Probability

Events
Algebra of Events
Theorems of Probability
Solved Examples

Introduction




Consider a random sampling process in which all the outcomes solely depend on the chance, i.e., each outcome is equally likely to happen. If the collection of all possible outcomes is U and the collection of desired outcomes is A, the probability of the desired outcomes is:






Random Experiment

It is an operation which can result in any one of its well-defined outcomes and the outcome cannot be predicted with certainity.

Example:
1 . Toss of a coin, which can result in either head or a tail
2 . Drawing a card from a well shuffled pack of cards
3 . throw of a die which can result in any one of the six faces


Sample Space And Sample Points


The set of all possible outcomes of a random experiment is called the sample space , associated with the random experiment.
Every possible outcomes i.e., every element of this set is a sample point .


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EVENT

An event is the outcome or a combination of outcomes of an experiment. In other words, an event is a subset of the sample space.

Simple Event

Each sample point in the sample space is called an elementary event or simple event.
Example : occurence of head in throw of a coin is a simple event.

Sure Event

The set containing all sample points is a sure event.
Example : as in the throw of a die the occurence of natural number less than 7, is a sure event.

Null Event

The set which does not contain any sample point.

Compound Event

Events obtained by combining two or more simple events together are called compound events or mixed event.

Complement of an Event

Let S be the sample space and E be an event then complement of E or Ec represents the complement of event E which is a subset containing all sample points in S which are not in E. It refers to non-occurence of event E .

Mutually Exclusive Events

If event A happens, then event B cannot, or vice-versa . The two events "it rained on Tuesday" and "it did not rain on Tuesday" are mutually exclusive events. When calculating the probabilities for exclusive events you add the probabilities .
Example : In the throw of a fair coin, occurence of a head and the occurence of a tail are mutually exclusive .

Exhaustive Number of Cases

Total number of possible outcomes of a random experiment in a trial is known as the exhaustive number of cases .
Example :
1) In throwing of a die exhaustive number of cases is 6.
2) In drawing two balls from a bag containing three white and two black balls, the exhaustive number of cases is 5C2 .

Equally Likely Events

The events are said to be equally likely, if none of them is expected to occur in preference to the other one .
Example : In throwing of a single die, each outcome is equally likely to happen .



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Algebra of Events

To calculate the probability of particular event, sometimes it becomes necessary to express the event in terms of algebraic combinations of some other simple events.

Types of such algebraic combination



(1) A or B (at least one of A or B) or A U B
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}

(2) Intersection of two events
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.

(3) Complementary event of A
It occurs when and only when A does not occur.

(4) Subset of an event
An event A is said to be a subset of B, if occurrence of A implies occurrence of B.
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Axiomatic Defination of Probability

Given a sample space S, with each event A of S, there is associated a number P(A), called the probability of A, such that the following axioms of probability are satisfied .

# For every A in S, 0 ≤ P(A) ≤ 1
# 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) .

Theorems of Probability

1 . Addition Rule of Probability : If A and B are any two events, then

                     P(A U B) = P(A) + P(B) - P(A ∩ B)

2 . P(AC) = 1 - P(A).

3 . P(f) = 0.

Conditional Probability

Suppose that A and B are two sets of outcomes. The probability of B under the condition that A has happened, denoted by P(B‌ A), can be expressed as:





Independent Events

Suppose that A and B are two sets of outcomes. If the probability of B was not affected by whether A has happened or not, A and B are two independent sets of outcomes.

                                               P(A ∩ B) = P(A)P(B)
This is also known as multiplication theorem for independent events .

Baye's theorem

Let S be the simple space and E1, E2, E3,….,En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any arbitrary event then
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Solved Examples


Example
: A die is rolled, find the probability that an even number is obtained.

Solution : Let us first write the sample space S of the experiment.
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
: Two coins are tossed, find the probability that two heads are obtained.

Solution : The sample space S is given by.
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
:What is the probability of getting at least two heads in a simultaneous throw of three coins?

Solution : The sample space S is given by.
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
:Two dice are rolled, find the probability that the sum is
a) equal to 1
b) equal to 4
c) less than 13

Solution : (a)The sample space S of two dice is shown below.

S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(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

(b) Three possible ouctcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.
P(E) = n(E) / n(S) = 3 / 36 = 1 / 12

(c) All possible ouctcomes, E = S, give a sum less than 13, hence.
b P(E) = n(E) / n(S) = 36 / 36 = 1


Example
: What is the probability of drawing either a king, a queen, or a jack from a deck of playing cards?

Solution :The total number of cards is 52. The individual probabilities are King = 4/52
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 bag contains 10 white and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that the first is white and the second is black?

Solution : Define events A and B such that

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
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