Here you will learn law of total probability theorem with examples.

Let’s begin –

## Total Probability Theorem (Law of Total Probability)

Let an event A of an experiment occurs with its n mutually exclusive and exhaustive events \(B_1,B_2,B_3….B_n\) then total probability of occurrence of event A is

P(A) = \(P(AB_1)\) + \(P(AB_2)\) +……..+ \(P(AB_n)\)

P(A) = P(\(B_1\))P(\(A/B_1\)) + P(\(B_2\))P(\(A/B_2\)) +……..+ P(\(B_n\))P(\(A/B_n\))

**Example** : A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

**Solution** : A red ball can be drawn in two mutually exclusive ways.

(I) Selecting bag I and then drawing a red ball from it.

(II) Selecting bag II and then drawing a red ball from it.

Let \(E_1\), \(E_2\) and A denote the events defined as follows :

\(E_1\) = Selecting bag I,

\(E_2\) = Selecting bag II

A = Drawing a red ball

Since one of the two bags is selected randomly.

\(\therefore\) P\(E_1\) = \(1\over 2\) and P\(E_2\) = \(1\over 2\)

Now, P(A/\(E_1\)) = Probability of drawing a red ball when the first bag has been chosen.

= \(4\over 7\) [ because First bag contains 4 red and 3 black balls ]

and, P(A/\(E_2\)) = Probability of drawing a red ball when the second bag has been chosen.

= \(2\over 6\) [ because Second bag contains 2 red and 4 black balls ]

Using the law of total probability, we have

Required Probabilty = P(A) = P(\(E_1\)) P(A/\(E_1\)) + P(\(E_2\)) P(A/\(E_2\))

= \(1\over 2\) \(\times\) \(4\over 7\) + \(1\over 2\) \(\times\) \(2\over 6\) = \(19\over 42\)

**Example** : A purse contains 4 copper and 3 silver coins and another purse contains 6 copper and 2 silver coins. One coin is drawn from any one of the these two purses. The probability that it is a copper coin is-

**Solution** : Let A = event of selecting first purse

B = event of selecting second purse

C = event of selecting copper coin

Then given event has two disjoint cases : AC and BC

P(C) = P(AC + BC) = P(AC) + P(BC) = P(A)P(C/A) + P(B)P(C/B)

= \(1\over 2\).\(4\over 7\) + \(1\over 2\).\(6\over 8\) = \(37\over 56\)