A) \[\frac{34}{55}\]
B) \[\frac{21}{55}\]
C) \[\frac{17}{55}\]
D) None of these
Correct Answer: A
Solution :
Mohan can gets one prize, 2 prizes or 3 prizes and his chance of failure means he get no prize. Number of total ways \[={}^{12}{{C}_{3}}=220\] Favourable number of ways to be failure \[={}^{9}{{C}_{3}}=84\] Hence required probability \[=1-\frac{84}{220}=\frac{34}{55}.\]
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