# Probability : Exercise 15.1 (Mathematics NCERT Class 9th)

Q.1       In a cricket match, batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Sol.

Since in a cricket match ,a batswoman hits a boundary 6 times out of 30 balls she plays i.e., she missed the boundary 30 – 6 = 24 times out of 30 balls.
Therefore, the probability that a batswoman does not hit a boundary.
$= {{Number\,of\,not\,hitting\,a\,boundary} \over {Total\,number\,of\,trials\,\left( {i.e.,balls} \right)}}$
$= {{24} \over {30}} = {\mkern 1mu} {4 \over 5}$

Q.2      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.

Let ${E_0},{E_1}\,and\,{E_2}$ be the event of getting no girl, 1girl and 2 girls.
(i) Therefore $P\left( {{E_2}} \right)$ = Probability of a family having 2 girls
$= {{Number\,of\,families\,having\,2\,girls} \over {Total\,number\,of\,girls}}$
$= {{475} \over {1500}} = {{19} \over {60}}$

(ii) Therefore $P\left( {{E_1}} \right)$ = Probability of a family having 1 girl
$= {{Number\,of\,families\,having\,1\,girls} \over {Total\,number\,of\,families}}$
$= {{814} \over {1500}} = {{407} \over {750}}$

(iii) Therefore $P\left( {{E_0}} \right)$ = Probability of a family having no girls
$= {{Number\,of\,families\,having\,no\,girls} \over {Total\,number\,of\,families}}$
$= {{211} \over {1500}}$
Therefore Sum of probabilities = $P\left( {{E_0}} \right) + P\left( {{E_1}} \right) + P\left( {{E_2}} \right)$
$= {{211} \over {1500}} + {{407} \over {750}} + {{19} \over {60}}$
$= {{211 + 814 + 475} \over {1500}} = {{1500} \over {1500}} = 1$

Q.3       Refer to Example 5, section 14.4 Chapter 14. Find the probability that a student of the class was born in August.
Sol.

Clearly from the histograph six students were born in the month of August out of 40 students of a particular section of class - IX.
Probability that a student of the class was born in August

$= {{Number\,of\,students\,born\,in\,August} \over {Total\,number\,of\,students}}$
$= {6 \over {40}} = {3 \over {20}}$

Q.4       Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes : If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Sol.

Since three coins are tossed 200 times, so the total number of trials is 200.
$= {{No.\,of\,outcomes\,having\,2\,heads} \over {Total\,no.\,of\,trials}}$
$= {{72} \over {200}}$
$= {9 \over {25}}$

Q.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 :

Suppose a family is chosen. Find the probability that the family chosen is
(i) earning Rs 10000-13000 per month and owning exactly 2 vehicles.
(ii) earning Rs 16000 or more per month and owning exactly 1 vehicles.
(iii) earning less than Rs 7000 per month and does not own any vehicle.
(iv) earning Rs 13000 -16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle
Sol.

The total number of families = 2400
(i) Number of families earning Rs 10000 - 13000 per month and owning exactly 2 vehicles = 29.
Therefore, P (families earning Rs 10000 - 13000 per month and owning exactly 2 vehicles)
$= {{29} \over {2400}}$

(ii) Number of families earning Rs 16000 or more per month and owning exactly 1 vehicle = 579.
Therefore P (Families earning Rs 16000 or more per month and owning exactly 1 vehicle)
$= {{579} \over {2400}}$
$= {{193} \over {800}}$

(iii) Number of families earning less than Rs. 7000 per month and does not own any vehicle = 10
Therefore, P (Families earning less than Rs 7000 per month and does not own any vehicle)
$= {{10} \over {2400}} = {1 \over {240}}$

(iv) Number of families earning Rs 13000 - 16000 per month and owning more than 2 vehicles = 25
Therefore, P (Families earning Rs 13000 - 16000 per month and owning more than two vehicles)
$= {{25} \over {2400}} = {1 \over {96}}$

(v) Number of families owning not more than 1 vehicle
= Families having no vehicle + Families having 1 vehicle
= (10 + 0 + 1 + 2 + 1) + (160 + 305 + 535 + 469 + 579)
= 14 + 2048 = 2062
P (Families owning not more than 1 vehicle)
$= {{2062} \over {2400}} = {{1031} \over {1200}}$

Q.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.
Sol.

Total number of students in mathematics is 90.
(i)    Clearly, from the given table, the number of student who obtained less than 20% marks in the mathematics test = 7.
P (a student obtaining less than 20% marks) = ${7 \over {90}}$

(ii)   Clearly, from the given table, number of students who obtained marks 60 or above.
= (students in 60 - 70) + (students above 70)
= 15 + 8 = 23
Therefore, P (a student obtaining marks 60 and above) = ${{23} \over {90}}$

Q.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. Find the probability that a student chosen at random
(i) likes statistics, (ii) does not like it.
Sol.

The total number of students = 200
(i) P (a student likes statistics)
$= {{Number\,\,of\,students\,who\,like\,statistics} \over {Total\,number\,of\,students}}$
$= {{135} \over {200}} = {{27} \over {40}}$

(ii) P (a student does not like statistics)
$= {{Number\,\,of\,students\,who\,does not\,like\,statistics} \over {Total\,number\,of\,students}}$
$= {{65} \over {200}} = {{13} \over {40}}$

Q.8      Refer to Q.2, Ex 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 ${1 \over 2}km$ from her place of work?
Sol.

Total number of engineers = 40
(i) Number of engineers living less than 7 km from their place of work = 9
Therefore P (an engineer lives less than 7 km from her place of work)
$= {9 \over {40}}$

(ii) Number of engineers living more than or equal to 7km from her place of work = 31
Therefore P (an engineer lives less than or equal to 7 km from her place of work)
$= {{31} \over {40}}$

(iii) Number of engineer lives within ${1 \over 2}$ km from her place of work = 0
Therefore P (an engineer lives with ${1 \over 2}$ km from her place of work) = ${0 \over {40}} = 0$

Q.9     Activity : Note the frequency of two- wheeler, three - wheeler and four - wheeler 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.
Sol.       Activity problem : Collect the data and find the desired probability.

Q.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.
Sol.       Activity problem : Do as directed and find the desired probability.

Q.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 5kg of flour.
Sol.

Total number of wheat bags = 11
Number of bags having more than 5 kg = 7
Therefore P(a bag contains more than 5kg ) = ${7 \over {11}}$

Q.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.
Sol.

Total number of days = 30
Concentration of sulphur dioxide in 0.12 - 0.16 on any day = 2
Therefore  ,   required probability $= {2 \over {30}} = {1 \over {15}}$

Q.13    In Q. 1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood group 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.
Sol.

Total number of students = 30
Number of students having blood group AB = 3
Therefore Required probability $= {3 \over {30}} = {1 \over {10}}$

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