Mathematics

Session

Probability - 1

Session Objectives

Experiment

Sample Space

Event

Types of Events

Probability of an Event

Class Exercise

Experiment

Experiment: An operation, which results in some well-defined outcomes is called an experiment.

Random Experiment: If we conduct an experiment and we do not know which of the possible outcome will occur this time, the experiment is called a random experiment.

For example:

1. Tossing a coin

2. Throwing a die

3. Drawing a card from a well shuffled pack of cards

Sample Space

Sample Space: The sample space of an random experiment is the setof all possible elementary outcomes. It is denoted by S.

For example:

1. When we toss a coin, the sample space

S = {H, T}

It is a random experiment, because when we toss a coin this time, we do not know whether we shall get head or tail.

2. When we throw a die, the sample space

S = {1, 2, 3, 4, 5, 6}

Event

Event : A subset ‘E’ of a sample space is called an event. An event is a combination of one or more of the possible outcomes of an experiment.

For example:In a single throw of a die, the event of getting a prime number is given

by E = {2, 3, 5} and the sample space S = {1, 2, 3, 4, 5, 6}.

E S

Algebra of Events

For any three events A, B and C with sample space S.

1 A B =either A or B or both

2 A B = both A and B

3 A = not A

4 A - B = A but not B

5 A B '= A B

Algebra of Events

8 A B C = A B C

9 A B C = A B A C

10 A B C = A B A C

6 A B '= A B

7 A B C = A B C

Types of Events (Sure Event)

Sure Event: In the throw of a die, sample space S = {1, 2, 3, 4, 5, 6}

S S S represents an event.

Each outcome of the experiment is a member of S.

S is called a sure event or certain event.

Types of Events (Impossible Event)

Impossible Event: In the throw of a die, sample space

S = {1, 2, 3, 4, 5, 6}.

Let E be the event of getting an ‘8’ on the die.

Clearly, no outcome can be a number 8.

E= is an impossible event.

CComplement of an event E or E =not E=S - E

Simple and Compound Event

Simple or Elementary EventAn event that contains only one element of the sample space is calleda simple or an elementary event.

Compound EventA subset of sample space which contains more than one element is called compound event or mixed event or composite event.

For example: In a simultaneous toss of two coins, the sample space is

S = {HT, TH, HH, TT}

1Event E ={TT} of getting both tails is an elementary event.

2Event E ={HT, TH, TT} of getting at least 1 tail is a compound event.

Equally Likely Outcomes

Equally Likely OutcomesThe outcomes are said to be equally likely, if none of them is expectedto occur in preference to the other or the chances of occurrence of all of them are same.

For example:In throwing of a single die, each outcome is equally likely to happen.

Mutually Exclusive Events

Mutually Exclusive EventsTwo or more events are said to be mutually exclusive if no two or moreof them can occur simultaneously in the same trial.

1 2 1 2Two events E and E are mutually exclusive, if E E = .

Facts:1. Elementary events related to an experiment are always mutually exclusive.

2. Compound events may or may not be mutually exclusive.

Example

In throwing a die, sample space S = {1, 2, 3, 4, 5, 6}

1

2

Event E =event of getting a number less than 3

and E = event of getting a number more than 4, then

1 2 1 2E = 1, 2 and E = 5, 6 E E =

1 2E and E are mutually exclusive events.

Exhaustive Events

Exhaustive EventsIn a random experiment, two or more events are exhaustive if their union is the sample space.

1 2 n

1 2 n

i.e. in a random experiment, events E , E , ...E

with sample space S are exhaustive if

E E ... E =S

Example

In throwing a die, sample space S = {1, 2, 3, 4, 5, 6}

1E : 1, 2 - the event of occurrence of '1' or '2'.

2E : 2, 3, 4 - the event of occurrence of '2' or '3' or '4'.

3E : 3, 4, 5, 6 - the event of occurrence of a number 3.

1 3 1 2 3 1 2E E S, E E E S but E E S

Example-1

A die is thrown twice. Each time the number appearing on it is recorded. Describe the events :

A: both numbers are odd.

B: sum of numbers is less than 6.

C: both numbers are even.

Describe . Which pairs of events are mutually exclusive.

A B, A B, A C, A C

Solution: A ={(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}

B = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

C ={(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}

Solution Cont.

A B {(1,1),(1,3),(3,1)}

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

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

A C {(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5)

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

A C

A and C are mutually exclusive

A B=

Example –2

Three coins are tossed . Describe

(1) two mutually exclusive events A and B.

(2) three mutually exclusive exhaustive events A, B and C.

(3) two events E and F which aren’t mutually exclusive.

Solution: Sample space for the toss of three coins is

S = {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T),

(T, H, T), (T, T, H), (T, T, T)}

Let A:{Event of getting three heads} ={(H, H, H)}

and B:{Event of getting three tails}={(T, T, T)}

A and B are mutually exclusive

Solution (Cont.)

Let C:{Event of getting one or two head}

={(H, H, T),(H, T, H),(T, H, H),(H, T, T),(T, H,T),(T,T,H)}

A, B and C are mutually exclusive exhaustive events.

Let E: {Event of getting two heads}

= {(H, H, T),(H, T, H),(T, H, H)} and

F: {Event of getting at least one tail}

= {(H, H,T), (H, T, H),(T, H, H), (H, T, T), (T, H, T), (T, T, H), (T,T,T)}

E and F aren’t mutually exclusive.

Probability of an Event

In a random experiment, the probability of happening of the event A with sample space S is defined as

n ANumber of outcomes favourable to AP A = =

Total number of possible outcomes n S

Probability of an Event (Cont.)

If m is the number of outcomes favourable to an event and n is the total number of possible outcomes. Then,

mP A =

n

Clearly, 0 m n

m0 1

n

0 P A 1

Hence, the probability of an event always lies between 0 and 1.

Probability of an Event (Cont.)

Sure Event: An event A is said to be sure or certain if P(A) = 1

Impossible Event: An event A is said to be impossible if P(A) = 0

Odds in favour of occurrence of an event A are defined as m : n - m, i.e. ratio of favorable outcomes to unfavorable ones.

Odds against occurrence of event A are defined as n - m : m , i.e. ratio of unfavourable outcomes to favourable ones.

Example-3

What is the probability of getting at least two heads in a simultaneous throw of three coins?

Solution: If three coins are tossed together possible outcomes are

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

Number of these exhaustive outcomes, n(S) = 8

At least two heads can be obtained in the following ways

E = {HHH, HHT, HTH, THH}

Solution Cont.

Number of favourable outcomes, n(E) = 4

n E 4 1

Required probability= = =n S 8 2

Example-4

In a single throw of two dice what is the probability of getting

(i) 8 as the sum (ii) a total of 9 or 11

Solution: In throwing of a pair of dice, total number of outcomes in

sample space = 6 × 6 = 36

(i) To get 8 as the sum favourable outcomes are

(2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).

Thus, the number of such outcomes = 5

Solution Cont.

(ii) Favourable outcomes to the event of getting the sum as 9 or 11

are (3, 6), (4, 5), (5, 4), (6, 3), (5, 6) and (6, 5).

Thus, the number of such outcomes = 6

6 1Required probability

36 6

Number of favourable outcomes 5Required probability

Number of total outcomes 36

Example-5

The letters of the word ‘SOCIETY’ are placed at random in a row. What

is the probability that three vowels come together.

Solution: There are 7 letters in the word ‘SOCIETY’.

These 7 letters can be arranged in a row in 7! ways.

O, I, E are three vowels in the word ‘SOCIETY’.

Assuming these three vowels as one letter, we get 5 letters which

can be arranged in a row in 5! ways.

Solution Cont.

The total number of arrangements in which three vowels come

together is 5! × 3!.

5!×3! 1

Required probability = =7! 7

But three vowels O, I, E can be arranged in 3! ways.

Example-6

A five-digit number is formed by the digits 1, 2, 3, 4, 5 without

repetition. Find the probability that the number is divisible by 4.

Solution: Total number of ways in which a five digit number can be

formed by digits 1, 2, 3, 4, 5 = 5!

Exhaustive number of cases=5!=120

A number is divisible by 4 if the numbers formed by last two digits are

divisible by 4.

Solution Cont.

Thus for an outcome to be favorable, the last two digits can be

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

The last two digits can have only these 4 arrangements. But the rest

of the three digits can be arranged in 3! ways.

Number of favourable outcome=3!×4

4 3! 1

5 4 3! 5

4×3!

Required probability =5!

Example-7

A bag contains 8 red and 5 white balls. Three balls are drawn at random.

Find the probability that one ball is red and two balls are white.

Solution: Total number of balls = 8 + 5 = 13

n(S) = number of ways of selecting 3 out of 13 balls

133

13×12×11= C = =286

3×2×1

Let A be the event of selecting one red and 2 white balls out of 8 red

and 5 white balls.

Solution Cont.

8 51 2n A = C × C =8×10=80

P getting one red and two white balls

n A 80 40=P A = = =

n S 286 143

Example-8

A bag contains 50 tickets numbers 1, 2, 3, …50 of which five are

drawn at random and arranged in ascending order of magnitude

30. CBSE 20021 2 3 5 3x <x <x <x . Find the probability that x

ways.505Solution: Five tickets out of 50 can be drawn in C

5 50Total number of elementary events = C

1 2 3 5 3x <x <x <x and x =30

291 2 2x , x <30 and this may happen in C ways.

Solution Cont.

204 5 2x , x >30 and this may happen in C ways.

29 292 2Favourable number of elementary events= C × C

29 292 250

5

C × C 551Required probability = =

C 15134

Example-9

Out of 9 outstanding students in a college, there are 4 boys and 5 girls.

A team of four students is to be selected for a quiz programme. Find the

probability that two are boys and two are girls.

Solution: Out of 9 students 4 students can be selected in 94C ways

Total number of events 94C

There are 4 boys and 5 girls out of which 2 boys and 2 girls can be

selected in 4 52 2= C × C

Solution Cont.

Favourable number of events 4 52 2= C × C

4 5

2 29

4

C × C 10Required probability = =

21C

Example-10

Four cards are drawn at random from a pack 52 playing cards.

Find the probability of getting

(i) all the four cards of the same suit (CBSE 1993)

(ii) all the four cards of the same number (CBSE 1993)

Solution:

(i) There are four suits: club, spade, heart and diamond, each of 13 cards.

Therefore, the total number of ways of getting all the four cards of

the same suit 4 13 13 13 13 134 4 4 4 4C + C + C + C C

Solution Cont.

134

524

4 C 198Required probability = =

C 20825

(ii) Four cards of the same number:

(1, 1, 1, 1), (2, 2, 2, 2), (3, 3, 3, 3), …(13, 13, 13, 13).

Favourable number of events = 13

524

13 13Required probability = =

C 270725

Example –11

Two dice are thrown. Find the odds in favour of getting the sum to be

(i) 4 (ii) 5 (iii) what are the odds against getting the sum to be six.

Solution: The sample space when two dice are thrown is

S = {(1, 1), (1, 2), ... (1, 6), (2, 1), (2, 2), ... (2, 6),

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

(5, 1), (5, 2), ... (5, 6), (6, 1), (6, 2), ... (6, 6)}

n(S) = 36

Solution (Cont.)

(i) Let A be the event of getting the sum on the pair of dice to be 4 A = {(1, 3), (2, 2), (3, 1)}

3 3 1Odds in favour of event A = = =

36- 3 33 11

(ii) Let B be the event of getting the sum on the pair of dice to be 5. B = {(1, 4), (2, 3) (3, 2) (4, 1)}

4 4 1Odds in favour of event B= = =

36- 4 32 8

Solution (Cont.)

(iii) Let C be the event of getting the sum to be six on the pair of dice C = {(1, 5), (2, 4) (3, 3) (4, 2), (5, 1)}

36- 5 31Odds against event C = =

5 5

Thank you