Maths Chapter-15: Probability
1. In an ODI match, a wicket keeper drops a catch 6 times out of 30 catches he gets. Find the probability of the wicketkeeper not dropping a catch.
Sol: Let E be the event of dropping a catch
P(E)= Probability of the wicketkeeper dropping a catch = \(\frac{6}{30}\) = \(\frac{1}{5}\)
Thus, Probability of not dropping a catch is 1-P(E) = \(1-\frac{1}{5}\) = \(\frac{4}{5}\)
2. 1500 families with 2 children were selected randomly, and the following data were recorded:
| No of girls in a family | 2 | 1 | 0 |
| No of families | 500 | 300 | 200 |
What is the probability that a family, chosen at random, has
i. 2 girls
ii. 1 girl
iii. No girl
Also check whether the sum of these probabilities is 1.
Sol: Total number of families = \(500+300+200\)= 1000
i. P(Probability of 2 girls) = \(\frac{500}{1000}\) = \(\frac{1}{2}\)
ii. P(Probability of 1 girl) = \(\frac{300}{1000}\) = \(\frac{3}{10}\)
iii. P(Probability of No girl) = \(\frac{200}{1000}\) = \(\frac{1}{5}\)
Sum of all Probabilities = \(P(Probability of 2 girls) + P(Probability of 1 girl)+ P(Probability of No girl)\) = \(\frac{1}{2}+\frac{3}{10}+\frac{1}{5}\) = 1
3. From the following table, Find the probability of a student selected at random being born in April.
| Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
| No of students | 5 | 6 | 4 | 2 | 8 | 1 | 5 | 4 | 4 | 3 | 7 | 1 |
Sol: Total number of students: \(5+6+4+2+8+1+5+4+4+3+7+1\) = 50
P( Student born in April) =\(\frac{2}{50}\) = \(\frac{1}{25}\)
4. Three coins are tossed simultaneously 210 times with the following frequencies of different outcomes:
| Outcome | 3 Heads | 2 Heads | 1 Head | No Head |
| Frequency | 45 | 37 | 55 | 73 |
If the three coins are simultaneously tossed again, What is the probability of not getting even a single head?
Sol: Number of times the coins were tossed = 210
P( Not getting even a single head) = \(\frac{73}{210}\)
5. An organization selected 1900 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:
Vehicles per family
| Income level | 0 | 1 | 2 | More than 2 |
| Less than 6000 | 10 | 60 | 25 | 0 |
| 6000 to 11000 | 0 | 205 | 27 | 2 |
| 11000 to 14000 | 1 | 435 | 29 | 1 |
| 14000 to 17000 | 2 | 369 | 59 | 25 |
| 17000 or more | 1 | 479 | 82 | 88 |
Find the probability that a family chosen at random is
(i) Earning 11000 – 14000 per month and owning exactly 1 vehicle.
(ii) Earning 17000 or more per month and owning exactly 2 vehicles.
(iii) Earning less than 6000 per month and does not own any vehicle.
(iv) Earning 6000-11000 per month and owning more than 2 vehicles
(v) Owning more than 2 vehicles
Sol: Total number of families: 1900
- P( Earning 11000-14000 and owning 1 vehicle) = \(\frac{435}{1900}\)
- P( Earning 17000 or more and owning 2 vehicles) = \(\frac{82}{1900}\)
- P( Earning less than 6000 and doesn’t own any vehicle) = \(\frac{10}{1900}\)
- P( Earning 6000-11000 and owning more than 2 vehicles) = \(\frac{2}{1900}\)
- P( Owning more than 2 vehicles) = \(\frac{116}{1900}\)
6. In a science test, the marks of 100 students of class VI are listed in the following table:
| Marks
(Out of 100) |
No of students |
| 0-20 | 18 |
| 20-30 | 17 |
| 30-40 | 15 |
| 40-50 | 8 |
| 50-60 | 14 |
| 60-70 | 21 |
| 70-100 | 7 |
| Total | 100 |
i. What is the probability that a student selected at random has scored less than 40?
ii. What is the probability that a student selected at random has scored more than 50?
Sol: Total number of students = 100
Number of students having scored less than 40% = \(18+17+15\) = 50
Number of students having scored more than 50% = \(14+21+7\) = 42
i. P(Less than 40%) = \(\frac{50}{100}\) = \(\frac{1}{2}\)
ii. P(More than 50%) = \(\frac{42}{100}\) = \(\frac{21}{50}\)
7. The distance (in km) of 20 doctors from their residence to their place of work were found as follows:
5, 6, 4, 7, 2, 9, 1, 6, 4, 3, 5, 32, 4, 6, 21, 15, 4,15,18,5
What is the probability that a doctor lives:
(i) Less than 6 km from her place of work?
(ii) More than or equal to 6 km from her place of work?
(iii) Within 0.5 km from her place of work?
Sol: Total number of doctors = 20
Number of doctors with travelling distance below 6km = 10
Number of doctors with travelling distance more than or equal to 6 km = 10
Number of doctors with travelling distance below 0.5 km = 0
(i) P( Less than 6 km) = \(\frac{10}{20}\) =\(\frac{1}{2}\)
(ii) P( More than or equal to 6) = \(\frac{10}{20}\) = \(\frac{1}{2}\)
(iii) P(Less than 0.5 km) = \(\frac{0}{20}\) = 0
8. A survey of 100 students was conducted to know the opinion of the students about the subject statistics which is recorded in the following table.
| Opinion | Number of students |
| Like | 37 |
| Dislike | 63 |
What is the probability that a student selected at random
i. Likes the subject?
ii. Dislikes the subject?
Sol: Total number of students = 100
Number of students liking the subject = 37
Number of students disliking the subject = 63
i. P(liking the subject) = \(\frac{37}{100}\)
ii. P(Disliking the subject) = \(\frac{63}{100}\)
9. Nine bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):
5.08 ,4.95, 5.00, 4.96, 5.08, 4.98, 5.04, 5.07 and 5.00
What is the probability that any of these bags chosen at random contains less than 5 kg of flour?
Sol: Total number of wheat bags = 9
Number of wheat bags weighing less than 5 kg = 3
P( Weighing less than 5 kg) = \(\frac{3}{9}\) = \(\frac{1}{3}\)
10. 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 10 students of Class VIII are recorded as follows:
O,AB,A,B,O,A,AB,AB,O and AB
Represent this data in the form of a frequency distribution table. Use this table to determine the probability that a student of this class, selected random, has blood group O.
Sol: Frequency Distribution Table:
| Blood Group | Frequency |
| A | 2 |
| B | 1 |
| AB | 4 |
| O | 3 |
Total number of students = 10
Number of students having blood group O = 3
P(Blood group O) = \(\frac{3}{10}\)