Question

If one ticket is selected at random from 30 tickets each with a distinct number from 1 to 30, then the probability that the number on the selected ticket is a multiple of 3 or 5 is

• A. \( \frac{14}{31} \)
• B. \( \frac{7}{30} \)
• C. \( \frac{14}{15} \)
• D. \( \frac{7}{15} \)

Hint:

Use the formula for the probability of the union of two events: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Identify the events (multiple of 3, multiple of 5). Count the number of outcomes for each event within the sample space (1 to 30). Also, count the number of outcomes for the intersection (multiple of both 3 and 5, i.e., multiple of 15). Calculate the individual probabilities and then use the formula.

Answer 1

The Correct Option is D

Total number of tickets = 30.
Let \( A \) be the event that the number on the selected ticket is a multiple of 3.
The multiples of 3 between 1 and 30 are \( 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 \).
Number of multiples of 3, \( n(A) = 10 \).
Let \( B \) be the event that the number on the selected ticket is a multiple of 5.
The multiples of 5 between 1 and 30 are \( 5, 10, 15, 20, 25, 30 \).
Number of multiples of 5, \( n(B) = 6 \).
We want to find the probability that the number on the selected ticket is a multiple of 3 or 5, which is \( P(A \cup B) \).
We use the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
First, we find the probability of event \( A \): $$ P(A) = \frac{n(A)}{\text{Total number of tickets}} = \frac{10}{30} = \frac{1}{3} $$ Next, we find the probability of event \( B \): $$ P(B) = \frac{n(B)}{\text{Total number of tickets}} = \frac{6}{30} = \frac{1}{5} $$ Now, we find the probability of \( A \cap B \), which is the event that the number on the selected ticket is a multiple of both 3 and 5, i.
e.
, a multiple of \( \text{lcm}(3, 5) = 15 \).
The multiples of 15 between 1 and 30 are \( 15, 30 \).
Number of multiples of 15, \( n(A \cap B) = 2 \).
The probability of \( A \cap B \) is: $$ P(A \cap B) = \frac{n(A \cap B)}{\text{Total number of tickets}} = \frac{2}{30} = \frac{1}{15} $$ Finally, we find \( P(A \cup B) \): $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15} $$ $$ P(A \cup B) = \frac{5}{15} + \frac{3}{15} - \frac{1}{15} = \frac{5 + 3 - 1}{15} = \frac{7}{15} $$