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)) =24/38=4/5`
Answer:
Probability of not hits `4/5`.
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
|
|
|
|
|
|
|
|
|
|
i) `2`girls
ii) `1` girl
iii) No girl.
Also, check whether the sum of these probabilities is `1`.
Sol. :
Given:
`n(E_1)=475, n(E_2)=814, n(E_3)=211`
To find:
i) `2`girls
ii) `1` girl
iii) No girl.
Solve:
i) `P(E_1)=(n(E_1))/(n(S))=475/1500=19/60`
Answer:
`19/60`
ii) `P(E_2)=(n(E_2))/(n(S))=814/1500=407/750`
Answer:
`407/750`
iii) `P(E_3)=(n(E_3))/(n(S))=211/1500`
Answer:
`211/1500`
Sum of the probability
`475/1500+814/1500+211/1500=1500/1500=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))=6/40=3/20`
Answer:
The probability that a student of the class was born on August `3/20`.
Q4. Three coins are tossed simultaneously `200` times with the following frequency of different outcomes
|
|
|
|
|
|
|
|
|
|
|
|
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))=72/200=9/25`
Answer:
The probability of `2` heads coming up is `9/25`.
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:
|
|
|
|||
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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(E_1)=29, n(E_2)=579`
`n(E_3)=10, n(E_4)=25,`
Solve:
i)
`P(E_1)=(n(E_1))/(n(S))=29/2400`
Answer:
Probability of `2`vehicles is`29/2400`.
ii)
`P(E_2)=(n(E_2))/(n(S))=579/2400=193/800`
Answer:
Probability of one vehicle is `193/800`.
iii)
`P(E_3)=(n(E_3))/(n(S))=10/2400=1/240`
Answer:
Probability does not own any vehicle is `1/240`
iv)
`P(E_4)=(n(E_4))/(n(S))=25/2400=1/96`
Answer:
The probability of more than `2` vehicles is `1/96`.
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(E_5)=14+2048=2062`
`P(E_5)=(n(E_5))/(n(S))=2062/2400=1031/1200`
Answer:
Probability of owning not more than `1` vehicle `1031/1200`
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, ...,60-70, 70-100`.
Then she formed the following table:
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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.
|
|
|
|
|
|
|
|
|
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?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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 by`3`? 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`.
Comments