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If Mohan has 3 tickets of a lottery containing 3 prizes and 9 blanks, then his chance of winning prize are:

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$\frac{34}{55}$

$\frac{21}{55}$

$\frac{17}{55}$

None of these

answer is A.

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Detailed 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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