The document Ex 15.1 NCERT Solutions- Probability Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.

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**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:** Here, the total number of trials = 30

âˆµ Number of times the ball touched boundary = 6

âˆ´ Number of times, the ball missed the boundary = 30 â€“ 6 = 24

Let the event not hitting the boundary be represented by E, then

= (24/30) = (4/5) = 0.8

Thus, the required probability = 0.8**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.

(i) âˆµ Number of families having 2 girls in a family = 475

âˆ´ Probability of a family having 2 girls in a family = (475/1500)= (19/60)

(ii) âˆµ Number of families having 1 girl = 814

âˆ´ Probability of a family having 1 girl in a family = (814/1500) = (407/750)

(iii) âˆµ Number of families having no girl in a family = 211

âˆ´ Probability of a family having no girl in a family = (211/1500)

Now, the sum of the obtained probabilities

=

i.e., Sum of the above probabilities is 1.**Question 3. Refer to Question. Find the probability that a student of the class was born in August.****Solution:** From the graph, we have:

Total number of students born in various months in a year = 40

Number of students born in August = 6

âˆ´ Probability of a student of the IX-Class who was born in August = (6/40)= (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 times the three coins are tossed = 200

Number of outcomes in which 2 heads coming up = 72

âˆ´ Probability of 2 heads coming up = (72/200) = (9/25)

Thus, the required probability = (9/25)

If the three coins are simultaneously tossed again, then the probability is (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:**

Monthly income On â‚¹) | Vehicles per family | |||

0 | 1 | 2 | Above 2 | |

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

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

10000-13000 | 1 | 535 | 29 | 1 |

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

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

(i) âˆµ Number of families having earning â‚¹ 10000â€“ â‚¹ 13000 per month and 2 vehicles = 29

âˆ´ Probability of a family (having earning â‚¹ 10000â€“13000 and 2 vehicles) = (29/2400)

(ii) âˆµ Number of families having earning â‚¹16000 or above and owning 1 vehicle = 579

âˆ´ Probability of a family (having earning â‚¹ 16000 and above and 1 vehicle) = (579/2400)

(iii) âˆµ Number of families having earning less than â‚¹7000 and does not own any vehicle = 10

âˆ´ Probability of a family (having earning less than â‚¹ 7000 and owning no vehicle) = (10/2400) = (1/240)

(iv) âˆµ Number of families having earning â‚¹13000â€“16000 and owing more than 2 vehicles = 25

âˆ´ Probability of a family (having earning â‚¹13000â€“10000 and owning no vehicle) = (25/2400) = (1/96)

(v) âˆµ Number of families owning not more than 1 vehicle

= [Number of families having no vehicle] + [Number of families having only 1 vehicle]

= [10 + 1 + 2+ 1] + [160 + 305 + 535 + 469 + 579]

= 14 + 2148

= 2162

âˆ´ Probability of a family (owning not more than one vehicle) = (2162/2400)

= (1031/1200)**Question 6. Refer to Table. ****(i) Find the probability that a student obtained less than 20% in Mathematics test. ****(ii) Find the probability that a student obtained 60 marks or above.Solution:** From the table 14.7, we have:

Marks | Number of students |

0â€“20 20â€“30 30â€“40 40â€“50 50â€“60 60â€“70 70 and above | 7 10 10 20 20 15 8 |

Total | 90 |

Total number of students = 90

(i) From the given table number of students who have obtained less than 20% marks = 7

â‡’ Probability of a student (obtaining less than 20% marks) = (7/90)

(ii) From the given table, number of students who obtained marks 60% or above

= [Number of students in class-interval 60â€“70] + [Number of students in the class interval 70 and above]

= 15 + 8 = 23

â‡’ Probability of a student (who obtained 60 marks and above) = (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 Dislike | 135 65 |

**Find the probability that a student chosen at random (i) likes statistics, (ii) does not like it.****Solution: **Total number of students whose opinion in obtained = 200

(i) âˆµ Number of students who like statistics = 135

âˆ´ Probability of a student (who likes statistics) = (135/200)

= (27/40)

(ii) âˆµ Number of students who do not like statistics = 65

âˆ´ Probability of a student (who dislike statistics) = (65/200) = (13/40)**Question 8. 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 (1/2) km from her place of work?****Solution: **Total number of engineers = 40

(i) âˆµ Number of engineers who are living within less than 7 km from their work place = 9

âˆ´ Probability of an engineer living within 7 km from work place = (9/40)

(ii) âˆµ Number of engineers living at a distance more than or equal to 7 km from their work place = 31

âˆ´ Probability of an engineer living at a distance more than or equal to 7 km = (31/40)

(iii) âˆµ The number of engineers living within (1/2) km from their work place = 0

âˆ´ Probability of an engineer who is living within (1/2) km from work place = (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:** It is an activity. Students can do it themselves.**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?****Solution:** A class room activity for students.**REMEMBER**

A number is divisible by 3, if the sum of its digits is divisible by 3.**Examples:**

(i) Number 45678 is divisible by 3 because 4 + 5 + 6 + 7 + 8 = 30 is divisible by 3

(ii) Number 10786 is not divisible by 3 because 1 + 0 + 7 + 8 + 6 = 22 is not divisible by 3.**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**

âˆµ Number of bags having more than 5 kg of flour = 7

âˆ´ Probability of a bag (Having more than 5 kg wheat flour) = (7/11)

âˆµ The number of days (on which the sulphur dioxide concentration is in the interval 0.12â€“0.16) = 2

âˆ´ Probability of a day (on which sulphur dioxide is in 0.12â€“0.16 interval) = (2/30)

= (1/15)

âˆµ Number of students having blood group as AB = 3

âˆ´ Probability of a student (whose blood group is AB) = (3/30) = (1/10)

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