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

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Solution

Number between $1$ and $100$ which are divisible by both $2$ and $3$ are $16 {6, 12, 18, . . . . . . ., 96}$

Hence, the probability $= {\frac{^{16} C_{3}}{^{100} C}}_{3} = \frac{560}{161700} = \frac{4}{1155}$

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

Number between 1 and 100 which are divisible by both 2 and 3 are 16{6,12,18,.......,96}Hence, the probability =16C3100C3=560161700=41155

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

Number between $1$ and $100$ which are divisible by both $2$ and $3$ are $16 {6, 12, 18, . . . . . . ., 96}$

Hence, the probability $= {\frac{^{16} C_{3}}{^{100} C}}_{3} = \frac{560}{161700} = \frac{4}{1155}$