**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:**

Data given in the question:

Total number of balls batswoman plays = 30

Numbers of boundary hit by batswoman = 6

To find number of time batswoman didn’t hit boundary , we will substract

⇒(Total number of balls batswoman plays) – (Numbers of boundary hit by batswoman)

⇒ 30 – 6 = 24

Probability of that she did’t hit a boundary = 24/30 = 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**

**Solution:**

According to question

Total numbers of families given in the question 1500

(i) Numbers of families having 2 girls = 475

Probability of choosen 2 girls = Numbers of families having 2 girls / Total numbers of families

= 475/1500 = 19/60

Probability of choosen 2 girls is 19/60(ii) Numbers of families having 1 girls = 814

Probability of choosen 1 girl = Numbers of families having 1 girl / Total numbers of families

= 814/1500 = 407/750

Probability of choosen 1 girl is 407/750(iii) Numbers of families having 2 girls = 211

Probability of choosen 0 girl = Numbers of families having 0 girls/Total numbers of families

= 211/1500Sum of the probability = (19/60)+(407/750)+(211/1500)

= (475+814+211)/1500

= 1500/1500 = 1Yes, the sum of these probabilities is 1.

### Question **3. Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.**

**Solution:**

According to questions:

Total number of students in the class in the given question = 40

Numbers of students born in August = 6

The probability that a student of the class was born in August = (Total numbers of students in the class) /

(Numbers of students born in August)

= 6/40 = 3/20

### Question **4. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:**

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

Frequency | 23 | 72 | 77 | 28 |

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

**Solution:**

Number of times 2 heads come up (in the given question) = 72

Total number of times the coins were tossed = 200

The probability of 2 heads coming up = (Number of times 2 heads come up) / (Total number of times the coins were tossed)

= 72/200 = 9/25

### Question **5. An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. **

**The information gathered is listed in the table below:**

| 0 | 1 | 2 | Above 2 |

Less than 7000 | 10 | 160 | 25 | 0 |

7000-10000 | 0 | 305 | 27 | 2 |

7000-10000 | 1 | 535 | 29 | 1 |

13000-16000 | 2 | 469 | 59 | 25 |

16000 or more | 1 | 579 | 82 | 88 |

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

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

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

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

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

**(v) owning not more than 1 vehicle. **

**Solution:**

Total number of families = 2400 (According to question)

(i)Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles = 29

The probability that the family chosen is earning ₹10000 – 13000 per month and owning exactly 2 vehicles =

(Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles) / (Total number of families)

= 29/2400

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

The probability that the family chosen is earning ₹16000 or more per month and owning exactly 1 vehicle =

(Number of families earning ₹16000 or more per month and owning exactly 1 vehicle) / (Total number of families)

=579/2400

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

The probability that the family chosen is earning less than ₹7000 per month and does not own any vehicle =

(Number of families earning less than ₹7000 per month and does not own any vehicle)/(Total number of families)

= 10/2400 = 1/240

(iv)Number of families earning ₹13000-16000 per month and owning more than 2 vehicles = 25

The probability that the family chosen is earning ₹13000 – 16000 per month and owning more than 2 vehicles =

(Number of families earning ₹13000-16000 per month and owning more than 2 vehicles ) / (Total number of families)

= 25/2400 = 1/96

(v)Number of families owning not more than 1 vehicle = 10+160+0+305+1+535+2+469+1+579 = 2062

The probability that the family chosen owns not more than 1 vehicle = 2062/2400 = 1031/1200

### Question **6. Refer to Table 14.7, Chapter 14.**

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

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

**Solution:**

Total number of students = 90

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

The probability that a student obtained less than 20% in the mathematics test =

( Number of students who obtained less than 20% in the mathematics test)/(Total number of students)

= 7/90

(ii)Number of students who obtained marks 60 or above = 15+8 = 23The probability that a student obtained marks 60 or above =

(Number of students who obtained marks 60 or above ) / (Total number of students)

= 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 = 135+65 = 200 (According to question)

(i) Number of students who like statistics = 135

The probability that a student likes statistics = (Number of students who like statistics) / (Total number of students)

= 135/200 = 27/40(ii) Number of students who do not like statistics = 65

The probability that a student does not like statistics =

(Number of students who do not like statistics) / (Total number of students)

= 65/200 = 13/40

### Question **8. Refer to Q.2, Exercise 14.2. 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 ½ km from her place of work?**

**Solution:**

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

5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 3 2

17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6

15 15 7 6 12

Total numbers of engineers = 40

(According to question)

(i)Number of engineers living less than 7 km from their place of work = 9The probability that an engineer lives less than 7 km from her place of work =

(Number of engineers living less than 7 km from their place of work) / (Total numbers of engineers )

= 9/40

(ii)Number of engineers living more than or equal to 7 km from their place of work = 40-9 = 31Probability that an engineer lives more than or equal to 7 km from her place of work =

(Number of engineers living more than or equal to 7 km from their place of work ) / (Total numbers of engineers)

= 31/40

(iii)Number of engineers living within ½ km from their place of work = 0The probability that an engineer lives within ½ km from her place of work =

(Number of engineers living within ½ km from their place of work) / (Total numbers of engineers)

=0/40 = 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:**

The question is an activity to be performed by the students.

### 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:**

The question is an activity to be performed by the students.

### 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:**

Data given in the question

Total number of bags present = 11

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

The probability that any of the bags chosen at random contains more than 5 kg of flour =

(Number of bags containing more than 5 kg of flour) / (Total number of bags present)

= 7/11

### Question **12. In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.**

**The data obtained for 30 days is as follows:**

**0.03 0.08 0.08 0.09 0.04 0.17 0.16 0.05 0.02 0.06 0.18 0.20 0.11 0.08 0.12 0.13 0.22 0.07 0.08 0.01 0.10 0.06 0.09 0.18 0.11 0.07 0.05 0.07 0.01 0.04**

**Solution:**

Total number of days in which the data was recorded = 30 days (According to the question)

Numbers of days in which sulphur dioxide was present in between the interval 0.12-0.16 = 2

The probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days =

(Numbers of days in which sulphur dioxide was present in between the interval 0.12-0.16) /

(Total number of days in which the data was recorded )

= 2/30 = 1/15

### Question **13. In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.**

**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.**

**Solution:**

Total numbers of students = 30 (according to questions)

Number of students having blood group AB = 3

The probability that a student of this class, selected at random, has blood groupAB = (Number of students having blood group AB) / (Total numbers of students)

= 3/30 = 1/10