Probability Ques 2
- If three distinct numbers are chosen randomly from the first $100$ natural numbers, then the probability that all three of them are divisible by both $2$ and $3$ , is
$(2004,1 \mathrm{m})$
(a) $\frac{4}{55}$
(b) $\frac{4}{35}$
(c) $\frac{4}{33}$
(d) $\frac{4}{1155}$
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Answer:
Correct Answer: 2.(d)
Solution: (d) Since, three distinct numbers are to be selected from first $100$ natural numbers.
$ \Rightarrow \quad n(S)={ }^{100} C_3 $
$E_{\text {(favourable events) }}=$ All three of them are divisible by both $2$ and $3$ .
$\Rightarrow \quad $ Divisible by $6$ i.e. $\{6,12,18, \ldots, 96\}$
Thus, out of $16$ we have to select $3$ .
$ \begin{aligned} & \therefore \quad n(E)={ }^{16} C_3 \\ & \therefore \quad \text { Required probability }=\frac{{ }^{16} C_3}{{ }^{100} C_3}=\frac{4}{1155} \\ \end{aligned} $
BETA
