# Total Probability Theorem – Definition and Examples

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)

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