Question

What is the probability of a randomly chosen 2-digit number being divisible by 3?

• A. \( \frac{2}{9} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{9} \)

Hint:

To find the probability that a randomly chosen number is divisible by a given number, count how many numbers in the range are divisible by the number and divide by the total number of numbers in the range.

Answer 1

The Correct Option is C

The two-digit numbers range from 10 to 99. Therefore, the total number of two-digit numbers is: \[ 99 - 10 + 1 = 90 \] Now, a number is divisible by 3 if the sum of its digits is divisible by 3. The two-digit numbers divisible by 3 are 12, 15, 18, 21, ..., 99. These numbers form an arithmetic sequence where the first term is 12, the common difference is 3, and the last term is 99. We can use the formula for the nth term of an arithmetic sequence to find the total number of two-digit numbers divisible by 3. The nth term of an arithmetic sequence is given by: \[ a_n = a_1 + (n - 1) \cdot d \] where \( a_1 = 12 \), \( d = 3 \), and \( a_n = 99 \). Substituting these values into the formula: \[ 99 = 12 + (n - 1) \cdot 3 \] \[ 99 - 12 = (n - 1) \cdot 3 \] \[ 87 = (n - 1) \cdot 3 \] \[ n - 1 = 29 \quad \Rightarrow \quad n = 30 \] Thus, there are 30 two-digit numbers divisible by 3. Therefore, the probability that a randomly chosen two-digit number is divisible by 3 is: \[ \frac{\text{Number of two-digit numbers divisible by 3}}{\text{Total number of two-digit numbers}} = \frac{30}{90} = \frac{1}{3} \] Thus, the correct answer is \( \frac{1{3}} \).