9th Maths 15.1
Chapter 15
Probability
NCERT Class 9th solution of Exercise 14.1
Exercise 15.1
Q1. 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.
Sol. :
Given:
Total number of balls n(S)=30
Number of hits 6
To find:
Probability of not hits.
Solve:
Number of not hits n(E)=30-6=24
Probability of not hits P(E)=n(E)n(S)=2438=45
Answer:
Probability of not hits 45.
Q2. 1500 families with 2 children were selected randomly, and the following data were recorded:
Compute the probability of a family, chosen at random, having
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i) 2girls
ii) 1 girl
iii) No girl.
Also, check whether the sum of these probabilities is 1.
Sol. :
Given:
n(E1)=475,n(E2)=814,n(E3)=211
To find:
i) 2girls
ii) 1 girl
iii) No girl.
Solve:
i) P(E1)=n(E1)n(S)=4751500=1960
Answer:
1960
ii) P(E2)=n(E2)n(S)=8141500=407750
Answer:
407750
iii) P(E3)=n(E3)n(S)=2111500
Answer:
2111500
Sum of the probability
4751500+8141500+2111500=15001500=1
Answer:
Yes, the sum of the probability is 1.
Q3. In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:
Find the probability that a student in the class was born in August.
Sol. :
Given:
n(S)=40,n(E)=6
To find:
P(E)
Solve:
P(E)=n(E)n(S)=640=320
Answer:
The probability that a student of the class was born on August 320.
Q4. Three coins are tossed simultaneously 200 times with the following frequency of different outcomes
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If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Sol. :
Given:
n(S)=200
n(E)=72
To find:
Compute the probability of 2 heads coming up P(E).
Solve:
P(E)=n(E)n(S)=72200=925
Answer:
The probability of 2 heads coming up is 925.
Q5. 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:
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Suppose a family is chosen. Find the probability that the family chosen is
i) earning ₹10000-13000 per month and owing exactly 2 vehicles.
ii) earning ₹16000 or more per month and owing exactly 1 vehicle.
iii) earning less than ₹7000 per month and does not own any vehicle.
iv) earning ₹13000-16000 per month and owing more than 2 vehicles.
v) owing not more than 1 vehicle.
Sol. :
Given:
n(S)=2400,
n(E1)=29,n(E2)=579
n(E3)=10,n(E4)=25,
Solve:
i)
P(E1)=n(E1)n(S)=292400
Answer:
Probability of 2vehicles is292400.
ii)
P(E2)=n(E2)n(S)=5792400=193800
Answer:
Probability of one vehicle is 193800.
iii)
P(E3)=n(E3)n(S)=102400=1240
Answer:
Probability does not own any vehicle is 1240
iv)
P(E4)=n(E4)n(S)=252400=196
Answer:
The probability of more than 2 vehicles is 196.
v)
No. of families owning no vehicle
=10+0+1+2+1=14
No. of families owning only one vehicle
=160+305+535+469+579=2048
n(E5)=14+2048=2062
P(E5)=n(E5)n(S)=20622400=10311200
Answer:
Probability of owning not more than 1 vehicle 10311200
Q6. A teacher wanted to analyze the performance of two sections of students in a mathematics test of 100 marks. Looking at their performance, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying size as follows:
0-20,20-30,....
Then she formed the following table:
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i) Find the probability that a student obtained less than 20% on the mathematics test.
ii) Find the probability that a student obtained marks of 60 or above.
Sol.
Given:
n(S)=90, n(E_1)=7,
n(E_2)=15+8=23
i)
Solve:
P(E_1)=(n(E))/(n(S))=7/90
Answer:
The probability is 7/90
ii)
Solve:
P(E_2)=(n(E_2))/(n(S))=23/90
Answer:
The probability is 23/90.
Q7. 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.
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Find the probability that a student chosen at random
i) likes statistics
ii) does not like it.
Sol. :
Given:
n(S)=200, n(E_1)=135,
n(E_2)=65
i)
Solve:
P(E_1)=(n(E_1))/(n(S))=135/200=27/40
Answer:
The probability is 27/40
ii)
Solve:
P(E_2)=(n(E_2))/(n(S))=65/200=13/40
Answer:
The probability is 13/40
Q8. The distance (in Km) of 40 engineers from their residence to their place of work was found as follows:
5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2,
7, 9, 7, 8, 3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12
Construction of a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
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i) less than 7 km from her place of work?
ii) more than or equal to 7 km from her place to work?
iii) within 1/2 km from her place of work?
Sol. :
Given:
n(S)=40, n(E_1)=9,
n(E_2)=31, n(E_3)=0
Solve:
i)
P(E_1)=(n(E_1))/(n(S))=9/40
Answer:
The probability is 9/40.
ii)
p(E_2)=(n(E_2))/(n(E))=31/40
Answer:
The probability is 31/40.
iii)
P(E_3)=(n(E_3))/(n(S))=0/40=0
Answer:
The probability is 0.
Q9. 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.
Q10. Activity: ask all the students in your class to write a 3-digit number. Choose any her/him is divisible by3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Q11. 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.
Sol. :
Given:
n(S)=11, n(E)=7
Solve:
P(E)=(n(E))/(n(S))=7/11
Answer:
The probability is 7/11
Q12. 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:
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
Find the probability of the concentration of sulphur dioxide in the interval 0.12-0.16
Sol. :
Given:
n(E)=2, n(S)=30
Solve:
P(E)=(n(E))/(n(S))=2/30=1/15
Answer:
The probability is 1/15.
Q13. : 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. Which is the most common, and which is the rarest, blood group among these students?
Sol. :
n(E)=3, n(S)=30
Solve:
P(E)=(n(E))/(n(S))=3/30=1/10
Answer:
The probability is 1/10.
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