{"id":4607,"date":"2023-03-02T05:14:39","date_gmt":"2023-03-02T05:14:39","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4607"},"modified":"2023-03-02T05:14:39","modified_gmt":"2023-03-02T05:14:39","slug":"ncert-solutions-class-9-maths-chapter-15-probability-ex-15-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-15-probability-ex-15-1\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 15 Probability Ex 15.1"},"content":{"rendered":"

NCERT Solutions Class 9 Maths\u00a0<\/span>
\nChapter – 15 (Probability)\u00a0<\/span><\/h2>\n

The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter – 15 Probability <\/strong>Exercise 15.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n

Exercise – 15.1<\/span><\/h2>\n

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.<\/strong><\/span><\/p>\n

Answer –
\n<\/strong>Number of balls played = 30
\nNumber of balls for which the batswoman hits boundary = 6
\nThus, number of balls for which the batswoman does not hit a boundary = 30 \u2013 6 = 24
\nProbability, P(E) = Number of instances of the event taking place \/ Total number of instances.
\n= 24\/30
\n= 4\/5<\/span><\/p>\n

2. 1500 families with 2 children were selected randomly, and the following data were recorded:<\/strong><\/span><\/p>\n

\n\n\n\n\n
Number of girls in a family<\/strong><\/span><\/td>\n2<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
Number of families<\/strong><\/span><\/td>\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 475\u00a0 \u00a0 \u00a0\u00a0 \u00a0\u00a0\u00a0<\/span><\/td>\n\u00a0\u00a0 \u00a0 \u00a0 \u00a0\u00a0 814\u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0<\/span><\/td>\n\u00a0 \u00a0 \u00a0 \u00a0\u00a0 211 \u00a0 \u00a0\u00a0 \u00a0\u00a0<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Compute the probability of a family, chosen at random, having
\n<\/strong>(i) 2 girls
\n(ii) 1 girl
\n(iii) No girl
\nAlso check whether the sum of these probabilities is 1.<\/strong><\/span><\/p>\n

Answer –
\n<\/strong>Total number of families = 1500<\/span><\/p>\n

(i)<\/strong> Number of families having 2 girls, P(2)= 475
\nProbability = Number of families having 2 girls\/Total number of families
\n= 475\/1500
\n= 19\/60<\/span><\/p>\n

(ii)<\/strong> Number of families having 1 girl, P(1) = 814
\nProbability = Number of families having 1 girl\/Total number of families
\n= 814\/1500
\n= 407\/750<\/span><\/p>\n

(iii)<\/strong> Number of families having 0 girls, P(0)= 211
\nProbability = Number of families having 0 girls\/Total number of families
\n= 211\/1500<\/span><\/p>\n

Now, sum of all the three probabilities = P(2) + P(1) + P(0) <\/span>
\n= 475\/1500 + 814\/1500 + 211\/1500<\/span>
\n= (475 + 814 + 211) \/ 1500<\/span>
\n= 1500\/1500<\/span>
\n= 1<\/span>
\nYes, the sum of these probabilities is 1.<\/span><\/p>\n

3. Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.<\/strong><\/span><\/p>\n

Answer –
\n\"NCERT
\n<\/strong>Total number of students in the class = 40
\nNumber of students born in August = 6
\nProbability of students born in August = Number of students born in August \/ Total number of students in class<\/span>
\n= 6\/40<\/span>
\n= 3\/20<\/span><\/p>\n

4. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:<\/strong><\/span><\/p>\n

\n\n\n\n\n
Outcome\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong><\/span><\/td>\n\u00a0\u00a0\u00a0\u00a0 3 heads\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/td>\n\u00a0\u00a0\u00a0\u00a0 2 heads\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/td>\n\u00a0\u00a0\u00a0 1 head\u00a0\u00a0\u00a0\u00a0<\/span><\/td>\n\u00a0\u00a0\u00a0\u00a0 No head\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/td>\n<\/tr>\n
Frequency<\/strong><\/span><\/td>\n23<\/span><\/td>\n72<\/span><\/td>\n77<\/span><\/td>\n28<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.<\/strong><\/span><\/p>\n

Answer –
\n<\/strong>Number of times 2 heads come up = 72
\nTotal number of times the coins were tossed = 200
\nProbability of 2 heads outcomes = Number of 2 heads outcomes \/ Total number of tosses<\/span>
\n= 72\/200<\/span>
\n= 9\/25<\/span><\/p>\n

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:<\/strong><\/span><\/p>\n

\n\n\n\n\n\n\n\n\n\n
Monthly income
\n(in \u20b9)<\/strong><\/span><\/td>\n		Vehicles per family<\/strong><\/span><\/td>\n<\/tr>\n
0<\/span><\/td>\n1<\/span><\/td>\n2<\/span><\/td>\nAbove 2<\/span><\/td>\n<\/tr>\n
Less than 7000<\/strong><\/span><\/td>\n10<\/span><\/td>\n160<\/span><\/td>\n25<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
7000-10000<\/strong><\/span><\/td>\n0<\/span><\/td>\n305<\/span><\/td>\n27<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
10000-13000<\/strong><\/span><\/td>\n1<\/span><\/td>\n535<\/span><\/td>\n29<\/span><\/td>\n1<\/span><\/td>\n<\/tr>\n
13000-16000<\/strong><\/span><\/td>\n2<\/span><\/td>\n469<\/span><\/td>\n59<\/span><\/td>\n25<\/span><\/td>\n<\/tr>\n
16000 or more<\/strong><\/span><\/td>\n1<\/span><\/td>\n579<\/span><\/td>\n82<\/span><\/td>\n88<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Suppose a family is chosen. Find the probability that the family chosen is
\n<\/strong>(i) <\/strong>earning \u20b910000 \u2013 13000 per month and owning exactly 2 vehicles.
\n<\/strong>(ii) <\/strong>earning \u20b916000 or more per month and owning exactly 1 vehicle.
\n<\/strong>(iii) <\/strong>earning less than \u20b97000 per month and does not own any vehicle.
\n<\/strong>(iv) <\/strong>earning \u20b913000 \u2013 16000 per month and owning more than 2 vehicles.
\n<\/strong>(v) <\/strong>owning not more than 1 vehicle.\u00a0<\/span><\/p>\n

Answer – <\/strong>Total number of families = 2400<\/span><\/p>\n

(i)<\/strong> Number of families earning \u20b910000 \u201313000 per month and owning exactly 2 vehicles = 29
\nProbability of family earning \u20b9 10000 \u2013 13000 per month and owning exactly 2 vehicles = 29\/2400<\/span><\/p>\n

(ii)<\/strong> Number of families earning \u20b916000 or more per month and owning exactly 1 vehicle = 579
\nProbability of family earning \u20b9 16000 or more per month and owning exactly 1 vehicle = 579\/2400<\/span><\/p>\n

(iii)<\/strong> Number of families earning less than \u20b97000 per month and does not own any vehicle = 10
\nProbability of family earning less than \u20b9 7000 per month and does not own any vehicle = 10\/2400 = 1\/240<\/span><\/p>\n

(iv)<\/strong> Number of families earning \u20b913000-16000 per month and owning more than 2 vehicles = 25
\nProbability of family earning \u20b9 13000 \u2013 16000 per month and owning more than 2 vehicles = 25\/2400 = 1\/96<\/span><\/p>\n

(v)<\/strong> Number of families owning not more than 1 vehicle = 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062
\nProbability of family owning not more than 1 vehicle = 2062\/2400 = 1031\/1200<\/span><\/p>\n

6. Refer to Table 14.7, Chapter 14.
\n<\/strong>(i) <\/strong>Find the probability that a student obtained less than 20% in the mathematics test.
\n(ii) <\/strong>Find the probability that a student obtained marks 60 or above.<\/span><\/p>\n

Answer –<\/strong><\/span><\/p>\n

\n\n\n\n\n\n\n\n\n\n\n\n
Marks<\/strong><\/span><\/td>\nNumber of students<\/strong><\/span><\/td>\n<\/tr>\n
0 \u2013 20<\/span><\/td>\n7<\/span><\/td>\n<\/tr>\n
20 \u2013 30<\/span><\/td>\n10<\/span><\/td>\n<\/tr>\n
30 \u2013 40<\/span><\/td>\n10<\/span><\/td>\n<\/tr>\n
40 \u2013 50<\/span><\/td>\n20<\/span><\/td>\n<\/tr>\n
50 \u2013 60<\/span><\/td>\n20<\/span><\/td>\n<\/tr>\n
60 \u2013 70<\/span><\/td>\n15<\/span><\/td>\n<\/tr>\n
70 \u2013 above<\/span><\/td>\n8<\/span><\/td>\n<\/tr>\n
Total<\/span><\/td>\n90<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Total number of students = 90<\/span><\/p>\n

(i)<\/strong> Number of students that obtained less than 20% marks = 7
\nProbability of students that obtained less than 20% marks = 7\/90<\/p>\n

(ii)<\/strong> Number of students that obtained 60 marks or above = 15 + 8 = 23
\nProbability of students that obtained 60 marks or above\u00a0<\/em>= 23\/90
\n<\/span><\/p>\n

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.<\/strong><\/span><\/p>\n

\n\n\n\n\n\n
Opinion<\/strong><\/span><\/td>\nNumber of students<\/strong><\/span><\/td>\n<\/tr>\n
like<\/span><\/td>\n135<\/span><\/td>\n<\/tr>\n
dislike<\/span><\/td>\n65<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Find the probability that a student chosen at random
\n(i) <\/strong>likes statistics,
\n(ii) <\/strong>does not like it.<\/span><\/p>\n

Answer –
\n<\/strong>Total number of students = 135+65 = 200<\/span><\/p>\n

(i)<\/strong> Number of students who like statistics = 135
\nProbability of students who like statistics = 135\/200 = 27\/40<\/span><\/p>\n

(ii)<\/strong> Number of students who do not like statistics = 65
\nProbability of students who dislike statistics = 65\/200 = 13\/40<\/span><\/p>\n

8. Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives:
\n<\/strong>(i) <\/strong>less than 7 km from her place of work?
\n(ii) <\/strong>more than or equal to 7 km from her place of work?
\n(iii) <\/strong>Within \u00bd km from her place of work?<\/span><\/p>\n

Answer – <\/strong>The distance (in km) of 40 engineers from their residence to their place of work were found as follows:<\/span><\/p>\n\n\n\n\n\n\n
5<\/span><\/td>\n3<\/span><\/td>\n10<\/span><\/td>\n20<\/span><\/td>\n25<\/span><\/td>\n11<\/span><\/td>\n13<\/span><\/td>\n7<\/span><\/td>\n12<\/span><\/td>\n31<\/span><\/td>\n<\/tr>\n
19<\/span><\/td>\n10<\/span><\/td>\n12<\/span><\/td>\n17<\/span><\/td>\n18<\/span><\/td>\n11<\/span><\/td>\n32<\/span><\/td>\n17<\/span><\/td>\n16<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
7<\/span><\/td>\n9<\/span><\/td>\n7<\/span><\/td>\n8<\/span><\/td>\n3<\/span><\/td>\n5<\/span><\/td>\n12<\/span><\/td>\n15<\/span><\/td>\n18<\/span><\/td>\n3<\/span><\/td>\n<\/tr>\n
12<\/span><\/td>\n14<\/span><\/td>\n2<\/span><\/td>\n9<\/span><\/td>\n6<\/span><\/td>\n15<\/span><\/td>\n15<\/span><\/td>\n7<\/span><\/td>\n6<\/span><\/td>\n12<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Total numbers of engineers = 40<\/span><\/p>\n

(i)<\/strong> Number of engineers living less than 7 km from their place of work = 9
\nProbability of an engineer who lives less than 7 km from their place of work = 9\/40<\/span><\/p>\n

(ii)<\/strong> Number of engineers living more than or equal to 7 km from their place of work = 40 – 9 = 31
\nProbability of an engineer who lives more than or equal to 7 km from their place of work = 31\/40<\/span><\/p>\n

(iii)<\/strong> Number of engineers living within \u00bd km from their place of work = 0
\nProbability of an engineer who lives within 1\/2 km from their place of work = 0\/40 = 0<\/span><\/p>\n

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.<\/strong><\/span><\/p>\n

Answer – <\/strong>The question is an activity to be performed by the students.
\nHence, perform the activity by yourself and note down your inference.<\/span><\/p>\n

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.<\/strong><\/span><\/p>\n

Answer – <\/strong>The question is an activity to be performed by the students.
\nHence, perform the activity by yourself and note down your inference.<\/span><\/p>\n

11. Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):
\n<\/strong>4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
\n<\/strong>Find the probability that any of these bags chosen at random contains more than 5 kg of flour.<\/strong><\/span><\/p>\n

Answer – <\/strong>Total number of bags present = 11
\nNumber of bags containing more than 5 kg of flour = 7
\nProbability of a bag containing more than 5 kg of flour = 7\/11<\/span><\/p>\n

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.<\/strong>The data obtained for 30 days is as follows:<\/strong><\/span><\/p>\n\n\n\n\n\n\n\n
0.03<\/span><\/td>\n0.08<\/span><\/td>\n0.08<\/span><\/td>\n0.09<\/span><\/td>\n0.04<\/span><\/td>\n0.17<\/span><\/td>\n<\/tr>\n
0.16<\/span><\/td>\n0.05<\/span><\/td>\n0.02<\/span><\/td>\n0.06<\/span><\/td>\n0.18<\/span><\/td>\n0.20<\/span><\/td>\n<\/tr>\n
0.11<\/span><\/td>\n0.08<\/span><\/td>\n0.12<\/span><\/td>\n0.13<\/span><\/td>\n0.22<\/span><\/td>\n0.07<\/span><\/td>\n<\/tr>\n
0.08<\/span><\/td>\n0.01<\/span><\/td>\n0.10<\/span><\/td>\n0.06<\/span><\/td>\n0.09<\/span><\/td>\n0.18<\/span><\/td>\n<\/tr>\n
0.11<\/span><\/td>\n0.07<\/span><\/td>\n0.05<\/span><\/td>\n0.07<\/span><\/td>\n0.01<\/span><\/td>\n0.04<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer –
\n<\/strong>Total number of days in which the data was recorded = 30 days
\nNumber of days in which sulphur dioxide was present in between the interval 0.12 – 0.16 = 2
\nProbability of the concentration of Sulphur dioxide in the interval 0.12 \u2013 0.16 = 2\/30 = 1\/15
\n<\/span><\/p>\n

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:
\nA, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
\nA, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. <\/strong><\/span><\/p>\n

Answer –
\n<\/strong>Total numbers of students = 30
\nNumber of students having blood group AB = 3
\nProbability of students having blood group AB = (Number of students having blood group AB) \/ Total number of students = 3\/30 = 1\/10<\/span><\/p>\n

\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions Class 9 Maths\u00a0 Chapter – 15 (Probability)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter – 15 Probability Exercise 15.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Exercise – 15.1 1. In a cricket match, a batswoman hits a boundary 6 times out […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[727,702,186,728,701,5],"class_list":["post-4607","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-14-statistics-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-14-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"\nNCERT Solutions Class 9 Maths Chapter 15 Probability Ex 15.1 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 15 (Probability)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 15 Probability Exercise 15.1 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 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