A) \[\frac{4}{105}\]
B) \[\frac{4}{33}\]
C) \[\frac{4}{35}\]
D) \[\frac{4}{1155}\]
E) \[\frac{3}{1155}\]
Correct Answer: D
Solution :
Let E= Events of numbers divisible by 2 and 3 (i.e., divisible by 6) \[=(6,12,...,96)\] \[n(E)=16\] \[\therefore \]Required probability\[=\frac{^{16}{{C}_{3}}}{{{100}_{{{C}_{3}}}}}\] \[=\frac{\frac{16\times 15\times 14}{3\times 2\times 1}}{\frac{100\times 99\times 98}{3\times 2\times 1}}\] \[=\frac{4}{1155}\]
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