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(C) Total ways to choose $4$ numbers from $40$ is $^{40}C_4 = \frac{40 	imes 39 	imes 38 	imes 37}{4 	imes 3 	imes 2 	imes 1} = 91390$. Number o

Vedclass · Accepted Answer

$\frac{2469}{2470}$

Vedclass · Answer

(C) Total ways to choose $4$ numbers from $40$ is $^{40}C_4 = \frac{40 	imes 39 	imes 38 	imes 37}{4 	imes 3 	imes 2 	imes 1} = 91390$. Number of ways to choose $4$ consecutive numbers is $37$ (i.e.,$\{1,2,3,4\}, \{2,3,4,5\}, \ldots, \{37,38,39,40\}$). Probability that the $4$ numbers are consecutive is $P(C) = \frac{37}{91390} = \frac{1}{2470}$. Probability that they are not consecutive is $1 - P(C) = 1 - \frac{1}{2470} = \frac{2469}{2470}$.

Four numbers are chosen at random from $\{1, 2, 3, \ldots, 40\}$. The probability that they are not consecutive is:

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