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

Answer: (B)

$\frac{21}{50}$

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: $\frac{21}{50}$