Algebra of Events

Theorems of Probability

Solved Examples

Consider a random sampling process in which all the outcomes solely depend on the chance, i.e., each outcome is

The set of all possible outcomes of a random experiment is called the

Every possible outcomes i.e., every element of this set is a

In a single throw of a die S = {1, 2, 3, 4, 5, 6}. Let A and B be 2 events such that

A = getting an even number = {2, 4, 6}

B = getting a multiple of 3 = {3, 6}

Then A or B = A U B

= {2, 3, 4, 6}

Let A and B be two events of a random experiment. Then there intersection is expressed as A ∩ B or

(A and B). It physically means that event A ∩ B is said to be occurred if both A and B occur simultaneously.

It occurs when and only when A does not occur.

An event A is said to be a subset of B, if occurrence of A implies occurrence of B.

# The entire sample space has the probability P(S) = 1

# For mutual exclusive events A and B (A ∩ B = Φ), P(A U B) = P(A) + P(B) .

P(A U B) = P(A) + P(B) - P(A ∩ B)

This is also known as

Example

S = {1,2,3,4,5,6} Let E be the event "an even number is obtained" .

E = {2,4,6}

We now use the formula .

P(E) = n(E) / n(S) = 3 / 6 = 1 / 2

Example

S = {(H,T),(H,H),(T,H),(T,T)} Let E be the event "two heads are obtained".

E = {(H,H)}

We use the formula of the classical probability.

P(E) = n(E) / n(S) = 1 / 4

Example

S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

Number of outcomes, S = 8 Let E be the event "at least two heads are obtained".

E = {HHH, HHT, HTH, THH }

Number of favourable outcomes, E= 4 We use the formula of the classical probability.

P(E) = n(E) / n(S) = 4 / 8

= 1/2

Example

a) equal to 1

b) equal to 4

c) less than 13

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) } Let E be the event "sum equal to 1". There are no outcomes which correspond to a sum equal to 1, hence

P(E) = n(E) / n(S) = 0 / 36 = 0

P(E) = n(E) / n(S) = 3 / 36 = 1 / 12

b P(E) = n(E) / n(S) = 36 / 36 = 1

Example

Queen = 4/52

Jack = 4/52 Therefore, the probability of success is

P = 4/52 + 4/52 + 4/52

= 12/52

= 3/13

Example

A = getting white ball in the first draw

B = getting black ball in the second draw

Required probability = Probability of getting white ball in the first draw and black in the second draw.

P(A ∩ B) = P(A)P(B/A) ........(i) Total number of balls = 10 white + 15 black = 25 balls

Total number of white balls = 10

Total number of balls present during the second draw = 9 white + 15 black = 24 balls

Number of black balls = 15

Required probability = P(A)P(B/A) [Using (i)]

(2/5)(5/8) = 2/8

= 1/4