Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis

Marks: 4M, 5M

Year: May 2015, Dec 2014

Total Probability Theorem:

Statement: If $B_1, B_2, ……….B_n$ be a set of exhaustive and mutually exclusive events and A is another event associated with (or caused by) $B_i$, then

$$P(A)= \sum_{i=1}^nP(B_i ).(\frac{A}B_i ) $$

Proof:

The inner circle represents the event A. A can occur along with (or due to) $B_1, B_2, ……….B_n$ that are exhaustive and mutually exclusive.

∴ $AB_1,AB_2,AB_3,AB_4…………………..AB_n$ are also mutually exclusive

∴ $A= AB_1+AB_2+ AB_3+ AB_4……+AB_n$ (By Addition Theorem)

$$∴P(A)=P(\sum_{i=1}^nAB_i)$$

$$=P(\sum_{i=1}^nPAB_i)$$

$∴P(A)= \sum_{i=1}^nP(B_i ).P(\frac{A}B_i ) $…(A) (Using conditional probability

$$P(AB)=P(A∩B)=P(B).P(B/A)=P(A).P(A/B))$$

Bayes’ Theorem or Theorem of Probability of causes

Statement: If $B_1, B_2, ……….B_n$ be a set of exhaustive and mutually exclusive events associated with a random experiment and A is another event associated with (or caused by) $B_i$, then

$$P(\frac{B_i}{A})=\frac{P(B_i ).P(\frac{A}{B_i })}{∑_{i=1}^nP{B_i }.P(\frac{A}{B_i } )}....... i=1,2..n$$

Proof:

We know Conditional Probability is given as:

$P(AB_i )=P(A∩B_i )=P(B_i ).P(A/B_i )=P(A).P(B_i/A)---------(1)$

$$P(\frac{B_i}{A})=\frac{P(B_i ).P(\frac{A}{B_i })}{P(A)}....... (2)$$

Now using Total Probability Theorem we have,

$$P(A)= \sum_{i=1}^nP{B_i }.P(\frac{A}{B_i }) ------ (3)$$

From equation (2) and equation (3)

$$P(\frac{B_i}{A})=\frac{P(B_i ).P(\frac{A}{B_i })}{∑_{i=1}^nP{B_i }.P(\frac{A}{B_i } )}.......$$ hence proved.