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:**

**i) a head?**

**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 (P_{1}) = numberoftimesheadappearstotalnumberoftrails = 285500 = 57100 = 0.57

**ii)** Number of times tail appears = 215

Hence, the probability of getting tail (P_{2}) = numberoftimestailappearstotalnumberoftrails = 215500 = 43100= 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:**

**i) 2 heads?**

**ii) 1 head?**

**iii) 0 head?**

**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 (P_{2})= numberoftimes2headsappearstotalnumberoftrails = 112400Â = 725= 0.28

**ii)** Number of times 1 head appears = 160 times

Hence, the probability of getting 1 head (P_{1}) = numberoftimes1headappearstotalnumberoftrails = 160400 = 25= 0.4

**iii)** Number of timesÂ 0 heads appears = 128 times

Hence, the probability of getting Â 2 heads (P_{2})= numberoftimes0headsappearstotalnumberoftrails = 128400 = 825= 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: **

**i) 3 heads?**

**ii) 2 heads?**

**iii) 1 head?**

**iv) 0 head?**

**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 (P_{3})= numberoftimes3headsappearstotalnumberoftrails = 39200 = 0.195

**ii)** Number of times 2 head appears = 58 times

Hence, the probability of getting 2 head (P_{2})= numberoftimes2headappearstotalnumberoftrails = 58200 = 29100= 0.335

**iii)** Number of times 1 heads appears = 67 times

Hence, the probability of getting 1 heads (P_{2})= numberoftimes1headsappearstotalnumberoftrails = 67200 = 0.18

**iv)** Number of times 0 head appears = 36 times

Hence, the probability of getting 0 head (P_{2})= numberoftimes2headappearstotalnumberoftrails = 36200 = 950= 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 (P_{3}) = numberoftimes3occurstotalnumberoftrails = 54300 = 0.18

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

Hence, the probability of getting 6 (P_{6})= numberoftimes6occurstotalnumberoftrails = 33300 = 0.11

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

Hence, the probability of getting 5 (P_{5})= numberoftimes5occurstotalnumberoftrails = 39300 = 0.13

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

Hence, the probability of getting 1 (P_{1})= numberoftimes1occurstotalnumberoftrails = 60300 = 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, L_{1}=lady likes coffee and L_{2 }= lady dislikes coffee

**i)** Number of ladies like coffee =Â 142

Hence, the probability of L_{1}= numberofladieslikecoffeetotalnumberofladies = 142200 = 0.71

**ii)** Number of ladies dislike coffeeÂ = Â 58

Hence, the probability of L_{2}= numberofladiesdislikecoffeetotalnumberofladies = 58200 = 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 = 26 = 13

**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 = 84240 = 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 (P_{5})= frequencyofoccurrenceof5totalnumberofphonenumber = 24200 = 0.12

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

Hence, probability (P_{8})= frequencyofoccurrenceof8totalnumberofphonenumber = 16200 = 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 (P_{0}) = numberofstudentshavingOastheirbloodgrouptotalnumberofstudents = 1440 = 0.35

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

Hence, probability (P_{AB})= numberofstudentshavingABastheirbloodgrouptotalnumberofstudents = 640 = 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 = 630 = 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 = 60360 = 16

**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 = 80360 = 29

**iii)** Patients aged less than 10 years = 0

Hence, the probability of getting Patients aged less than 10 years = numberofoutcomestotalnumberofoutcomes = 0360 = 0

**iv)**Â Patients aged 10 years or more =360

Hence, the probability of getting Patients aged 10 years or more= numberofoutcomestotalnumberofoutcomes = 360360 = 1