Haryana State Board HBSE 9th Class Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 15 Probability Exercise 15.1

Question 1.

In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Solution:

Number of balls thrown = 30

∴ Total number of possible outcomes = 30

Number of times, the boundary is hit by the ball = 6

Number of times, the boundary is not hit by the ball= 30 – 6 = 24

∴ Number of favourable outcomes = 24

Let E be event that the boundary is not hit by the ball

∴ P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{24}{30}\) = \(\frac{4}{5}\)

Hence, P(E) = \(\frac{4}{5}\)

Question 2.

1500 families with 2 children were selected randomly, and the following data were recorded :

Number of girls in a family | 2 | 1 | 0 |

Number of families | 475 | 814 | 211 |

Compute the probability of a family, chosen at random, having :

(i) 2 girls

(ii) 1 girl

(iii) No girl.

Also check whether the sum of these probabilities is 1.

Solution:

Total number of families = 1500

∴ Total number of possible outcomes = 1500

(i) Number of families having 2 girls = 475

∴ Number of favourable outcomes = 475

Let E_{1} be event that the selected family has 2 girls, then

P(E_{1}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{475}{1500}\) = \(\frac{19}{60}\)

(ii) Number of families having 1 girl = 814

∴ Number of favourable outcomes = 814

Let E_{2} be the event that the selected family has 1 girl, then

P(E_{2}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{814}{1500}\) = \(\frac{407}{750}\)

(iii) Number of families having no girl = 211

∴ Number of favourable outcomes = 211

Let E_{3} be the event that the selected family has no girl; then

P(E_{3}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{211}{1500}\)

(iv) Sum of the probabilities

= \(\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}\)

= \(\frac{475+814+211}{1500}\)

= \(\frac{1500}{1500}\) = 1

Hence, (i) P(E_{1}) = \(\frac{19}{60}\), (ii) P(E_{2}) = \(\frac{407}{750}\) (iii) P(E_{3}) = \(\frac{211}{1500}\) (iv) Sum of the probabilities = 1.

Question 3.

In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data so obtained :

Find the probability that a student of the class was born in August.

Solution :

Total number of students = 40

∴ Total number of possible outcomes = 40

Number of students was born in August = 6

∴ Number of favourable outcomes = 6

Let E be the event that a student of the class was born in August, then

P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{6}{40}\) = \(\frac{3}{20}\)

Hence, P(E) = \(\frac{3}{20}\)

Question 4.

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes :

Outcome | 3 heads | 2 heads | 1 head | No head |

Frequency | 23 | 72 | 77 | 28 |

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Solution :

Total number of tosses of three coins simultaneously = 200

∴ Total number of possible outcomes = 200

Number of outcomes of 2 heads = 72

∴ Number of favourable outcomes = 72

Let E be the event of 2 heads coming up, then

P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{72}{200}\) = \(\frac{9}{25}\)

Hence, P(E) = \(\frac{9}{25}\)

Question 5.

An organisation selected 2400 families at random and surveyed them to determine their income level and the number of vehicles in a family. The information gathered is listed in the table below :

Suppose a family is chosen. Find the probability that the family chosen is :

(i) earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles.

(ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle.

(iii) earning less than Rs. 7000 per month and does not own any vehicle.

(iv) earning Rs. 13000 – 16000 per month and owning more than 2 vehicles.

(v) owning not more than 1 vehicle.

Solution:

The total number of families = 2400

∴ Total number of possible outcomes= 2400

(i) Number of families earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles = 29

∴ Number of favourable outcomes = 29

Let E_{1} be the event that a family’s earning 10000 – 13000 per month and owning exactly 2 vehicles, then

P(E_{1}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{29}{2400}\)

(ii) Number of families earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579

∴ Number of favourable outcomes = 579

Let E_{2} be the event that a family’s earning Rs. 16000 or more per month and owning exactly 1 vehicle, then

P(E_{2}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{579}{2400}\)

(iii) Number of families earning less than Rs. 7000 per month and does not own any vehicle = 10

∴ Number of favourable outcomes = 10

Let E_{3} be the event that a family’s earning less than Rs. 7000 per month and does not own any vehicle, then

P(E_{3}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{10}{2400}\) = \(\frac{1}{240}\)

(iv) Number of families earning Rs. 13000 – 16000 per month and owning more than 2 vehicles = 25

∴ Number of favourable outcomes = 25

Let E_{4} be the event that a family’s earning Rs. 13000 – 16000 per month and owning more than 2 vehicles, then

P(E_{4}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{25}{2400}\) = \(\frac{1}{96}\)

(v) Number of families owning not more than 1 vehicle = 10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579

= 2062

∴ Number of favourable outcomes= 2062

Let E_{5} be the event that a family owning not more than 1 vehicle, then

P(E_{5}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{2062}{2400}\) = \(\frac{1031}{1200}\)

Hence,

(i) P(E_{1}) = \(\frac{29}{2400}\);

(ii) P(E_{2}) = \(\frac{579}{2400}\)

(i) P(E_{3}) = \(\frac{1}{240}\)

(iv) P(E_{4}) = \(\frac{1}{96}\)

(v) P(E_{5}) = \(\frac{1031}{1200}\)

Question 6.

A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. A data of their performances is given below in the table :

Marks | Number of students |

0 – 20 | 7 |

20 – 30 | 10 |

30 – 40 | 10 |

40 – 50 | 20 |

50 – 60 | 20 |

60 – 70 | 15 |

70 – above | 8 |

Total | 90 |

(i) Find the probability that a student obtained less than 20% marks in the mathematics test.

(ii) Find the probability that a student obtained marks 60 or above.

Solution:

Total number of students = 90

∴ Total number of possible outcomes = 90

(i) Number of students who obtained less than 20% marks in the test = 7

∴ Number of favourable outcomes = 7

Let E_{1} be the event of a student obtaining less than 20% marks in the test.

P(E_{1}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{7}{90}\)

(ii) Number of students who obtained marks 60 or above in the test = 15 + 8 = 23

∴ Number of favourable outcomes = 23

Let E_{2} be the event of a student obtaining 60 or above marks in the test.

P(E_{2}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{23}{90}\)

Hence, (i) P(E_{1}) = \(\frac{7}{90}\),

(ii) P(E_{2}) = \(\frac{23}{90}\)

Question 7.

To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table :

Opinion | Number of students |

like | 135 |

dislike | 65 |

Find the probability that a student chosen at random :

(i) likes statistics,

(ii) does not like it.

Solution:

Total number of students = 200

∴ Total number of possible outcomes = 200

(i) Number of students who likes statistics = 135

∴ Number of favourable outcomes = 135

Let E_{1} be the event of a student likes statistics, then

P(E_{1}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{135}{200}\) = \(\frac{27}{40}\)

(ii) Number of students who dislikes statistics = 65

∴ Number of favourable outcomes = 65

Let E_{2} be the event of a student dislikes statistics, then

P(E_{2}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{65}{200}\) = \(\frac{13}{40}\)

Hence, (i) P(E_{1}) = \(\frac{27}{40}\) (ii) P(E_{2}) = \(\frac{13}{40}\)

Question 8.

The distance (in km) of 40 engineers from their residence to their place of work were found as follows:

Construct a grouped frequency distribution table with class size 7 for the data given above taking the first interval as 0-7 (7 is not included).

What is the empirical probability that an engineer lives :

(i) less than 7 km from her place of work?

(ii) more than or equal to 7 km from her place of work ?

(iii) within \(\frac{1}{2}\) km from her place of work?

Solution:

First we construct the frequency distribution table taking class interval 0 – 7, 7 – 14, 14 – 21, … as follows :

Distance (in km) | No. of Engineers |

0 – 7 | 9 |

7 – 14 | 17 |

14 – 21 | 11 |

21 – 28 | 1 |

28 – 35 | 2 |

Total | 40 |

Total number of engineers = 40

∴ Total number of possible outcomes = 40

(i) Number of engineers who live at a distance less than 7 km from their place of work = 9

∴ Number of favourable outcomes = 9

Let E_{1} be the event of a engineer lives at a distance less than 7 km from their place, then

P(E_{1}) = \(\frac{9}{40}\)

(ii) Number of engineers who live at a distance more than or equal to 7 km from their place of work = 17 + 11 + 1 + 2 = 31

∴ Number of favourable outcomes = 31

Let E_{2 }be the event of a engineer lives at a distance more than or equal to 7 km from their place, then

P(E_{2}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{31}{40}\)

(iii) Number of engineers who live within \(\frac{1}{2}\) km from their place of work = 0

Number of favourable outcomes = 0

Let E_{3} be the event of a engineer lives within \(\frac{1}{2}\) km from their place of work.

P(E_{3}) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{0}{40}\) = 0

Hence, (i) P(E_{1}) = \(\frac{9}{40}\), P(E_{2}) = \(\frac{31}{40}\), (iii) P(E_{3}) = 0.

Question 9.

Activity: Note the frequency of two wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.

Solution :

Collect the data according to activity and find the required probability.

Question 10.

Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Solution :

According to activity, collect the data and find required probability

Question 11.

Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):

4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00

Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Solution:

Total number of bags = 11

∴ Total number of possible outcomes = 11

Number of bags containing more than 5 kg of flour = 7

∴ Number of favourable outcomes = 7

P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{7}{11}\)

Hence, P(E) = \(\frac{7}{11}\)

Question 12.

A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows :

make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0·08 and so on. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.

Solution :

First we construct frequency distribution table taking class intervals as 0.00 – 0.04, 0.04 – 0.08, …, as follows:

Concentration of SO_{2} (in ppm) |
Frequency |

0.00 – 0.04 | 4 |

0.04 – 0.08 | 9 |

0.08 – 0.12 | 9 |

0.12 – 0.16 | 2 |

0.16 – 0.20 | 4 |

0.20 – 0.24 | 2 |

Total | 30 |

Total number of days = 30

∴ Total number of possible outcomes = 30

Concentration of sulphur dioxide in the interval 0.12 – 0.16 on any day = 2

∴ Number of favourable outcomes = 2

Let E be the event of concentration of sulphur dioxide in the interval 0·12 – 0.18 on any day. Then,

P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{2}{30}\) = \(\frac{1}{15}\)

Hence, P(E) = \(\frac{1}{15}\)

Question 13.

The blood groups of 30 students of class VIII are recorded as follows:

A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,

A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

Represent this data in the form of a frequency distribution table. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.

Solution :

First prepare the frequency distribution table as follows :

Blood groups | Number of students |

A | 9 |

B | 6 |

AB | 3 |

O | 12 |

Total | 30 |

Total number of students = 30

∴ Total number of possible outcomes = 30

Number of students having blood group AB = 3

∴ Number of favourable outcomes = 3

Let E be event of a student having blood group AB, then

P(E) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{3}{30}\) = \(\frac{1}{10}\)

Hence, P(E) = \(\frac{1}{10}\)