An integer is chosen at random from the integers `1, 2, 3, …, 50`. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is

Options

A. `8/(25)`
B. `(21)/(50)`
C. `9/(50)`
D. `(14)/(25)`

Answer

To find the probability that a randomly chosen integer from the set {1, 2, 3, ..., 50} is a multiple of at least one of the numbers 4, 6, or 7, we can use the principle of inclusion-exclusion.

### Step-by-Step Solution:

1. **Define Events**:
Let:
- \( A \): the event that a number is a multiple of 4.
- \( B \): the event that a number is a multiple of 6.
- \( C \): the event that a number is a multiple of 7.

2. **Count the Total Outcomes**:
The total number of integers from 1 to 50 is 50.

3. **Count the Favorable Outcomes**:
- **Multiples of 4**: The multiples of 4 from 1 to 50 are \( 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48 \). There are \( \frac{50}{4} = 12.5 \) which gives us 12 multiples.
- **Multiples of 6**: The multiples of 6 from 1 to 50 are \( 6, 12, 18, 24, 30, 36, 42, 48 \). There are \( \frac{50}{6} = 8.33 \) which gives us 8 multiples.
- **Multiples of 7**: The multiples of 7 from 1 to 50 are \( 7, 14, 21, 28, 35, 42, 49 \). There are \( \frac{50}{7} = 7.14 \) which gives us 7 multiples.

4. **Count the Intersections**:
- **Multiples of both 4 and 6 (LCM = 12)**: The multiples of 12 from 1 to 50 are \( 12, 24, 36, 48 \). There are \( \frac{50}{12} = 4.16 \) which gives us 4 multiples.
- **Multiples of both 6 and 7 (LCM = 42)**: The multiples of 42 from 1 to 50 are \( 42 \). There is \( \frac{50}{42} = 1.19 \) which gives us 1 multiple.
- **Multiples of both 4 and 7 (LCM = 28)**: The multiples of 28 from 1 to 50 are \( 28 \). There is \( \frac{50}{28} = 1.79 \) which gives us 1 multiple.
- **Multiples of 4, 6, and 7 (LCM = 84)**: There are no multiples of 84 from 1 to 50, so this count is 0.

5. **Apply the Inclusion-Exclusion Principle**:
Using the formula:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C) \]
We substitute the values:
\[ P(A) = \frac{12}{50}, \quad P(B) = \frac{8}{50}, \quad P(C) = \frac{7}{50} \]
\[ P(A \cap B) = \frac{4}{50}, \quad P(B \cap C) = \frac{1}{50}, \quad P(A \cap C) = \frac{1}{50}, \quad P(A \cap B \cap C) = 0 \]
Thus,
\[ P(A \cup B \cup C) = \frac{12}{50} + \frac{8}{50} + \frac{7}{50} - \frac{4}{50} - \frac{1}{50} - \frac{1}{50} + 0 \]
Simplifying this gives:
\[ P(A \cup B \cup C) = \frac{12 + 8 + 7 - 4 - 1 - 1}{50} = \frac{21}{50} \]

6. **Final Probability**:
The probability that the chosen integer is a multiple of at least one of 4, 6, or 7 is:
\[ \frac{21}{50} \]