NCERT solutions class x

Probability

Exercise 15.1

Question 1

Complete the following sentence:

(1) Probability of an event E + Probability of an event 'not E' =

Answer; 1

(ii) The probability of an event that cannot happen is such as event is

Answer: 0, Impossible event

(iii) The probability of an event that is certain to happen is such as event is called

Answer: 1, sure event or certain event

(iv) The sum of the probabilities of all the elementary events of an experiment is

Answer: 1

(v) The probability of an event is greater than or equal to and less than or equal to

Answer: 1

Question 2

Which of the following experiments have equally likely ,outcomes? Explain

(i) A driver attempts to start a car .The car starts or does not start.

(ii) A player attempts to shoot a basket ball. She or he shoots or misses the shot.

(iii) A trial is made to answer a True or False question. The answer is Right or Wrong

(iv) A baby is born. It is a boy or girl.

Answer

(i) It is not equally likely event as it depends on various factors such as whether the car will start or

not. And factors for both conditions are not the same.

(ii) It is not an equally likely event ,as it depends on the players ability and there is no information

given about that.

(iii) It is an equally likely event.

( iv) I t is an equally likely event

Question 3

Why is tossing a coin considered to be a fair way of deciding which team

should get the ball at the beginning of a football game?

Answer

When we toss a coin, the possible outcomes are only two ,head or tail, which are

equally likely outcomes .Therefore ,the result of an individual toss is completely

unpredictable.

Question 4

Which of the following cannot be the probability of an event?

( A) 2/3 (B) -1.5 (C) 15% (D) 0.7

Answer

Probability of an event (E) is always greater than or equal to 0 . Also, it is always

less than or equal to one .This implies that the probability of an event cannot be

negative or greater than 1. Therefore, out of these alternatives, -1.5 cannot be a

probability of an event.

Question 5

If p(E) = 0.05. What is the probability of not E?

Answer

We know that

P( not E) = 1- P(E)

P(not E) =1- 0.05

=0.95

Therefore, the probability of not E is 0.95

Question 6

A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag.

What is the probability that she takes out

(i) an orange flavored candy?

(ii) a lemon flavored candy?

Answer

(i) The bag contains lemon flavored candies only. It does not contains any

orange flavored candies

This implies that every time, she will take out only lemon flavored candies .

Therefore, event that Malini will takeout an orange flavored candy is an

impossible event .

Hence P (an orange flavored candy ) = 0

(ii) As the bag has lemon flavored candies, Malini will take out only lemon

flavored candies. Therefore ,event that Malini will take out a lemon flavored candy

is a sure event.

P(a lemon flavored candy) = 1

Question 7

It is given that in a group of 3 students ,the probability of 2 students not

having the same birthday is 0.992. What is the probability that the 2 students

have the same birthday?

Answer

Probability that two students are not having the same birthday p(not E) =0.992

Probability that two students are having same birthday p(E) = 1-- P(not E)

= 1- 0.992

= 0.008

Question 8

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from

the bag. What is the probability that ball is

(i) red

(ii) not red?

Answer

Total number of balls in the bag = 8

(i) Number of red ball=5

probability of getting a red ball = No. of favourable outcomes

No. of total possible outcomes

= 3/8

(ii) probability of not getting a red ball = 1 - p(getting a red ball)

= 1- 3/8

= 5/8

Question 9

A box contains 5 red marbles, 8 white marbles and 4 green marbles . One

marble is taken out of the box at random. What is the probability that the

marble taken out will be

(i) red

(ii) white

(iii) not green?

Answer

Total number of marbles = 5+8+4

= 17

(i) Number of red marbles =5

probability of getting a red marble = No. of favorable outcomes

No.of total possible outcomes

= 5/17

(ii) probability of getting white marbles = No of favorable outcomes

No of total possible outcomes

= 8 /17

(iii)probability of getting green marbles = No of favorable outcomes

No of total possible outcomes

= 4/17

so ,

No of not getting a green marbles = 1- p(getting a green marbles)

=1 - 4/17

= 13/17

Question 10

A piggy bank contains hundred 50 paise coins , fifty Rs 1 coins, twenty Rs 2

coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall

out when the bank is turned upside down, what is the probability that the coin

(I) Will be a 50 p coin

(ii) Will not be a Rs 5 coin

Answer

Total no. of coins in a piggy bank = 100+ 50+ 20+ 10

=180

(i) Number of 50 paise coins = Number of favorable outcomes

Number of total possible outcomes

=100/ 180

=5/9

(ii)Number of Rs 5 coins = 10

probability of getting a Rs 5 coin = Number of favorable outcomes

Number of total possible outcomes

=10/180

= 1/18

probability of getting a Rs 5 coin = 1 - 1/18

= 17 /18

Question 11

Gobi buys a fish from a shop for his aquarium .The shopkeeper takes out

one fish at random from a tank containing 5 male fish and 8 female fish (see

the given figure ). What is the probability that the fish taken out is male fish?

Answer

Total no .of fishes in a tank = No of male fishes + No. of female fishes

= 5 +8

= 13

Probability of getting a male fish = No. of favorable outcomes

No .of total possible outcomes

= 5/13

Question 12

A game of chance consisting of spinning an arrow which comes to rest

pointing at one of the numbers1,2,3,4,5,6,7,8 ( see the given figure) and these

are equally likely outcomes . What is the probability that it will point at

(i)8 ?

(ii) an odd number

(iii) a number greater than 2?

(iv) a number less than 9 ?

Answer

Total no. of possible outcomes = 8

(i) probability of getting 8 = No. of favorable outcomes

No. of total possible outcomes

=1/8

(ii)Total no. of odd numbers on spinner =4

probability of getting an odd number = No. of favorable outcomes

No. of total possible outcomes

= 4/8

(iii) Total no. of getting a number greater than 2 = 6

probability of getting a number greater than 2 = 6/8

=3/4

(iv) Total no. of getting a number less than 9 =8

probability of getting a number less than 9 = No. of favorable outcomes

No .of total possible outcomes

= 8/8

Question 13

A die is thrown once . Find the probability of getting

(i) a prime number

(ii)a number lying between 2 and 6

(iii) an odd number

Answer

The possible outcomes when a die is thrown = [1,2,3,4,5,6 ]

Number of possible outcomes of a dice = 6

(i) prime numbers on dice are 2, 3 and 5

so total prime numbers are 3

probability of getting a prime number = No. of favorable outcomes

No. of total possible outcomes

= 3/6

= 1/2

(ii) Numbers lying between 2 and 6 are 3,4,5

so total numbers between 2 and 6 are 3

probability of getting a number lying between 2 and 6 = No. of favorable outcome

No. of total possible

= 3/6

=1/2

(iii) odd numbers on dice are 1, 3 and 5

so total odd numbers are 3

probability of getting odd numbers = No. of favorable outcomes

No. of total possible outcomes

= 3/6

= 1/2

Question 14

One card is drawn from a well shuffled deck of 52 cards. Find the probability of

getting

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

Answer

Total no. of cards in a well shuffled deck =52

(i) Total no. of kings of red colour =2

p(getting a king of red colour) =No .of favorable outcomes

No . of total possible outcomes

=2/52

=1/26

(ii) Total no. of face cards =12

p(getting a face card) = No. of favorable outcomes

No. of total possible outcomes

=12/52

=3/13

(iii) Total no. of red face cards =6

p(getting a red face card) No. of favorable outcomes

No. of total possible outcomes

=6/52

=3/26

(iv) Total no. of jack of hearts =1

p(getting a jack of hearts) = No.of favorable outcomes

No. of total possible outcomes

=1 /52

(v) Total no. of spades = 13

p (getting a spade card) = No. of favorable outcomes

No. of total possible outcomes

=13/52

=1/4

(vi) Total no. queen of diamonds =1

p(getting a queen of diamonds) = No. of favorable outcomes

No .of total possible outcomes

=1/52

Question 15

Five cards, the ten, jack, queen, king and ace of diamonds, are well-shuffled

with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the

second card picked up is

(a) an ace?

(b) a queen?

Answer

Total number of cards = 5

(i) Number of queens = 1

P (picking a queen) = 1/5

= 0.2

(ii) If the queen is drawn and put aside, the total number of cards left is (5-4) = 4

(a) Total numbers of ace = 1

P (picking an ace) = ¼ = 0.25

(b) Total number of queens = 0

P (picking a queen) = 0/4 = 0

Question 16

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

taken out is a good one.

Answer

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

∴ Total number of pens = 132+12

= 144 pens

P(picking a good pen) = No .of favorable outcomes

Total number of outcomes

= 132/144

= 11/12

= 0.916

Question 17

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random

from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now

one bulb is drawn at random from the rest. What is the probability that this

bulb is not defective?

Answer

(i) Number of defective bulbs = 4

The total number of bulbs = 20

P(getting a defective bulb) =No. of favorable outcome

Total number of outcomes

= 4/20

= 1/5= 0.2

(ii) Since 1 non-defective bulb is drawn, then the total number of bulbs left is 19

So, the total number of events (or outcomes) = 19

No. of non-defective bulbs = 19-4 = 15

Therefore, the probability that the bulb is not defective = 15/19 = 0.789

Question 18

A box contains 90 discs which are numbered from 1 to 90. If one disc

is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5

Answer

The total number of discs = 90

(i) No. of discs having two digit numbers = 81

Since 1 to 9 are single-digit numbers and

so, total 2-digit numbers are 90-9 = 81

P (bearing a two-digit number) = No. of favorable outcomes

Total no. of outcomes

=81/90

= 9/10 = 0.9

(ii) No of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)

P (getting a perfect square number) = No. of favorable outcomes

Total no. of outcomes

=9/90

= 1/10 = 0.1

(iii) Numbers which are divisible by 5 = 18

(5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) = No. of favorable outcomes

Total no . of outcomes

=18/90

= 1/5 = 0.2

Question 19

A child has a die whose six faces show the letters as given below:

Ncert solutions class 10 chapter 15-3

The die is thrown once. What is the probability of getting

(i) A?

(ii) D?

Answer

The total number of possible outcomes (or events) = 6

(i) No. of faces having A on it = 2

P (getting A) = No. of favorable outcomes

Total no. of outcomes

=2/6

=1/3

= 0.33

(ii) The total number of faces having D on it = 1

P (getting D) =No. of favorable outcomes

Total no. of outcomes

=1/ 6

= 0.166

Question 20

Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with a diameter 1m?

Answer

First, calculate the area of the rectangle and the area of the circle

Area of the rectangle = (3×2) m2

= 6 m2

Area of the circle = πr2

= π(½)2 m2

= π/4 m2 = 0.78

∴ The probability that die will land inside the circle = [(π/4)/6]

= π/24

= 0.78/6 = 0.13

Question 21

A lot consists of 144 ball pens, of which 20 are defective, and the others are

good. Nuri will buy a pen if it is good, but will not buy it if it is defective. The

shopkeeper draws one pen at random and gives it to her. What is the

probability that

(i) She will buy it?

(ii) She will not buy it?

Answer

The total number of pens = 144

No. of defective pens = 20

∴ The No. of non defective pens = 144-20 = 124

(i) No .of of events in which she will buy them = 124

P (buying) = No .of favorable outcomes

Total no. of outcomes

=124/144

= 31/36 = 0.86

(ii) Number of events in which she will not buy them = 20

P (not buying) = No. of favorable outcomes

Total no. of outcomes

=20/144

= 5/36 = 0.138

Question 22

Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9,

10, 11 and 12. Therefore, each of them has a probability of 1/11. Do you agree

with this argument? Justify your Solution:

Answer

If 2 dice are thrown, the possible events are:

(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)

Total number of events: 6×6 = 36

(i) It is given that to get the sum as 2, the only possible outcomes = (1,1)

p(getting sum 2) = No. of favorable outcomes

Total no. of outcomes

= 1/36

Getting the sum as 3, the possible events (or outcomes) = (1,2) and (2,1)

P( Getting sum 3) =No. of favorable outcomes

Total no. of outcomes

= 2/36

Similarly,

Getting the sum as 4= (1,3), (3,1), and (2,2)

P (getting sum 4) =No. of favorable outcomes

Total no. of outcomes

=3/36

Getting sum as 5 =(1,4), (4,1), (2,3), and (3,2)

P (getting the sum 5) = No. of favorable outcomes

Total no. of outcomes

=4/36

Getting sum as 6 = (1,5), (5,1), (2,4), (4,2), and (3,3)

P (getting the sum 6) = No. of favorable outcomes

Total no. of outcomes

= 5/36

Getting the sum as 7 = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

P (getting the sum 7) = No. of favorable outcomes

Total no. of outcomes

= 6/36

Getting the sum as 8 = (2,6), (6,2), (3,5), (5,3), and (4,4)

P ( getting the sum 8) =No. of favorable outcomes

Total no. of outcomes

= 5/36

Getting the sum as 8 = (3,6), (6,3), (4,5), and (5,4)

P (getting the sum 9) =No. of favorable outcomes

Total no. of outcomes

=4/36

Getting the sum as 10 = (4,6), (6,4), and (5,5)

P ( getting the sum 10) = No. of favorable outcomes

Total no. of outcomes

= 3/36

Getting the sum as 11= (5,6), and (6,5)

P ( getting the sum 11) = No. of favorable outcomes

Total no. of outcomes

= 2/36

Getting the sum as 12= (6,6)

P ( getting the sum 12) =No. of favorable outcomes

Total no. of outcomes

= 1/36

So, the table will be as:

Event: Sum on 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36

(ii) The argument is not correct as it is already justified in (i) that the number of

all possible outcomes is 36 and not 11.

Question 23

A game consists of tossing a one-rupee coin 3 times and noting its outcome

each time. Hanif wins if all the tosses give the same result, i.e. three heads or

three tails, and loses otherwise. Calculate the probability that Hanif will lose

the game.

Answer

The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH,

HTT, THT)

P (losing the game) = No. of favorable outcomes

Total no. of outcomes

= 6/8

=3/4 = 0.75

Question 24

A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the

same experiment]

Answer

Outcomes are:

(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)

So, the total number of outcomes = 6×6 = 36

(i)Total no. of outcomes when 5 comes up on either time are; (5,1),(5,2),(5,3),(5,4)

(5,5),(5,6)

Hence total no. of favourable cases =11

P(5 will come up either time) =11/36

P(5 will not come up either time) =1 -11/36

=25/36

(ii) Number of events when 5 comes at least once = 11

∴ The required probability = 11/36

Question 25

Which of the following arguments are correct and which are not correct?

Give reasons for your solution:

(i) If two coins are tossed simultaneously, there are three possible outcomes

– two heads, two tails or one of each. Therefore, for each of these outcomes,

the probability is 1/3

(ii) If a die is thrown, there are two possible outcomes – an odd number or an

even number. Therefore, the probability of getting an odd number is 1/2

Answer

(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)

P (getting two heads) = 1/4

P (getting one of each) = 2/4

=1/2

∴ This statement is incorrect.

(ii) Since the two outcomes are equally likely, this statement is correct.

Exercise: 15.2 (Page No: 311)

Question 1

Two customers, Shyam and Ekta, are visiting a particular shop in the same

week (Tuesday to Saturday). Each is equally likely to visit the shop on any

day as on another day. What is the probability that both will visit the shop on

(i) the same day?

(ii) consecutive days?

(iii) different days?

Answer

Since there are 5 days and both can go to the shop in 5 ways each so,

The total number of possible outcomes = 5×5 = 25

(i) The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu.,

Thu.), (Fri., Fri.), (Sat., Sat.)

P (both visiting on the same day) = No. of favorable outcomes

Total no. of outcomes

=5/25

= 1/5

(ii) The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu.,

Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)

P(both visiting on the consecutive days) = No. of favorable outcomes

Total no. of outcomes

= 8/25

(iii) P (both visiting on different days) = 1-P (both visiting on the same day)

P (both visiting on different days) = 1-(⅕)

=4/5

Question 2

A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3,

6. It is thrown two times, and the total score in two throws is noted. Complete

the following table, which gives a few values of the total score on the two

throws:

What is the probability that the total score is

(i) even?

(ii) 6?

(iii) at least 6?

Answer

The table will be as follows:

+	1	2	2	3	3	6
1	2	3	3	4	4	7
2	3	4	4	5	5	8
2	3	4	4	5	5	8
3	4	5	5	6	6	9
3	4	5	5	6	6	9
6	7	8	8	9	9	12

So, the total number of outcomes = 6×6 = 36

(i) Getting an even = 18

P (Even) = No. of favorable outcomes

Total no. of outcomes

=18/36

= ½

(ii) Getting the sum 6= 4

P (sum is 6) = No. of favorable outcomes

Total no. of outcomes

= 4/36 = 1/9

(iii) Getting the sum is atleast 6 = 15

P (sum is atleast 6) = No. of favorable outcomes

Total no. of outcomes

=15/36 = 5/12

Question 3

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, determine the number of blue balls in

the bag.

Answer

It is given that the total number of red balls = 5

Let the total number of blue balls = x

∴ P (drawing a blue ball) =No. of favorable outcomes

Total no. of outcomes

= [x/(x+5)] ——–(i)

Similarly,

P (drawing a red ball) = No. of favorable outcomes

Total no. of outcomes

= [5/(x+5)] ——–(i)

From equations (i) and (ii)

x = 10

So, the total number of blue balls = 10

Question 4

A box contains 12 balls, out of which x are black. If one ball is drawn at

random from the box, what is the probability that it will be a black ball?

If 6 more black balls are put in the box, the probability of drawing a black ball

is now double of what it was before. Find x

Answer

Number of black balls = x

Total number of balls = 12

P (getting black balls) = No. of favorable outcomes

Total no. of outcomes

= x/12 ——————-(i)

Now, when 6 more black balls are added,

Total balls become = 18

∴ Total number of black balls = x+6

Now, P (getting black balls) =No. of favorable outcomes

Total no. of outcomes

= (x+6)/18 ——————-(ii)

It’s given that the probability of drawing a black ball now is double of

what it was before.

(x+6)/18 = 2 × (x/12)

x + 6 = 3x

2x = 6

∴ x = 3

Question 5

A jar contains 24 marbles, some are green, and others are blue. If a marble is

drawn at random from the jar, the probability that it is green is ⅔. Find the

number of blue balls in the jar.

Answer

Total marbles = 24

Let the green marbles = x

So, the blue marbles = 24-x

P(getting green marble) = No. of favorable outcomes

Total no. of outcomes

= x/24

From the question, x/24 = 2/3

So, the total green marbles = 16

And, the total blue marbles = 24-16 = 8

Introduction

Probability is a branch of mathematics that deals with the likelihood or chance of events occurring. In the Central Board of Secondary Education (CBSE) Class 10 curriculum, probability is introduced as an essential topic in the field of statistics. This article aims to provide an overview of probability for Class 10 CBSE students, covering key concepts, methods of calculation, and practical applications.

Understanding Probability

Sample Space: In probability, a sample space refers to the set of all possible outcomes of an experiment. It is denoted by the symbol "S." For example, when tossing a fair coin, the sample space consists of two outcomes: heads (H) and tails (T). Understanding the sample space is crucial for calculating probabilities accurately.

Events:

An event is a subset of the sample space, representing one or more outcomes of interest. It is denoted by a capital letter. For example, in the coin-tossing experiment, the event of getting a head can be denoted as "E," and the event of getting a tail can be denoted as "F."

Probability of an Event:

The probability of an event is a numerical measure of the likelihood of that event occurring. It is denoted by P(E), where E represents the event. The probability of an event lies between 0 and 1, inclusive.

Methods of Calculating Probability

Theoretical Probability:

Theoretical probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes in a sample space. It is based on assumptions and ideal conditions. For example, the theoretical probability of getting a head when tossing a fair coin is 1/2.

Experimental Probability:

Experimental probability is determined by conducting an experiment and observing the frequency of the desired event occurring. It is calculated by dividing the number of times the event occurs by the total number of trials. As the number of trials increases, the experimental probability tends to approach the theoretical probability.

Addition Rule of Probability:

The addition rule is used to calculate the probability of the union of two or more events. If E and F are two events, the probability of either E or F occurring is given by P(E or F) = P(E) + P(F) - P(E and F).

Multiplication Rule of Probability: The multiplication rule is used to calculate the probability of the intersection of two or more events. If E and F are two events, the probability of both E and F occurring is given by P(E and F) = P(E) × P(F|E), where P(F|E) represents the probability of F given that E has already occurred.

Practical Applications of Probability

Games of Chance:

Probability is fundamental in analyzing and predicting outcomes in games of chance, such as rolling dice, drawing cards, or spinning a roulette wheel. Understanding probability can help students make informed decisions and strategies while playing such games.

Risk Assessment:

Probability is used in risk assessment to quantify the likelihood of different events occurring. It is applied in various fields, including insurance, finance, and safety analysis, to estimate and manage potential risks.

Weather Forecasting:

Weather forecasting relies on probabilistic models to predict the likelihood of specific weather conditions. By understanding probability, students can appreciate the uncertainties involved in weather predictions and interpret probabilistic forecasts effectively.

Opinion Polls and Surveys:

Probability is used to analyze data from opinion polls and surveys. It helps in estimating the likelihood of certain responses, predicting outcomes, and drawing conclusions about a larger population based on a sample.

Conclusion

Probability is a powerful tool for analyzing and quantifying the likelihood of events. By understanding the concepts of sample space,