Unit 13: Probability

Exercise 13.2 Solutions

Step-by-step solved textbook exercises for relative frequencies, empirical probabilities, and mathematical expectations of events and game payoffs.

Question 1 Relative Frequency
A researcher collected data on number of deaths from Horse-Kicks in Prussian Army corps over 20 years. The table is as follows:
No. of deaths 0 1 2 3 4 5 6
Frequency 60 50 87 40 32 15 10
Find the relative frequency of the data.
Solution

The relative frequency (r.f.) is calculated using the formula:

Relative Frequency = Frequency of specific event (f)Total frequency (Σf)

First, we calculate the sum of all frequencies:

Σf = 60 + 50 + 87 + 40 + 32 + 15 + 10 = 294

No. of Deaths Frequency (f) Relative Frequency calculation Relative Frequency (r.f.)
0 60 60294 30147 ≈ 0.204
1 50 50294 25147 ≈ 0.170
2 87 87294 2998 ≈ 0.296
3 40 40294 20147 ≈ 0.136
4 32 32294 16147 ≈ 0.109
5 15 15294 598 ≈ 0.051
6 10 10294 5147 ≈ 0.034
Total Σf = 294 — 1.000
Question 2 Defective Products
The frequency of defective products in 750 samples are shown in the following table. Find the relative frequency for the given table.
No. of defectives per sample 0 1 2 3 4 5 6 7 8
No. of samples 120 140 94 85 105 50 40 66 50
Solution

The total frequency is already given as 750 (verify: 120 + 140 + 94 + 85 + 105 + 50 + 40 + 66 + 50 = 750).

Using the relative frequency formula, we obtain:

Defectives per Sample No. of Samples (f) Relative Frequency calculation Relative Frequency (r.f.)
0 120 120750 425 ≈ 0.160
1 140 140750 1475 ≈ 0.187
2 94 94750 47375 ≈ 0.125
3 85 85750 17150 ≈ 0.113
4 105 105750 750 ≈ 0.140
5 50 50750 115 ≈ 0.067
6 40 40750 475 ≈ 0.053
7 66 66750 11125 ≈ 0.088
8 50 50750 115 ≈ 0.067
Total Σf = 750 — 1.000
Question 3 Quiz Performance
A quiz competition on general knowledge is conducted. The number of correct answers out of 5 questions for 100 sets of questions is given below.
Correct answers 0 1 2 3 4 5
Frequency (f) 10 23 15 25 18 9
Find the relative frequencies for the given data.
Solution

Total sets tested (Σf) = 100.

We divide each class frequency by 100 to find the relative frequencies:

Correct Answers Frequency (f) Relative Frequency calculation Relative Frequency (r.f.)
0 10 10100 110 = 0.10
1 23 23100 23100 = 0.23
2 15 15100 320 = 0.15
3 25 25100 14 = 0.25
4 18 18100 950 = 0.18
5 9 9100 9100 = 0.09
Total Σf = 100 — 1.00
Question 4 Correction Applied
A survey was conducted of the students of a class and they were asked about their favorite food. The responses are as under:
Favorite Food Item Biryani Fresh Juice Chicken Bar. B.Q Sweets
No. of students (f) 40 7 21 15 25
Answer the following:
  • (i) What percentage of students like Biryani?
  • (ii) What percentage of students like Chicken?
  • (iii) Which food is least liked by the students?
  • (iv) Which food is most preferred by the students?
Solution

First, we calculate the sum of student counts in the table:

Σf = 40 + 7 + 21 + 15 + 25 = 108 students

We calculate the percentage for each food item by dividing its frequency by 108 and multiplying by 100:

Food Item No. of Students (f) Relative Frequency calculation Percentage (%)
Biryani 40 40108 37.0% (rounded to 37%)
Fresh Juice 7 7108 6.5%
Chicken 21 21108 19.4%
Bar. B.Q 15 15108 13.9% (rounded to 14%)
Sweets 25 25108 23.1% (rounded to 23%)
Total 108 — 100%
  • (i) Percentage of students liking Biryani: 37%
  • (ii) Percentage of students liking Chicken: 19.4%
  • (iii) Least liked food: Fresh Juice (only 6.5% of students).
  • (iv) Most preferred food: Biryani (preferred by 37.0% of students).
Mathematical Note on Q.4 textbook typos The textbook text claims "50 students" were surveyed, but the frequencies sum to 108. The book's solved key computes the percentages using 108 as the denominator, which is correct for the data. Additionally, the textbook contains a copy-paste error where it prints "(6.5% only)" next to Biryani in part (iv) (this percentage belongs to Fresh Juice). We corrected the text values and calculations here to ensure logical clarity.
Question 5 Expected Frequency
In 500 trials of a throw of two dice, what is expected frequency that the sum will be greater than 8?
Solution

We are given:

  • Total trials (N) = 500

When two fair dice are rolled, the total number of outcomes is:

n(S) = 36

Let A be the event that the sum of the dots is greater than 8. The sum can be 9, 10, 11, or 12. Listing all outcomes where sum > 8:

A = {(3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)}

The number of favorable outcomes is:

n(A) = 10

The probability of getting a sum greater than 8 in a single trial is:

P(A) = 1036 = 518

The expected frequency in 500 trials is:

Expected Frequency = N × P(A) = 500 × 518 = 250018 ≈ 138.89

Rounding to the nearest whole number, we expect the sum to be greater than 8 approximately 139 times.

Question 6 Mathematical Expectation
What is the expectation of a person who is to get Rs. 120 if he obtains at least 2 heads in single toss of three coins?
Solution

We are given:

  • Payoff value (X) = Rs. 120

When three fair coins are tossed, the sample space S is:

S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT},   n(S) = 8

Let A be the event of getting at least 2 heads (meaning 2 or 3 heads):

A = {HHH, HHT, HTH, THH}

The number of favorable outcomes is:

n(A) = 4

The probability of this event is:

P(A) = 48 = 12

The mathematical expectation of the payoff is calculated as:

Expectation = Payoff × Probability = 120 × 12 = Rs. 60

Thus, the mathematical expectation is Rs. 60.

Question 7 Probability Distribution
Find the expected frequencies of the given data if the experiment is repeated 200 times.
x 0 1 2 3 4 5 6
P(x) 0.11 0.21 0.17 0.18 0.09 0.17 0.07
Solution

We are given:

  • Total repetitions (N) = 200

Expected frequency for each outcome is calculated by: Expected Frequency = N × P(x)

Outcome (x) Probability (P(x)) Expected Frequency calculation Expected Frequency
0 0.11 200 × 0.11 22
1 0.21 200 × 0.21 42
2 0.17 200 × 0.17 34
3 0.18 200 × 0.18 36
4 0.09 200 × 0.09 18
5 0.17 200 × 0.17 34
6 0.07 200 × 0.07 14
Total 1.00 — 200
Question 8 Dice Roll Expectation
The probability of getting 5 sixes while tossing dice is 25. The dice is rolled 200 times. How many times would you expect it to show 5 sixes?
Solution

We are given:

  • Probability of the event (P(A)) = 25
  • Number of trials (N) = 200

The expected frequency is computed using the formula:

Expected Frequency = N × P(A) = 200 × 25 = 40 × 2 = 80

In 200 rolls, we expect it to show 5 sixes exactly 80 times.

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