Question

If a two digit number is choosen at random then the probability that number choosen is a multiple of 3

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{1}{5}\)

Answer 1

The Correct Option is B

To solve this problem, we need to find the probability that a randomly chosen two-digit number is a multiple of 3.

1. Total Number of Two-Digit Numbers:
Two-digit numbers range from 10 to 99.
So, the total number of two-digit numbers is:
\[ 99 - 10 + 1 = 90 \]

2. Counting Multiples of 3 Among Two-Digit Numbers:
The smallest two-digit multiple of 3 is 12, and the largest is 99.
We form an arithmetic sequence: 12, 15, 18, ..., 99 with common difference 3.
Using the formula for the nth term: \[ a_n = a + (n - 1)d \Rightarrow 99 = 12 + (n - 1) \cdot 3 \Rightarrow 87 = (n - 1) \cdot 3 \Rightarrow n - 1 = 29 \Rightarrow n = 30 \] So, there are 30 two-digit multiples of 3.

3. Calculating the Probability:
\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{30}{90} = \frac{1}{3} \]

Final Answer:
The correct answer is (B) \( \frac{1}{3} \).