(B)=\frac {n(B)}{n(s)}, \; \; \; \; P(A \cap B)=\frac{n(A \cap B)}{n(S)} \; \; \; \; \; ....................................(i)$$

For the conditional event \(\frac{B}{A}\) ,the reduced sample space is A ,because ,the sample points of the event\(\frac{B}{A}\)are include in the event A. So, the number of exhaustive case is n(a) out of which \(n(A \cap B\)) cases are favourable to\(\frac{B}{A}\). Thus,

$$P(\frac{B}{A})=\frac {n(A \cap B)}{n(A)} \; \; \; \; \; .........................(ii)$$

Now, (i) can be expressed as

$$ P(A \cap B)=\frac {n(A)}{n(S)} \times \frac {n(A \cap B)}{n(A)}=P(A).P(\frac{B}{A}) \; \; \; \; \; \; \; \; [From \; \; (ii)]$$

similarly, we get,

$$ P(A \cap B)=\frac {n(B)}{n(S)} \times \frac {n(A \cap B)}{n(B)}=P(B).P(\frac{A}{B})$$

Hence ,the theorem 11 is proved.

Theorem12 :

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

$$ P(A \cap B) =P(A).P(B)$$

Proof:

For any two events A and B ,the multiplication theorem ofprobability is

$$ P(A \cap B)=P(A).P(\frac {B}{A})\; \; ;P (A)> 0$$

$$=P(B).P(\frac {A}{B})\; \; ;P (B)> 0\; \; \; \; \; \; \; ..........................................(i)$$

If A and B are independent ,then we have \(P(\frac {A}{B}\))=P(A) and \(P(\frac {B}{A}\))=P(B).................(ii)

From (i) and (ii) ,we get

$$P(A \cap B)=P(A).P(B)$$

Conversely, if \(P(A \cap B\))=P(A).P(B)

$$ \frac {P(A \cap B)}{P(A)}=P(B) \Rightarrow \; P(\frac {B}{A})=P(B)$$

$$ and \; \;\frac {P(A \cap B)}{P(B)}=P(A) \Rightarrow \; P(\frac {A}{B})=P(A) \; \; \; ...........................................(iii)$$

The equation (iii) implise that A and B are independent events.

Note: Two events A and B are independent if and only if \(P(A \cap B\))=P(A).(B)

Extension of Multiplication Theorem of Probability to n Events

Theorem 13.

Let \(A_1,A_2,A_3,.....A_n\) be n dependent events in a sample space S, then

$$ P(A_1 \cap A_2 \cap ..... \cap A_n)=P(A_1)P(\frac {A_2}{A_1})P(\frac {A_3} {A_1 \cap A_2}).......P(\frac {A_n}{A_1 \cap A_2 \cap ...... \cap A_{n-1}}) \; \; \; \; \; \; .............(i)$$

Proof : For two events \(A_1\) and \(A_2\) , we have

$$P(A_1 \cap A_2)=P(A_1)P(\frac {A_2}{A_1})\; \; \; \; \; .........(a) $$

thus equation (i) is true for n=2.

For three events \(A_1,A_2 \; \;and \; \; A_3\)

$$P(A_1 \cap A_2 \cap A_3)=P[A_1 \cap ( A_2 \cap A_3)]$$

$$=P(A_1)[\frac {P(A_2 \cap A_3)}{A_1}] \; \; \; \; [ using \; \; (a)]$$

$$=P(A_1)P(\frac {A_2}{A_1})P[\frac {A_3}{(A_1 \cap A_2)}] \; \; \; \; [ using \; \; (a)]$$

Thus the equation (i) is true for n=3.

Proceeding in this way, the equation (i) is true for all positive integral values of n. Hence by the principle of mathematical induction , the multiplication theorem of probability can be extended (or generalised) to n events as

$$ P(A_1 \cap A_2 \cap .... \cap A_n)=P(A_1) P(\frac {A_2}{A_1})P(\frac {A_3}{A_1 \cap A_2}) ..... P(\frac {A_n}{A_1 \cap A_2 \cap ....\cap A_{n-1}})$$

$$ i.e. \; \; P(\bigcap_{i=1}^{n}A_i)=P(\bigcap_{i=1}^{{n-1}}A_i)P(\frac{A_n}{\bigcap_{i=1}^{{n-1}}A_i})$$

Theorem 14.

The n events \(A_1,A_2, ....A_n\) are independent of each other if and only if the probability of their simultaneous occurrence is equal to the product of their respective probabilities i.e. iff

$$P(A_1 \cap A_2 \cap ...... \cap A_n)=P(A_1)P(A_2)...P(A_n)\; \; \; \; \; .........(ii)$$

Proof:

If \( A_1,A_2,....,A_n\) are n independent events ,then

$$P(\frac {A_2}{A_1})=P(A_2), P(\frac {A_3}{A_1 \cap A_2})=A_3,......,P(\frac {A_n}{A_1 \cap A_2 \cap .... \cap A_{n-1}})=P(A_n)$$

Hence ,from equation (I) ,we get

$$P(A_1 \cap A_2 \cap ..... \cap A_n)=P(A_1).P(A_2)....P(A_n)$$

Conversely, if equation (ii) holds, then from (i) and(ii) , we get

$$ P(A_2)=P(\frac {A_2}{A_1}), \; \; \; P(A_3)=P(\frac {A_3}{A_1 \cap A_2}),....., P(A_n)=P(\frac {A_n}{A_1 \cap A_2 \cap .... \cap A_{n-1}})$$

These equations simply that \(A_1,A_2...,A_n\) are independent events.

Hence the theorem 14 is proved.

Theorem 15.

If A and B are independent events ,then A and \( \bar B\) are also independent events.

Proof:

If A and B are independent events,then P(\(A \cap B\))=P(A).P(B)

Since (\(A \cap \bar B\)) \(\cup\) (\(A \cap B\))=A then by axiom (iii) of axiomatic definition of probability,we get

$$P(A)=P(A \cap \bar B)+P(A \cap B)$$

$$ \Rightarrow P(A \cap \bar B)=P(A)-P(A \cap B)$$

$$=P(A)-P(A).P(B)$$

$$=P(A)[1-P(B)]$$

$$=P(A)P(\bar B)$$

$$ i.e. \; \; \; P(A \cap \bar B)= P(A)P(\bar B)$$

$$ A \; and \;\bar B \; are \; indepandent \; events.$$

Similarly ,it can be proved that,if A and B are independent events,then \( \bar A\) and B are also independent.

Theorem 16.

If A and B are independent events, then \(\bar A\) and \(\bar B\) are also independent events.

Proof:

Since A and B are independent events.

$$ P(A \cap B)=P(A).P(B)$$

$$ Now, \; \; P(\bar A \cap \bar B)=P \bar {(A \cup B)}= 1-P( A \cup B)$$

$$=1-[P(A)+P(B)-P(A\cap B)]$$

$$=1-P(A)-P(B)+P(A).P(B)$$

$$=1-P(B)-P(A)[1-P(B)]$$

$$=[1-P(A)][1-P(B)]$$

$$=P(\bar A)P(\bar B)$$

$$ therefore \; \; \; \; \bar A \; \; and \; \; \bar B \; \; are \; \; independent \; \; events.$$

Thus, if A and B are independent events, then the probability that none of them
(A and B) will happen is given by

$$P(\bar A \cap \bar B)=P(\bar A)P(\bar B)$$

This is the probability of non-occurrence of both A and B.

This theorem can be extended to n independents events \(A_1,A_2 A_3.....,A_n\).

In general ,the probability of non occurrence of all the n independent events \(A_1,A_2,A_3.....,A_n\). i.e P(nither \(A_1\) nor \(A_2\).....nor \(A_n\)) is given by

$$P(\bar A_1 \cap \bar A_2 \cap ..... \cap \bar A_n)=P(\bar A_1)P(\bar A_2).....P( \bar A_n)$$

Moreover ,if \(A_1,A_2,.....,A_n\) are n independent events ,then the probability that at leat one of these events will happen is given by

$$P(A_1 \cup A_2 \cup .... \cup A_n)=1-P(\bar A_1 \cap \bar A_2 \cap .... \cap \bar A_n)$$

$$=1-P(\bar A_1)P(\bar A_2).....P(\bar A_n) \; \; \; \; \; \; \; ................................(iii)$$

References

Sukubhattu,Narendra Prasad. Probability and Inference-I. Asmita Books Publishers & Distributors (P)Ltd. 2013