Introduction Probability Theory was first used to

solve problems in gambling Blaise Pascal (1623-1662) - laid the

foundation for the Theory of Probability

Now this theory is used in business, Science and industry

Next we define-Experiment and Event

Simple Probability An Experiment is an operation or a

process with a result or an outcome which is determined by , or depends on, Chance.

Examples: (1) Tossing a coin. (2) Tossing a die An Event is the outcome of an

experiment Example: Tossing a HEAD , getting a

SIX are events.

Definition of Probability In an experiment resulting in n equally

likely outcomes, if m of these outcomes are favour the occurrence of an event E

Then the Probability of event E happening,written as P(E), is defined as

No. of outcomes favourable to the occurance of E

Total number of equally likely outcomesP(E) =

=mn

A two-digit number is written down at random. Find the probability that the number will be(i) smaller than 20(ii) even (iii) a multiple of 5

Example 1:

How many two digit numbers are there?Is it 100 or 90 or 91The correct answer is 90

Now we will find the probabilities.

How many numbers are less than 20 ?

How many are even numbers ?

How many are multiples of 5 ?

10

45

18

Example 1:

ProbabilityThe possible outcomes is called the Sample space (S)

Hence, the probability of an Event E, P(E) = n(E)

n(S)

for any event E, 0 P(E) 1

If P(E) = 0, then the event cannot possibly occurIf P(E) = 1, then the event will certainly occur.

In the Probability Theory, an event is any subset of aSample Space

Sample SpacePossible outcomes of the following experiments

1. Tossing a Coin: S = [ H , T ]

2. Tossing a die : S = [ 1, 2, 3, 4, 5, 6 ]

3. Tossing two coins: S = [ HH, HT, TH, TT ]

4. Tossing two dice : S = [ (1,1),(1,2) ….. (6,5),(6,6)]

We can draw sample space for the above experiments

Example 2 :Two dice are thrown together. Find the probability that the sum

of the resulting numbers is(a) odd (b) even (c) a prime number(d) a multiple of 4 (e) at least 7First we draw the sample space, then

using that we can find the probability .

All the possible sums are displayed in the above diagramThis is called the sample space of this experiment

+

First die

1 2 3 4 5 6S

econ

d d

ie

1

2

3

4

5

6 127 8 9 10 11

2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

1198 1076

98 105 76

We define the following:A: the sum is oddB: the sum is evenC: the sum is a prime numberD: the sum is a multiple of 4E: the sum is at least 7

Total possible outcomes are 36. Hence n(S) = 36n(A) = 18A and B are complementary events. n(B ) = 36 - n(A)n(C) = 15n(D) = 9n(E) = 21

Now, it is very easy to calculate the probabilities.

Example 2 :

Answers:

( )

( )

( )

( )

( )

a

b

c

d

e

P(A) = 18

36

P(B) = 1 - P(A) = 1 - 1

2

P(C) = 15

36

P(D) = 9

36

P(E) = 21

36

1

21

25

121

47

12

Hence P(E’) = 1 - P(E) , where E’ is the complement of E

Box A contains 4 pieces of paper numbered 1,2,3,4Box B contains 2 pieces of paper numbered 1,2.One piece of paper is removed at random from each boxThe sample space is as follows

1 2 3 4

1

2

Box A

Box

B

Another way to illustrate the possible outcome

Example 3 :

TREE DIAGRAM

Box A

Box B Outcome

1

1

2

2

1

2

3

1

2

41

2

(1,1)

(1,2)

(2,1)

(2,2)

(3,1)

(3,2)

(4,1)

(4,2)

Hence n (S) = 8

A coin is tossed three times.Display all the outcomes using a tree diagramfind the probability of getting (i) three heads (ii) exactly two heads (iii) at least two heads

T

H

H

[T,H,H]

T

H [H,T,H]H

H

H [H,H,H]

T [H,H,T]

T [H,T,T]

T [T,H,T]

T

H [T,T,H]

T [T,T,T]

n (S) = 8

Now it is very easy to find the probabilities.

Adding Probabilities - Mutually Exclusive Events

Two events A and B are said to be Mutually Exclusive(ME)if the occurance of one event will not affect the occurance of the other event.

Set theoritically A B Hence these two events A and B cannot occur simultaneously.

If you want to calculate the probability of A or B, thenP(A or B )= P(AUB) = P(A) + P(B)

Example: A - getting an odd number B - getting an even number, while tossing a die once

Also, P(AUBUC) = P(A) + P(B) + P(C)

The probabilities of three teams L, M and N, winning a football competition are 1/4 , 1/8 and 1/10 respectively.

Calculate the probability that (i) either L or M wins, (ii) neither L nor N wins.

Example 4 :

We assume that only one team can win, so the events are mutually exclusive.

(i) P( L or M wins) = P(Lwins) + P(M wins) = 1/4 + 1/8 = 3/8

(ii) P(L or N wins) = 1/4 + 1/10 = 7/20 P(neither L nor N wins ) = 1 - P(L or N wins) = 1 - 7/20 = 13/20Note:

“Branches” of a “Probability Tree” represent outcomes which are mutually exclusive

Example 4 :

Example 5 Consider the experiment,

Tossing a die oncelet A - getting an odd number [ 1,3,5] B - getting a prime number [2,3,5]Here, A intersection B is not empty

P(A) = 3/6 = 1/2, P(B) = 3/6 = 1/2A B = [3,5] , P(A B) = 2/6 = 1/3P(AUB) = P(A) + P(B) - P(A B) = 3/6 + 3/6 - 2/6 = 4/6 = 2/3

TREE DIAGRAMBox A

Box B Outcomes

3

1

2

2

1

2

1

1

2

4

1

2

(1,1)

(1,2)

(2,1)

(2,2)

(3,1)

(3,2)

(4,1)

(4,2)Hence n (S) = 8

Now, we go back to the same example

If we replace the numbers 1,2,3..by the corresponding probabilities, we get

Box A contains 4 pieces of paper numbered 1,2,3,4Box B contains 2 pieces of paper numbered 1,2.One piece of paper is removed at random from each box

TREE DIAGRAM

Box A

Box B

Probability

P(1,1)= 1/8

)4

1(

)2

1(

1

1

P(3,2)=1/8)2

1(

)2

1( P(2,1)=1/8)

4

1(

P(3,1)=1/8

)4

1(

)2

1(

P(2,2)=1/8)

2

1(

P(4,2)=1/8)2

1(

P(1,2)=1/8)2

1(

P(4,1)=1/8

)4

1(

)2

1(

2

21

3 14

1

2

2

2

To find P(1,2), we multiply along the branches

Multiplying Probabilities - Probability Tree

The probability that two events, A and B, will both occur, written as P(A occurs and B occurs) orsimply P(A and B), is given by

P(A and B ) = P(A) x P(B)

Note: we have to multiply the probabilities along the branch,

Consider the following example:Suppose in a bag, there are 5 blue and 3 yellow marbles.A marble is drawn at random from the bag, the colourin noted and the marble is replaced. A second marbleis then drawn.

Y

B

Y

Y

B

B

5

85

85

83

8

3

8

3

8

1 st draw

2 st draw

This shows the Probability Tree

Since the first marble drawn is replaced, the total numberof marbles in the bag remains the same for the second draw.

Hence, we can say that the results of the two draws are independent and the two outcomes from each of the two draws are independent events.

If the marble is not replaced, then the probability of selecting a marble from the second draw is affected. This kind of events are called dependent events

Probability Treediagram for dependent events 5

7

B

Y

Y

Y

B

B

5

75

8

3

8

3

7

3

7

A garden has three flower beds. The first bed has 20 daffodils and 20 tulips, the second has 30daffodils and 10 tulips and the third has 10daffodils and 20 tulips.

A flower bed is to be chosen by throwing a die which has its six faces numbered 1,1,1,2,2,3. If the die shows a ‘1’, the first flower bed is chosen, if it shows a ‘2’ the second bed is chosen and so on.

A flower is then to be picked at random from the chosen bed.

Copy and complete the Probability Tree:

Bed 1

Bed 2

Bed 3

daffodil

daffodil

daffodil

Tulip

Tulip

Tulip

( )

( )

( )

( )

( )

( )

( )

( )

( )

2

2

2

4

6

1

1

1

1

Sample Space S = [ 1,1,1,2,2,3]

[20]

[20]

[30]

[10]

[10]

[20]

[40]

[40]

[30]

1 1

334

13

23

Prob. of picking a daffodil =1

2

1

2

1

3

3

4

1

6

1

3

5

9x x x

SUMMARY In an experiment which results in n equally likely outcomes, if m of these outcomes favour the occurrence of an event E, then probability of the event E is given by P(E) = m / n, also 0 P(E) 1

The Sample Space refers to the set of all the possible outcomes of an experiment.

An event is any subset of the sample space

If A and B are Mutually Exclusive, then P(A or B) = P(A) + P(B)

If A and B are independent, then P(A and B) = P(A) x P(B)

SUMMARY