Question

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

• A. \( \frac{8}{25} \)
• B. \( \frac{21}{50} \)
• C. \( \frac{9}{50} \)
• D. \( \frac{14}{25} \)

Answer 1

The Correct Option is B

To find the probability that an integer chosen at random from the integers 1 to 50 is a multiple of at least one of 4, 6, and 7, we will use the principle of Inclusion-Exclusion. Let's go through the steps: 

1. Identify the set of integers divisible by each number:
   • Multiples of 4: The smallest multiple of 4 within 1 to 50 is 4, and the largest is 48. The multiples are 4, 8, 12, ..., 48. This forms an arithmetic sequence where the nth term is 4n. 
     The sequence's number of terms, \( n \), can be found from the equation \( 4n = 48 \), giving \( n = 12 \).
   • Multiples of 6: Similarly, the sequence is 6, 12, 18, ..., 48, and there are 8 terms (since \( 6n = 48 \) gives \( n = 8 \)).
   • Multiples of 7: The sequence is 7, 14, 21, ..., 49, and has 7 terms (since \( 7n = 49 \) gives \( n = 7 \)).
2. Count the multiples of the least common multiples (LCM) of pairs:
   • LCM(4, 6) = 12: The sequence is 12, 24, 36, 48, so there are 4 terms (as \( 12n = 48 \) gives \( n = 4 \)).
   • LCM(6, 7) = 42: The sequence is 42, yielding 1 term.
   • LCM(4, 7) = 28: The sequence is 28, also yielding 1 term.
3. Count the multiples of the LCM of all three numbers:
   • LCM(4, 6, 7) = 84: Since 84 is greater than 50, it does not contribute any terms within 1 to 50.
4. Apply the Inclusion-Exclusion principle to count the number of integers that are multiples of at least one of 4, 6, or 7: 
   Number of multiples = \( 12 + 8 + 7 - 4 - 1 - 1 + 0 = 21 \). 
   Thus, there are 21 integers that are multiples of at least one of 4, 6, or 7.
5. Calculate the probability: 
   Total number of integers from 1 to 50 is 50. 
   Probability \( P \) = \(\frac{21}{50}\).

By following these steps, we determined that the correct answer is \(\frac{21}{50}\).

Answer 2

Define the events:
- \( P(A) \): Probability that the number is a multiple of 4.  
- \( P(B) \): Probability that the number is a multiple of 6.  
- \( P(C) \): Probability that the number is a multiple of 7.  

Step 1. Calculate \( P(A) \), \( P(B) \), and \( P(C) \): 
  \(P(A) = \frac{12}{50}, \, P(B) = \frac{8}{50}, \, P(C) = \frac{7}{50}\)
Step 2. Calculate the probabilities of intersections:
 \(P(A \cap B) = \frac{4}{50}, \, P(B \cap C) = \frac{1}{50}, \, P(A \cap C) = \frac{1}{50}\)
 \(P(A \cap B \cap C) = 0\)
Step 3. Apply the formula for the union of three events:
  \(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)\)
  Substituting values:  
    \(P(A \cup B \cup C) = \frac{12}{50} + \frac{8}{50} + \frac{7}{50} - \frac{4}{50} - \frac{1}{50} - \frac{1}{50} + 0\)
  \(= \frac{21}{50}\)
Thus, the probability that the chosen integer is a multiple of at least one of 4, 6, or 7 is \( \frac{21}{50} \).

The Correct Answer is: \( \frac{21}{50} \)