Three distinct numbers are selected from $100$ natural number. Find the probability that all the three numbers are divisible by $2$ and $3$ .

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Solution

The numbers should be divisible by $6$ . Thus the number of favourable ways is (as there are $16$ numbers in first $100$ natural numbers, divisible by $6$ ).
Required probability is $= \frac{^{16} C_{3}}{^{100} C_{3}}$
$= \frac{16 \times 15 \times 14}{100 \times 99 \times 98}$
$= \frac{2 \times 2}{5 \times 33 \times 7}$
$= \frac{4}{1155}$

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Three distinct numbers are selected from 100 natural number. Find the probability that all the three numbers are divisible by 2 and 3.

The numbers should be divisible by 6. Thus the number of favourable ways is (as there are 16 numbers in first 100 natural numbers, divisible by 6). Required probability is=16C3100C3=16×15×14100×99×98=2×25×33×7=41155

Three distinct numbers are selected from $100$ natural number. Find the probability that all the three numbers are divisible by $2$ and $3$ .