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(B) Step $1$: Total number of ways to choose $3$ numbers from $30$ is $\binom{30}{3} = \frac{30 	imes 29 	imes 28}{3 	imes 2 	imes 1} = 4060$.Step

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$\frac{3}{812}$

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(B) Step $1$: Total number of ways to choose $3$ numbers from $30$ is $\binom{30}{3} = \frac{30 	imes 29 	imes 28}{3 	imes 2 	imes 1} = 4060$.Step $2$: For the minimum to be $10$ and the maximum to be $26$, one number must be $10$, one must be $26$, and the third number $x$ must satisfy $10 Step $3$: The numbers between $10$ and $26$ are $\{11, 12, \dots, 25\}$. The count of these numbers is $26 - 10 - 1 = 15$.Step $4$: The number of favorable outcomes is $15$.Step $5$: The probability is $\frac{15}{4060} = \frac{3}{812}$.

Three numbers are chosen from $1$ to $30$. The probability that the minimum number is $10$ and the maximum number is $26$ is

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