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

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14.Probability

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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

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

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