If three distinct numbers are chosen randomly from the first 100 natural numbers, the probability that all three of them are divisible by both 2 and 3 is:
Given that the total numbers divisible by both 2 and 3 are \(6, 12, 18, \dots, 96\), the total number of such numbers is 16. \[\] The total number of natural numbers is 100. \[\] The probability \( P(E) \) is given by: \[ P(E) = \frac{{}^{16}C_3}{{}^{100}C_3} \] Using the formula for combinations: \[ P(E) = \frac{16 \times 15 \times 14}{100 \times 99 \times 98} \] Simplifying the expression: \[ P(E) = \frac{4}{1155} \]
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