A) \[\frac{16}{81}\]
B) \[\frac{144}{145}\]
C) \[\frac{80}{145}\]
D) \[\frac{65}{81}\]
Correct Answer: B
Solution :
The total number of ways in which \[3\] numbers can be chosen out of \[30\] numbers\[{{=}^{30}}{{C}_{3}}=4060\]. The number of ways of choosing \[3\] consecutive numbers is\[28\]. Therefore, the number of ways in which the three numbers chosen are not consecutive is\[4060-28=4032\]. Hence, the required probability\[=\frac{4032}{4060}=\frac{144}{145}\]
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