Question

If a number $ n $ is chosen at random from the set \{11, 12, 13, ..., 30\, then, the probability that $ n $ is neither divisible by 3 nor divisible by 5, is}

• A. \( \frac{7}{20} \)
• B. \( \frac{9}{20} \)
• C. \( \frac{11}{20} \)
• D. \( \frac{13}{20} \)

Hint:

To solve probability problems with divisibility, use the principle of inclusion and exclusion to account for overlaps in divisible sets.

Answer 1

The Correct Option is C

We are asked to find the probability that a number chosen from the set \( \{11, 12, 13, \dots, 30\} \) is neither divisible by 3 nor divisible by 5. First, let us determine the total number of elements in the set and those that are divisible by 3 or 5. The total number of elements in the set \( \{11, 12, 13, \dots, 30\} \) is: \[ \text{Total numbers} = 20 \quad \text{(i.e., 11, 12, 13, ..., 30)} \] Now, let’s find how many numbers in this set are divisible by 3 or 5: - Numbers divisible by 3: \( 12, 15, 18, 21, 24, 27, 30 \) (7 numbers) - Numbers divisible by 5: \( 15, 20, 25, 30 \) (4 numbers) Note that the number 15 and 30 appear in both sets, so we need to subtract those to avoid double counting: - Numbers divisible by both 3 and 5 (i.e., divisible by 15): \( 15, 30 \) (2 numbers) By the principle of inclusion and exclusion, the total number of numbers divisible by either 3 or 5 is: \[ \text{Numbers divisible by 3 or 5} = 7 + 4 - 2 = 9 \] Thus, the number of numbers neither divisible by 3 nor 5 is: \[ \text{Numbers neither divisible by 3 nor 5} = 20 - 9 = 11 \] The probability is then the ratio of favorable outcomes to the total number of outcomes: \[ P = \frac{\text{Numbers neither divisible by 3 nor 5}}{\text{Total numbers}} = \frac{11}{20} \] Thus, the probability that \( n \) is neither divisible by 3 nor divisible by 5 is \( \frac{11}{20} \).