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 :-
Let E be the event that she doesn't hit a boundry.
Total number of balls played = 30
Number of times the batswoman hits a boundary = 6
Number of times that the batswoman does not hit a boundary = 30 − 6 = 24
P(E) = 24/30 = 4/5
Question-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.
Let E, F and G are be the event that 2 girls, 1 girl and no girl respectively.
Total number of families = 475 + 814 + 211 = 1500
(i) 2 girls
P(E) = 475/1500 = 19/60
(ii) 1 girl
P(F) = 814/1500 = 407/750
(iii) No girl
P(G) = 211/1500
Verification :
Sum of all Probabilities
= P(E) + P(F) + P(G)
= 19/60 + 407/750 + 211/1500
= (475 + 814 + 211)/1500
= 1500/1500
= 1
Question-3 :- 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 of the class was born in August.
Let E be the event that students born in the month of August.
Number of studen ts born in the month of August = 6
Total number of students = 40
P(E) = 6/40 = 3/20
Question-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.
Let E be the event that 2heads will comes up.
Number of times 2 heads come up = 72
Total number of times the coins were tossed = 200
P(E) = 72/200 = 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:
Suppose a family is chosen. Find the probability that the family chosen is
(i) earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not own any vehicle.
(iv) earning ₹ 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Number of total families surveyed = 10 + 160 + 25 + 0 + 0 + 305 + 27 + 2 + 1 + 535 + 29 + 1 + 2 + 469 + 59 + 25 + 1 + 579 + 82 + 88 = 2400
(i) Let E be the event that earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles.
Number of families earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles = 29
P(E) = 29/2400
(ii) Let F be the event that earning ₹ 16000 or more per month and owning exactly 1 vehicle.
Number of families earning ₹ 16000 or more per month and owning exactly 1 vehicle = 579
P(E) = 579/2400
(iii) Let G be the event that earning less than ₹ 7000 per month and does not own any vehicle.
Number of families earning less than ₹ 7000 per month and does not own any vehicle = 10
P(G) = 10/2400 = 1/240
(iv) Let H be the event that earning ₹ 13000 – 16000 per month and owning more than 2 vehicles.
Number of families earning ₹ 13000 – 16000 per month and owning more than 2 vehicles = 25
P(H) = 25/2400 = 1/96
(v) Let I be the event that owning not more than 1 vehicle.
Number of families owning not more than 1 vehicle = 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062
P(I) = 2062/2400 = 1031/1200
Question-6 :- A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks.
Looking at their performances, 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 sizes 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% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Total number of students = 90
(i) Let E be the event that a student obtained less than 20% in the mathematics test.
Number of students getting less than 20 % marks in the test = 7
P(E) = 7/90
(ii) Let F be the event that a student obtained marks 60 or above.
Number of students obtaining marks 60 or above = 15 + 8 = 23
P(F) = 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:
Find the probability that a student chosen at random
(i) likes statistics,
(ii) does not like it
Total number of students = 135 + 65 = 200
(i) Let E be the event that likes statistics.
Number of students liking statistics = 135
P(E) = 135/200 = 27/40
(ii) Let F be the event that doesn't like statistics.
Number of students who do not like statistics = 65
P(F) = 65/200 = 13/40
Question-8 :- The distance (in km) of 40 engineers from their residence to their place of work were 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
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?
Total number of engineers = 40
(i) Let E be the event that less than 7 km from her place of work.
Number of engineers living less than 7 km from their place of work = 9
P(E) = 9/40
(ii) Let F be the event that more than or equal to 7 km from her place of work.
Number of engineers living more than or equal to 7 km from their place of work = 40 − 9 = 31
P(F) = 31/40
(iii) Let G be the event that within 1/2 km from her place of work.
Number of engineers living within 1/2 km from her place of work = 0
P(G) = 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 :-
Do Activity Yourself.
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? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Solution :-
Do Activity Yourself.
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
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Let E be the event that any of these bags chosen at random contains more than 5 kg of flour.
Number of total bags = 11
Number of bags containing more than 5 kg of flour = 7
P(E) = 7/11
Question-12 :- 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
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.
Let E be the event that the concentration of sulphur dioxide was in the interval of 0.12 − 0.16.
Total number of days = 30
Number days for which the concentration of sulphur dioxide was in the interval of 0.12 − 0.16 = 2
P(E) = 2/30 = 1/15
Question-13 :- 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.
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.
Let E be the event that students having blood group AB.
Number of students having blood group AB = 3
Total number of students = 30
P(E) = 3/30 = 1/10
