Page 1 of 13

BISHOP SCOTT BOYS’ SCHOOL

(Affiliated to CBSE, New Delhi) Affiliation No.: 330726, School Campus: Chainpur, Jaganpura, By-Pass, Patna 804453. Phone Number: 7061717782, 9798903550. , Web: www.bishopscottboysschool.comEmail: info@bishopscottboysschool.com

STUDY COURSE MATERIAL

MATHEMATICS

SESSION-2020-21

CLASS- X

TOPIC: PROBABILITY

DAY-1

NCERT MATERIAL

http://www.ncert.nic.in/ncerts/l/iemh105.pdf

TEACHING MATERIAL

PROBABILITY: -

Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed between zero and one. Probability has been introduced in Maths to predict how likely events are to happen.

The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory which is also used in the probability distribution, where you will learn the possibility of outcomes for a random experiment. To find the probability of a single event to occur, first we should know the total number of possible outcomes.

Types of Probability

There are three major types of probabilities:

Theoretical Probability

Experimental Probability

Axiomatic Probability

Theoretical Probability

Page 2 of 13

It is based on the possible chances of something to happen. The theoretical probability is mainly based on the reasoning behind probability. For example, if a coin is tossed, the theoretical probability of getting head will be ½.

Experimental Probability

It is based on the basis of the observations of an experiment. The experimental probability can be calculated based on the number of possible outcomes by the total number of trials. For example, if a coin is tossed 10 times and heads is recorded 6 times then, the experimental probability for heads is 6/10 or, 3/5.

Axiomatic Probability

In axiomatic probability, a set of rules or axioms are set which applies to all types. These axioms are set by Kolmogorov and are known as Kolmogorov’s three axioms. With the axiomatic approach to probability, the chances of occurrence or non-occurrence of the events can be quantified. The axiomatic probability lesson covers this concept in detail with Kolmogorov’s three rules (axioms) along with various examples.

Some terms related to probability

EXPERIMENT: An operation which can produce some well defined outcomes is known as an experiment.

RANDON EXPERIMENT: An experiment whose all possible outcomes are known, but exact outcome cannot be predicted in advance.

Ex: - Tossing a coin, throwing a die, taking a card from a pack of playing card etc.

TRIAL: Performing of an experiment is called Trial. For example : Tossing a coin, throwing a dice.

EVENT: The outcomes of an experiment are called events. For example

Getting a head or tail tossing a coin is an Event.

EQUALLY LIKELY EVENT: Outcomes of trial are said to be equally likely if taking into consideration all the relevant evidences, there is no reason to expect one in preference to the others. For example,

(a) In throwing an unbiased die, all the six faces are equally likely to come.

ELEMENTARY EVENT: An event having only one outcome is called an elementary event.

REMARK: The sum of the probabilities of all the elementary events of an experiment is 1.

VIDEO-LINKS

https://youtu.be/hXeNr5GjicM

Page 3 of 13

DAY-2

TEACHING MATERIAL

Sample Space

A sample space is the set of all possible outcomes in the experiment. It is usually denoted by the letter S. Sample space can be written using the set notation, { }.

An outcome is a possible result of an experiment.

Total no. of outcomes in different trials = (no. of outcomes in one trial) no. of trials

NOTE: - Event is the part of sample space

Sample space of some common experiments

Experiment 1: Tossing a coin

Possible outcomes are head or tail.

Sample space, S = {head, tail}.

n(s) = 2

Experiment 2: Tossing of two coins

Possible outcomes are {(H, T)(T, H)(T, T)(H,H)}.

n(s) = 22 = 4

Experiment 3: Tossing of three coins

Possible outcomes are {(H, T, T)(T, T, H)(T, H, T)(H, H, T)

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

n(s) = 23 = 8

Page 4 of 13

Experiment 4: Throwing a die

Possible outcomes are the numbers 1, 2, 3, 4, 5, and 6

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

n(s) = 6

Experiment 5: Throwing two dice

Possible outcomes are the pair of numbers

(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)

n(s) = 62 = 36

Page 5 of 13

VIDEO-LINKS

https://youtu.be/CO95EP4vYI0

DAY-3

TEACHING MATERIAL

Experiment 5: - Description Playing cards

Page 6 of 13

There are 52 cards in a pack of Playing cards. It is in two colours, Red and Black. There are 26 red coloured and 26 black coloured cards. Red coloured card divided into two suits i.e

Diamond and Heart

Black coloured cards also divided into two suits i.e

Spades and Cub

So, there are 13 cards each of Diamond, Heart, Spades and Club

They are: - Ace, King, Queen, Jack, 2, 3, 4, 5, 6, 7, 8, 9 and 10

Face cards:- King, Queen and Jack are called face card as some picture is made over there. There are 12 face cards in a pack of playing card.

Honor cards: - Ace, King, Queen and Jack are called nonor card as they have some special

values in some games. There are 16 honor cards in a pack of playing card.

VIDEO-LINKS

1. https://youtu.be/6UwszxZhvHc

2. https://youtu.be/XLKC3N-rfkI

DAY-4

TEACHING MATERIAL

Tree Diagram

Page 7 of 13

Tree diagram of an experiments is used to write the sample space when an experiment is repeated several times.

Some example of tree diagram are

1. When two coins are tossed

Sample space are {(H, T)(T, H)(T, T)(H,H)}

2. When three coins are tossed simultenously.

Sample space are {(H, T, T)(T, T, H)(T, H, T)(H, H, T) (H, T, H)(T, H, H)(H, H, H)(T, T, T)}

Suppose a closet has three pairs of pants (black, white, and green), four shirts (green, white,

purple, and yellow), and two pairs of shoes (black and white). How many different outfits can be

made? There are 3 choices for pants, 4 choices for shirts, and 2 choices for shoes. For our tree

diagram, let's use B for black, W for white, G for green, P for purple, and Y for yellow.

Page 8 of 13

Total combinations are (B,G,W)(B,G,B)(B,W,W)(B,W,B)(B,P,W)(B,P,B)(B,Y,W)(B,Y,B)

(W,G,W)(W,G,B)(W,W,W)(W,W,B)(W,P,W)(W,P,B)(W,Y,W)(W,Y,B)

(G,G,W)(G,G,B)(G,W,W)(G,W,B)(G,P,W)(G,P,B)(G,Y,W)(G,Y,B)

Total no. of combinations = 3×4×2 = 24

Empirical formula of probability

Where n = total number of sample space

m = number of favorable event

NOTE

Favourable outcomes are those outcomes in the sample space that are favourable to the occurrence of an event.

If P(E) = 1, then it is called a ‘Certain Event’ or ‘Sure Event’. If P(E) = 0, then it is called an ‘Impossible Event’. The probability of an event E is a number P(E) such that: 0 ≤ P(E) ≤ 1 An event having only one outcome is called an elementary event. The sum of the

probabilities of all the elementary events of an experiment is 1.

For any event E, P(E) + P( ) = 1, where stands for ‘not E’. E and are called complementary events.

Some Solved example

Page 9 of 13

Question 1: What is the probability of each outcome when a coin is tossed?

Solution: - The sample space of is: {head, tail}

n(s) = 2

Probabilities:

P(head) = 1 2

P(tail) = 1 2

Question 2: 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen is taken out is a good one.

Solution:

Numbers of pens = Numbers of defective pens + Numbers of good pens

∴ Total number of pens = 132 + 12 = 144 pens

P(E) = (Number of favourable outcomes) / (Total number of outcomes)

P(picking a good pen) = 132/144 = 11/12 = 0.916

Question 3: A dice is thrown once. What is the probability of getting a number greater than 4?

Solution

Total possible outcome = 6

Favourable outcome (5,6)

No. of favourable outcome = 2

Probability = 2/6= 1/3

Question 4: Cards with numbers 2 to 101 are placed in a box. A card selected at random from the

box. Find the probability that the card which is selected has a number which is a perfect square.

Solution

Total number = 100

Favourable outcome ( 4,9,16,25,36,49,64,81,100)

No. of favourable outcome = 9

Probability = 9/100

Question 5: A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting

an ace.

Solution

Page 10 of 1

3

Total number = 52

Favourable outcome = 4

Probability = 4/52 = 1/13

Question 6: Find the probability that a leap year selected randomly will have 53 Sundays?

Solutionp

No. of days in a leap year = 366 days = 52 weeks + 2 days

It implies a leap year will have 52 Sundays. In remaining 2 days, possible outcomes are:

(Sun, Mon) (Mon, Tue) (Tue, Wed) (Wed, Thu) (Thu, Fri) (Fri, Sat) (Sat, Sun)

Total out comes = 7

Favourable outcomes that Sunday will come in these two days = 2

Required probability = 27

Question 7: Two dice are thrown simultaneously. Find the probability of: (i) getting six as a product (ii) getting sum ≤ 3 (iii) getting sum ≤ 10 (iv) getting a doublet (v) getting a sum of 8 (vi) getting sum divisible by 5 (vii) getting sum of at least 11 (viii) getting a total of at least 10 (x) getting a prime number as the sum (xii) getting a multiple of 2 on one die and a multiple of 3 on the other die

Solution:

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36.

(i) getting six as a product:

Let E1 = event of getting six as a product. The number whose product is six will be E1= [(1, 6), (2, 3), (3, 2), (6, 1)] = 4 Therefore, probability of getting ‘six as a product’

Number of favorable outcomes P(E1) = Total number of possible outcome = 4/36 = 1/9 (ii) getting sum ≤ 3:

Let E2 = event of getting sum ≤ 3. The number whose sum ≤ 3 will be E2 = [(1, 1), (1, 2), (2, 1)] = 3 Therefore, probability of getting ‘sum ≤ 3’

Number of favorable outcomes P(E2) = Total number of possible outcome = 3/36

Page 11 of 1

3

= 1/12 (iii) getting sum ≤ 10:

Let E3 = event of getting sum ≤ 10. The number whose sum ≤ 10 will be E3 = [(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),

(6, 1), (6, 2), (6, 3), (6, 4)] = 33

Therefore, probability of getting ‘sum ≤ 10’

Number of favorable outcomes P(E3) = Total number of possible outcome = 33/36 = 11/12 (iv) getting a doublet: Let E4 = event of getting a doublet. The number which doublet will be E4 = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6 Therefore, probability of getting ‘a doublet’

Number of favorable outcomes P(E4) = Total number of possible outcome = 6/36 = 1/6 (v) getting a sum of 8:

Let E5 = event of getting a sum of 8. The number which is a sum of 8 will be E5 = [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)] = 5 Therefore, probability of getting ‘a sum of 8’

Number of favorable outcomes P(E5) = Total number of possible outcome = 5/36 (vi) getting sum divisible by 5:

Let E6 = event of getting sum divisible by 5. The number whose sum divisible by 5 will be E6 = [(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)] = 7 Therefore, probability of getting ‘sum divisible by 5’

Number of favorable outcomes P(E6) = Total number of possible outcome = 7/36

Page 12 of 1

3

(vii) getting sum of atleast 11:

Let E7 = event of getting sum of atleast 11. The events of the sum of atleast 11 will be E7 = [(5, 6), (6, 5), (6, 6)] = 3 Therefore, probability of getting ‘sum of atleast 11’

Number of favorable outcomes P(E7) = Total number of possible outcome = 3/36 = 1/12 (viii) getting a total of atleast 10:

Let E9 = event of getting a total of atleast 10. The events of a total of atleast 10 will be E9 = [(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)] = 6 Therefore, probability of getting ‘a total of atleast 10’

Number of favorable outcomes P(E9) = Total number of possible outcome = 6/36 = 1/6 (x) getting a prime number as the sum:

Let E11 = event of getting a prime number as the sum. The events of a prime number as the sum will be E11 = [(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)] = 15 Therefore, probability of getting ‘a prime number as the sum’

Number of favorable outcomes P(E11) = Total number of possible outcome = 15/36 = 5/12 (xii) getting a multiple of 2 on one die and a multiple of 3 on the other die:

Let E13 = event of getting a multiple of 2 on one die and a multiple of 3 on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will be E13 = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11 Therefore, probability of getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’

Number of favorable outcomes

P(E13) = Total number of possible outcome

= 11/36

VIDEO-LINKS

1. https://youtu.be/d6G-mAKQ_uo

Page 13 of 1

3

2. https://youtu.be/mkDzmI7YOx0

DAY-5

HOME ASSINGMENT

1. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is

double that of a red ball, find the number of blue balls in the bag.

2. Find the probability of getting a number less than 5 in a single throw of a die.

3. A card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. What

are the odds against his winning this bet?

4. A bag contains 12 balls out of which x are white.(i) If one ball is drawn at random, what is

the probability that it will be a white ball?(ii) If 6 more white balls are put in the bag, the

probability of drawing a white ball will be double than that in case (i). Find x.

5. Five male and three female candidates are available for selection as a manager in a

company. Find the probability that male candidate is selected.

6. If P(E) = 0.05, what is the probability of ‘not E’?

7. An integer is chosen between 0 and 100. What is the probability that it is

(i) divisible by 7?

(ii) not divisible by 7?

8. A coin is tossed two times. Find the probability of getting at most one head.

9. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that

the card will