Chapter 14: Probability (Exercise 14.1 - Questions 1 to 3)

Chapter 14: Probability (Exercise 14.1 - Questions 1 to 3)

Question 1:
A coin is tossed 100 times with the following frequencies:
* Head: 45
* Tail: 55
Compute the probability for each event.

Solution:
* Total number of trials = 100

1. Probability of getting a Head, P(H):
P(H) = 45 / 100 = 9 / 20 (or 0.45)

2. Probability of getting a Tail, P(T):
P(T) = 55 / 100 = 11 / 20 (or 0.55)

* Verification: P(H) + P(T) = 0.45 + 0.55 = 1

Question 2:
Two coins are tossed simultaneously 500 times, and we get the following outcomes:
* Two heads: 105 times
* One head: 275 times
* No head: 120 times
Find the probability of occurrence of each of these events.

Solution:
* Total number of tosses = 500

1. Probability of getting 2 heads, P(E₁):
P(E₁) = 105 / 500 = 21 / 100 (or 0.21)

2. Probability of getting 1 head, P(E₂):
P(E₂) = 275 / 500 = 11 / 20 (or 0.55)

3. Probability of getting no head, P(E₃):
P(E₃) = 120 / 500 = 6 / 25 (or 0.24)

* Verification: P(E₁) + P(E₂) + P(E₃) = 0.21 + 0.55 + 0.24 = 1

Question 3:
A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 given in the following table:
* Outcome 1: 179
* Outcome 2: 150
* Outcome 3: 157
* Outcome 4: 149
* Outcome 5: 175
* Outcome 6: 190
Find the probability of getting each outcome.

Solution:
* Total number of throws = 1000

1. P(Getting 1): 179 / 1000 = 0.179
2. P(Getting 2): 150 / 1000 = 0.15
3. P(Getting 3): 157 / 1000 = 0.157
4. P(Getting 4): 149 / 1000 = 0.149
5. P(Getting 5): 175 / 1000 = 0.175
6. P(Getting 6): 190 / 1000 = 0.19

* Verification: Sum = 0.179 + 0.15 + 0.157 + 0.149 + 0.175 + 0.19 = 1

NCERT Chapter 14: Statistics (Exercise 14.2 - Questions 1 to 3)

Question 1:
The blood groups of 30 students of Class IX 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?

Solution:
Frequency Distribution Table:
* Blood Group A: 9 students
* Blood Group B: 6 students
* Blood Group O: 12 students
* Blood Group AB: 3 students
* Total: 30 students

* Most common blood group: Group O (12 students)
* Rarest blood group: Group AB (3 students)

Question 2:
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.
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0–5 (5 not included).

Solution:
* 0 – 5: 5 engineers
* 5 – 10: 11 engineers
* 10 – 15: 11 engineers
* 15 – 20: 9 engineers
* 20 – 25: 1 engineer
* 25 – 30: 1 engineer
* 30 – 35: 2 engineers
* Total: 40 engineers

* Observation: Most of the engineers (36 out of 40) reside within a distance of 20 km from their workplace.

Question 3:
The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1, 98.6, 99.2, 90.3, 86.5, 95.3, 92.9, 96.3, 94.2, 95.1, 89.2, 92.3, 97.1, 93.5, 92.7, 95.1, 97.2, 93.3, 95.2, 97.3, 96.2, 92.1, 84.9, 90.2, 95.7, 98.3, 97.3, 96.1, 92.1, 89.0.
1. Construct a grouped frequency distribution table with classes 84–86, 86–88, etc.
2. Which month or season do you think this data is about?
3. What is the range of this data?

Solution:
1. Frequency Table:
* 84 – 86: 1 day
* 86 – 88: 1 day
* 88 – 90: 2 days
* 90 – 92: 2 days
* 92 – 94: 7 days
* 94 – 96: 6 days
* 96 – 98: 7 days
* 98 – 100: 4 days
* Total: 30 days

2. Season: The relative humidity is very high (above 90% for most days), so the data appears to be from the rainy (monsoon) season.

3. Range: Maximum value - Minimum value = 99.2 - 84.9 = 14.3%

Chapter 14: Probability (Exercise 14.1 - Remaining Questions)

Question 4:
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 data is summarized below:
* Less than ₹7000: 10 (0 vehicles), 160 (1 vehicle), 25 (2 vehicles), 0 (>2 vehicles)
* ₹7000 – ₹10000: 0 (0 vehicles), 305 (1 vehicle), 27 (2 vehicles), 2 (>2 vehicles)
* ₹10000 – ₹13000: 1 (0 vehicles), 535 (1 vehicle), 29 (2 vehicles), 1 (>2 vehicles)
* ₹13000 – ₹16000: 2 (0 vehicles), 469 (1 vehicle), 59 (2 vehicles), 25 (>2 vehicles)
* ₹16000 or more: 1 (0 vehicles), 579 (1 vehicle), 82 (2 vehicles), 88 (>2 vehicles)

Suppose a family is chosen at random. Find the probability that the family chosen is:
1. Earning ₹10000 – ₹13000 per month and owning exactly 2 vehicles.
2. Earning ₹16000 or more per month and owning exactly 1 vehicle.
3. Earning less than ₹7000 per month and does not own any vehicle.
4. Earning ₹13000 – ₹16000 per month and owning more than 2 vehicles.
5. Owning not more than 1 vehicle.

Solution:
Total number of families surveyed = 2400

1. Earning ₹10000 – ₹13000 & exactly 2 vehicles:
Favourable outcomes = 29
P = 29 / 2400

2. Earning ₹16000 or more & exactly 1 vehicle:
Favourable outcomes = 579
P = 579 / 2400 = 193 / 800

3. Earning < ₹7000 & 0 vehicles:
Favourable outcomes = 10
P = 10 / 2400 = 1 / 240

4. Earning ₹13000 – ₹16000 & > 2 vehicles:
Favourable outcomes = 25
P = 25 / 2400 = 1 / 96

5. Owning not more than 1 vehicle (0 or 1 vehicle across all income groups):
Families with 0 vehicles = 10 + 0 + 1 + 2 + 1 = 14
Families with 1 vehicle = 160 + 305 + 535 + 469 + 579 = 2048
Total favourable families = 14 + 2048 = 2062
P = 2062 / 2400 = 1031 / 1200

Question 5:
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:
* Likes Statistics: 135 students
* Dislikes Statistics: 65 students

Find the probability that a student chosen at random:
1. likes Statistics
2. does not like Statistics

Solution:
Total number of students = 200

1. P(Student likes Statistics):
P = 135 / 200 = 27 / 40 (or 0.675)

2. P(Student does not like Statistics):
P = 65 / 200 = 13 / 40 (or 0.325)

Question 6:
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:
1. less than 7 km from her place of work?
2. at least 7 km from her place of work?
3. within ½ km from her place of work?

Solution:
Total number of engineers = 40

1. Distance less than 7 km:
Engineers with distance < 7 km (5, 3, 2, 3, 5, 3, 2, 6, 6) = 9 engineers
P = 9 / 40

2. Distance at least 7 km (≥ 7 km):
Engineers with distance ≥ 7 km = 40 - 9 = 31 engineers
P = 31 / 40

3. Distance within ½ km (≤ 0.5 km):
Engineers with distance ≤ 0.5 km = 0
P = 0 / 40 = 0

Question 7:
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.

Solution:
* Total number of bags = 11
* Bags containing more than 5 kg (> 5.00 kg):
5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07 → 7 bags

P(Bag contains > 5 kg) = 7 / 11

Question 8:
The concentration of sulphur dioxide in the air (in parts per million, i.e., ppm) in a city was recorded for 30 days. The number of days in which concentration lay in the interval 0.12 - 0.16 ppm was 2 days.

Find the probability that the concentration of sulphur dioxide on a day chosen at random lies in the interval 0.12 - 0.16 ppm.

Solution:
* Total number of days = 30
* Number of days in interval 0.12 - 0.16 ppm = 2

P(Concentration in 0.12 - 0.16) = 2 / 30 = 1 / 15

Question 9:
The blood groups of 30 students of Class IX 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.

Find the probability that a student of this class, chosen at random, has blood group AB.

Solution:
* Total number of students = 30
* Frequency distribution of blood groups:
- Group A = 9 students
- Group B = 6 students
- Group O = 12 students
- Group AB = 3 students

P(Student has blood group AB) = 3 / 30 = 1 / 10 (or 0.1)