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14.Probability
medium
If Mohan has $3$ tickets of a lottery containing $3$ prizes and $9$ blanks, then his chance of winning prize are
A
$\frac{{34}}{{55}}$
B
$\frac{{21}}{{55}}$
C
$\frac{{17}}{{55}}$
D
None of these
Solution
(a) 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}}.$
Standard 11
Mathematics
