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(A) Total number of ways to draw $3$ tickets out of $20$ is given by $^{20}C_3 = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 1140$.We want to

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$\frac{3}{190}$

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(A) Total number of ways to draw $3$ tickets out of $20$ is given by $^{20}C_3 = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 1140$.We want to find the probability that the tickets marked $7$ and $11$ are included in the selection.Since $2$ tickets ($7$ and $11$) are already fixed,we only need to choose $1$ more ticket from the remaining $18$ tickets $(20 - 2 = 18)$.The number of ways to choose the $3^{rd}$ ticket is $^{18}C_1 = 18$.Therefore,the required probability is $\frac{18}{1140} = \frac{18}{1140} = \frac{3}{190}$.

Twenty tickets are marked with the numbers $1, 2, \dots, 20$. If three tickets are drawn at random,what is the probability that those marked $7$ and $11$ are among them?

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