If Mohan has 3 tickets of a lottery containin | vedclass

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(A) Total number of tickets $= 3 + 9 = 12$.Mohan selects $3$ tickets out of $12$. The total number of ways to select $3$ tickets is ${}^{12}C_3 = \fra

Vedclass · Accepted Answer

$\frac{34}{55}$

Vedclass · Answer

(A) Total number of tickets $= 3 + 9 = 12$.Mohan selects $3$ tickets out of $12$. The total number of ways to select $3$ tickets is ${}^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.The probability of winning at least one prize is $1 - P(	ext{winning no prize})$.Winning no prize means all $3$ tickets are selected from the $9$ blanks.The number of ways to select $3$ blanks out of $9$ is ${}^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.The probability of winning no prize is $\frac{84}{220} = \frac{21}{55}$.Therefore,the probability of winning at least one prize is $1 - \frac{21}{55} = \frac{34}{55}$.

If Mohan has $3$ tickets of a lottery containing $3$ prizes and $9$ blanks,then his chance of winning at least one prize is

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