Relative Frequency Approach in Probability
Data Science and A.I. Lecture Series
Author: Bindeshwar Singh Kushwaha
Relative Frequencies and Probabilities
So, in general, if \( X \) is a variable having the values \( x_1, x_2, \dots, x_n \) with frequencies \( f_1, f_2, \dots, f_n \), respectively, then:
\[
\frac{f_1}{\sum f_i}, \frac{f_2}{\sum f_i}, \dots, \frac{f_n}{\sum f_i}
\]
are the relative frequencies of \( x_1, x_2, \dots, x_n \), respectively, and hence the probabilities of \( X \) taking the values \( x_1, x_2, \dots, x_n \).
Example 1: Age of Couples
| Age of Husband Age of Wife |
15-25 | 25-35 | 35-45 | 45-55 |
|---|---|---|---|---|
| 6 | 3 | 0 | 0 | |
| 20-30 | 3 | 16 | 10 | 0 |
| 30-40 | 0 | 10 | 15 | 7 |
| 40-50 | 0 | 0 | 7 | 10 |
| 50-60 | 0 | 0 | 4 | 5 |
Probability of wife’s age \( 20-50 \):
\[
P = \frac{3 + 16 + 10 + 10 + 15 + 7}{100} = 0.82
\]
Probability of wife’s age \( 20-40 \) and husband’s age \( 35-45 \):
\[
P = \frac{10 + 15}{100} = 0.25
\]
Example 2: Student Ages
| Age (Years) | Frequency (\( f \)) | Relative Frequency (\( \frac{f}{15} \)) |
|---|---|---|
| 14 | 2 | \( \frac{2}{15} \) |
| 15 | 1 | \( \frac{1}{15} \) |
| 16 | 2 | \( \frac{2}{15} \) |
| 17 | 3 | \( \frac{3}{15} \) |
| 18 | 1 | \( \frac{1}{15} \) |
| 19 | 2 | \( \frac{2}{15} \) |
| 20 | 3 | \( \frac{3}{15} \) |
| 21 | 1 | \( \frac{1}{15} \) |
Probability of age divisible by 3 (\( 15, 18, 21 \)):
\[
P = \frac{1 + 1 + 1}{15} = 0.2
\]
Probability of age \( >16 \):
\[
P = \frac{3 + 1 + 2 + 3 + 1}{15} = 0.6667
\]
Probability of age \( \geq 18 \):
\[
P = \frac{1 + 2 + 3 + 1}{15} = 0.4667
\]
Example 3: Tyre Durability
| Distance (km) | Frequency (\( f \)) | Relative Frequency (\( \frac{f}{2000} \)) |
|---|---|---|
| Less than 4000 | 20 | \( \frac{20}{2000} \) |
| 4001-10000 | 100 | \( \frac{100}{2000} \) |
| 10001-20000 | 200 | \( \frac{200}{2000} \) |
| 20001-40000 | 1500 | \( \frac{1500}{2000} \) |
| More than 40000 | 180 | \( \frac{180}{2000} \) |
Probability of \( \geq 4001 \, \text{km} \):
\[
P = \frac{100 + 200 + 1500 + 180}{2000} = 0.99
\]
Probability of \( \leq 20000 \, \text{km} \):
\[
P = \frac{20 + 100 + 200}{2000} = 0.16
\]
Probability of \( >20000 \, \text{km} \):
\[
P = \frac{1500 + 180}{2000} = 0.84
\]