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

To determine the probability that a randomly chosen number \( n \) from the set \{11, 12, 13, \ldots, 30\} is neither divisible by 3 nor divisible by 5, we can follow these steps:

Step 1: Determine the Total Number of Outcomes

The set contains numbers from 11 to 30, inclusive. To find the total number of outcomes, compute:

\( 30 - 11 + 1 = 20 \)

Thus, there are 20 numbers in the set.

Step 2: Determine Numbers Divisible by 3 or 5

Numbers Divisible by 3:

The numbers divisible by 3 are identified by solving \( n \equiv 0 \pmod{3} \) within the range. Numbers satisfying this are 12, 15, 18, 21, 24, 27, 30.

Count: 7 numbers.

Numbers Divisible by 5:

The numbers divisible by 5 are identified by solving \( n \equiv 0 \pmod{5} \) within the range. Numbers satisfying this are 15, 20, 25, 30.

Count: 4 numbers.

Step 3: Use Inclusion-Exclusion Principle

To find numbers that are either divisible by 3 or by 5, apply the inclusion-exclusion principle:

\( |A \cup B| = |A| + |B| - |A \cap B| \)

where:

\(|A|\) = Numbers divisible by 3 = 7

\(|B|\) = Numbers divisible by 5 = 4

\(|A \cap B|\) = Numbers divisible by both, i.e., 15, 30

Count: 2 numbers.

Applying the principle:

\( |A \cup B| = 7 + 4 - 2 = 9 \)

This means there are 9 numbers divisible by either 3 or 5.

Step 4: Calculate the Probability

Numbers that are neither divisible by 3 nor by 5 are given by the total numbers minus those divisible by 3 or 5:

Total numbers = 20

Numbers neither divisible by 3 nor by 5 = 20 - 9 = 11

Thus, the probability \( P \) is:

\( P = \frac{11}{20} \)

Therefore, the probability that a number \( n \) is neither divisible by 3 nor divisible by 5 is \( \frac{11}{20} \).

Answer 2

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 \( \frac{11}{20} \).