Question

If three distinct numbers are chosen randomly, three of them are divisible by both 2 and 3 from the first 100 natural numbers, then the probability that all three are divisible by both 2 and 3 is:

• A. \( \frac{4}{33} \)
• B. \( \frac{4}{25} \)
• C. \( \frac{4}{1155} \)
• D. \( \frac{4}{35} \)

Hint:

When calculating probabilities with combinations, carefully consider the total number of favorable outcomes and divide by the total possible outcomes.

Answer 1

The Correct Option is C

The numbers divisible by both 2 and 3 are divisible by 6. The number of such numbers from 1 to 100 is: \[ \left\lfloor \frac{100}{6} \right\rfloor = 16. \] Thus, there are 16 numbers divisible by both 2 and 3. The probability of selecting 3 numbers divisible by 6 out of 100 is given by: \[ \frac{\binom{16}{3}}{\binom{100}{3}}. \] This simplifies to: \[ \frac{4}{1155}. \] Thus, the correct answer is: \[ \boxed{\frac{4}{1155}}. \]