Three distinct numbers are selected from first 100 natural numbers. probability that all three numbe

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14.Probability

medium

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

$4/25$

$4/35$

$4/55$

$4/1155$

(IIT-2004)

Solution

(d) The numbers should be divisible by $6$. Thus, the number of favourable ways is $^{16}{C_3}$ (as there are $16$ numbers in first $100$ natural numbers, divisible by $6$).
Required probability is $\frac{{^{16}{C_3}}}{{^{100}{C_3}}} = \frac{{16 \times 15 \times 14}}{{100 \times 99 \times 98}} = \frac{4}{{1155}}$.

Standard 11

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(JEE MAIN-2017)