# RS Aggarwal Solutions Class 9 Probability

The likelihood of an event occurring is called Probability and it is always represented quantifiably between 0 and 1. Using a coin is a prime example of Probability, Heads and Tails are the two outcomes which may occur and the probability of getting one or the other is 50/50. This can be represented mathematically as either a percentage which is 50% or as a fraction ‘½’. Some of the major applications of Probability are Behavioral finance, Financial Regulation as well as Natural Language Processing.

The different types of events which are related to Probability are:

• Independent events
• Not mutually exclusive events
• Mutually exclusive events

Check out the RS Aggarwal Class 9 Solutions Chapter 15 Probability available below:

## RS Aggarwal Solutions Class 9 Chapter 15

Q.1: A coin is tossed 500 times and we get, Head =285 times, Tail =215 times.

When a coin is tossed at random, what is the probability of getting:

ii) a tail?

Sol:

When a coin is tossed the outcomes are either head or tail

Total number of trails is 500

i.e T=500

i) Number of times head appears = 285

Hence, the probability of getting head (P1) = numberoftimesheadappearstotalnumberoftrails$\frac{number\, of\, times\, head\, appears}{total\, number\, of\, trails}$ = 285500$\frac{285}{500}$ = 57100$\frac{57}{100}$ = 0.57

ii) Number of times tail appears = 215

Hence, the probability of getting tail (P2) = numberoftimestailappearstotalnumberoftrails$\frac{number\, of\, times\, tail\, appears}{total\, number\, of\, trails}$ = 215500$\frac{215}{500}$ = 43100$\frac{43}{100}$= 0.43

Q.2: Two coins are tossed 400 times and we get:

Two heads: 112 times; one head: 160 times; 0 head:128 times.

When two coins tossed at random, what is the probability of getting:

Sol:

Total number of trails is 400 i.e. T = 400

i) Number of timesÂ 2 heads appears = 112 times

Hence, the probability of getting 2 heads (P2)= numberoftimes2headsappearstotalnumberoftrails$\frac{number\, of\, times\, 2heads\, appears}{total\, number\, of\, trails}$ = 112400$\frac{112}{400}$Â = 725$\frac{7}{25}$= 0.28

ii) Number of times 1 head appears = 160 times

Hence, the probability of getting 1 head (P1) = numberoftimes1headappearstotalnumberoftrails$\frac{number\, of\, times\, 1head\, appears}{total\, number\, of\, trails}$ = 160400$\frac{160}{400}$ = 25$\frac{2}{5}$= 0.4

iii) Number of timesÂ 0 heads appears = 128 times

Hence, the probability of getting Â 2 heads (P2)= numberoftimes0headsappearstotalnumberoftrails$\frac{number\, of\, times\, 0heads\, appears}{total\, number\, of\, trails}$ = 128400$\frac{128}{400}$ = 825$\frac{8}{25}$= 0.32

Q.3: Two coins are tossed 200 times and we get:

Three heads: 39 times; Two heads: 58 times; one head :67 times; 0 head: 36 times.

When two coins are tossed at random, what is the probability of getting:

Sol:

Total number of trails is 200 i.e. T=200

i). Number of timesÂ 3 heads appears = 39 times

Hence, the probability of getting Â 3 heads (P3)= numberoftimes3headsappearstotalnumberoftrails$\frac{number\, of\, times\, 3heads\, appears}{total\, number\, of\, trails}$ = 39200$\frac{39}{200}$ = 0.195

ii) Number of times 2 head appears = 58 times

Hence, the probability of getting 2 head (P2)= numberoftimes2headappearstotalnumberoftrails$\frac{number\, of\, times\, 2head\, appears}{total\, number\, of\, trails}$ = 58200$\frac{58}{200}$ = 29100$\frac{29}{100}$= 0.335

iii) Number of times 1 heads appears = 67 times

Hence, the probability of getting 1 heads (P2)= numberoftimes1headsappearstotalnumberoftrails$\frac{number\, of\, times\, 1heads\, appears}{total\, number\, of\, trails}$ = 67200$\frac{67}{200}$ = 0.18

iv) Number of times 0 head appears = 36 times

Hence, the probability of getting 0 head (P2)= numberoftimes2headappearstotalnumberoftrails$\frac{number\, of\, times\, 2head\, appears}{total\, number\, of\, trails}$ = 36200$\frac{36}{200}$ = 950$\frac{9}{50}$= 0.29

Q.4: A die is thrown 300 times and the outcomes are noted as given below.

 Outcome 1 2 3 4 5 6 Frequency 60 72 54 42 39 33

When a die is thrown at random, what is the probability of getting:

i) 3? Â Â Â Â Â Â ii) 6? Â Â Â iii) 5? Â Â Â Â Â Â iv) 1? Â Â

Sol:

When a die is thrown then outcomes are (1, 2, 3, 4, 5, 6)

Total number of trails is 300 i.e. T=300

i) Frequency for occurrence of 3 = 54 times

Hence, the probability of getting 3 (P3) = numberoftimes3occurstotalnumberoftrails$\frac{number\, of\, times\, 3\,occurs }{total\, number\, of\, trails}$ = 54300$\frac{54}{300}$ = 0.18

ii) Frequency for occurrence of Â 6 Â = 33 times

Hence, the probability of getting 6 (P6)= numberoftimes6occurstotalnumberoftrails$\frac{number\, of\, times\, 6\,occurs }{total\, number\, of\, trails}$ = 33300$\frac{33}{300}$ = 0.11

iii) Frequency for occurrence of Â 5 Â = 39 times

Hence, the probability of getting 5 (P5)= numberoftimes5occurstotalnumberoftrails$\frac{number\, of\, times\, 5\,occurs }{total\, number\, of\, trails}$ = 39300$\frac{39}{300}$ = 0.13

iv) Frequency for occurrence of Â 1 Â = 60 times

Hence, the probability of getting 1 (P1)= numberoftimes1occurstotalnumberoftrails$\frac{number\, of\, times\, 1\,occurs }{total\, number\, of\, trails}$ = 60300$\frac{60}{300}$ = 0.2

Q.5: In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.

Find the probability that a lady chosen at random

i) Like coffee Â Â Â Â Â Â Â ii) dislike coffee

Sol:

Total number of ladies = 200

Let, L1=lady likes coffee and L2 = lady dislikes coffee

i) Number of ladies like coffee =Â 142

Hence, the probability of L1= numberofladieslikecoffeetotalnumberofladies$\frac{number\, of\, ladies\, like\,coffee }{total\, number\, of\, ladies}$ = 142200$\frac{142}{200}$ = 0.71

ii) Number of ladies dislike coffeeÂ = Â 58

Hence, the probability of L2= numberofladiesdislikecoffeetotalnumberofladies$\frac{number\, of\, ladies\, dislike\,coffee }{total\, number\, of\, ladies}$ = 58200$\frac{58}{200}$ = 0.29

Q.6: The percentage of marks obtained by a student in six unit tests are given below:

 Unit test I II III IV V VI Percentage of marks obtained 53 72 28 46 67 59

A unit test is selected at random. What is the probability that the student gets more than 60% marks in the test?

Sol:

Total number of unit tests = 6

Hence, the probability that the student gets more than 60% marks in the test = numberoftestsstudentgotmorethan60totalnumberoftests$\frac{number\, of\, tests\, student\,got\, more\, than\, 60% }{total\, number\, of\, tests}$ = 26$\frac{2}{6}$ = 13$\frac{1}{3}$

Q.7: On a particular day, at the Crossing in a city the various types of 240 vehicles going past during a time interval were observed as under:

 Types of vehicle Two-wheelers Three-wheelers Four-wheelers Frequency 84 68 88

Out of these vehicles, one is chosen at random. What is the probability that the chosen vehicle is a Two-wheeler?

Sol:

Total number of vehicles = 240

Hence, the probability that the chosen vehicle is a Two-wheeler = numberofTwoâˆ’wheelertotalnumberofvehicles$\frac{number\, of\, Two-wheeler }{total\, number\, of\, vehicles}$ = 84240$\frac{84}{240}$ = 0.35

Q.8: On one page of a telephone dictionary there are 200 phone numbers. The frequency distribution of their units digits is given below:

 Unit digit 0 1 2 3 4 5 6 7 8 9 Frequency 19 22 23 19 21 24 23 18 16 15

One of the numbers is chosen at random from the page. What is the probability that the unitâ€™s digit of the chosen number is

(i). 5 Â Â Â Â Â Â Â Â Â Â

(ii). 8

Sol:

Total number of phone numbers = 200

i) Frequency of occurrence of Â 5 as unit digit Â = 24

Hence, probability (P5)= frequencyofoccurrenceof5totalnumberofphonenumber$\frac{frequency\, of\, occurrence\, of\,5 }{total\, number\, of\, phone\, number }$ = 24200$\frac{24}{200}$ = 0.12

ii) Frequency of occurrence of Â 8 as unit digit Â = 24

Hence, probability (P8)= frequencyofoccurrenceof8totalnumberofphonenumber$\frac{frequency\, of\, occurrence\, of\,8 }{total\, number\, of\, phone\, number }$ = 16200$\frac{16}{200}$ = 0.08

Q.9: The following table shows the blood groups of 40 students of a class:

 Blood Group A B O AB Number of students 11 9 14 6

One student of the class is chosen at random. What is the probability that the chosen student has blood group:

(i). OÂ

(ii). AB

Sol:

Total number of students = 40

i) Number of students having O as their blood group = 14

Hence, probability (P0) = numberofstudentshavingOastheirbloodgrouptotalnumberofstudents$\frac{number\, of\, students\, having\, O\, as\, their\, blood\, group }{total\, number\, of\, Â students }$ = 1440$\frac{14}{40}$ = 0.35

ii) Number of students having AB as their blood group = 6

Hence, probability (PAB)= numberofstudentshavingABastheirbloodgrouptotalnumberofstudents$\frac{number\, of\, students\, having\, AB\, as\, their\, blood\, group }{total\, number\, of\, Â students }$ = 640$\frac{6}{40}$ = 0.15

Q.10: The table given below shows the marks obtained by 30 students in a test:

 Marks (class interval) Number of students (Frequency) 1-10 7 11-20 10 21-30 6 31-40 4 41-50 3

Out of the students, one is chosen at random. What is the probability that his marks lie in the intervalÂ 21-30?

Sol:

Total number of students = 30

So, Number of students having marks in the range 21-30 = 6

Hence, probability (P) = numberofstudentshavingmarksbetween21âˆ’30totalnumberofstudents$\frac{number\, of\, students\, having\, marks\, between\, 21-30 }{total\, number\, of\, Â students }$ = 630$\frac{6}{30}$ = 0.2

Q.10: The following are the ages of 360 patients getting medical treatment in a hospital:

 Age(in years) 10-20 20-30 30-40 40-50 50-60 60-70 Number of patients 90 50 60 80 50 30

One of the patients is selected at random. Find the probability that his age is

(i). 30 years or more but less than 40 years

(ii). 50 years or more but less than 70 years

(iii). less than 10 years

(iv). 10 years or more.

Sol:

Total number of patient = 360 i.e T=360

i) Patients aged 30 years or more but less than 40 years = 60

Hence, the probability of getting Patients having aged 30 years or more but less than 40 years = numberofoutcomestotalnumberofoutcomes$\frac{number\, of\, outcomes }{total\, number\, of\, outcomes}$ = 60360$\frac{60}{360}$ = 16$\frac{1}{6}$

ii) Patients aged 50 years or more but less than 70 years = 80

Hence, the probability of getting Patients aged 50 years or more but less than 70 years = numberofoutcomestotalnumberofoutcomes$\frac{number\, of\, outcomes }{total\, number\, of\, outcomes}$ = 80360$\frac{80}{360}$ = 29$\frac{2}{9}$

iii) Patients aged less than 10 years = 0

Hence, the probability of getting Patients aged less than 10 years = numberofoutcomestotalnumberofoutcomes$\frac{number\, of\, outcomes }{total\, number\, of\, outcomes}$ = 0360$\frac{0}{360}$ = 0

iv)Â Patients aged 10 years or more =360

Hence, the probability of getting Patients aged 10 years or more= numberofoutcomestotalnumberofoutcomes$\frac{number\, of\, outcomes }{total\, number\, of\, outcomes}$ = 360360$\frac{360}{360}$ = 1

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Bromine is the only non-metal which is liquid at room temperature.