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

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: \( \frac{21}{50} \)