A) \[\frac{3}{190}\]
B) \[\frac{1}{19}\]
C) \[\frac{1}{190}\]
D) None of these
Correct Answer: A
Solution :
\[7,\,\,11\] have always to be in that group of three, therefore 3rd ticket may be chosen in 18 ways. Hence required probability is \[\frac{18}{{}^{20}{{C}_{3}}}=\frac{18\,.\,3\,.\,2}{20\,.\,19\,.\,18}=\frac{3}{190}\].
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