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(C) The total number of ways to choose $2$ numbers from $8$ is given by ${}^8C_2 = \frac{8 	imes 7}{2} = 28$. Let the two chosen numbers be $x$ and $

Vedclass · Accepted Answer

$\frac{9}{14}$

Vedclass · Answer

(C) The total number of ways to choose $2$ numbers from $8$ is given by ${}^8C_2 = \frac{8 	imes 7}{2} = 28$. Let the two chosen numbers be $x$ and $y$ such that $x Case $I$: If $x = 1$, then $y$ can be any of the remaining $7$ numbers $(\{2, 3, 4, 5, 6, 7, 8\})$. Number of ways $= 7$. Case $II$: If $x = 2$, then $y$ can be any of the remaining $6$ numbers $(\{3, 4, 5, 6, 7, 8\})$. Number of ways $= 6$. Case $III$: If $x = 3$, then $y$ can be any of the remaining $5$ numbers $(\{4, 5, 6, 7, 8\})$. Number of ways $= 5$. Total favourable cases $= 7 + 6 + 5 = 18$. Therefore, the required probability $= \frac{18}{28} = \frac{9}{14}$.

Two numbers are chosen at random from $\{1, 2, 3, 4, 5, 6, 7, 8\}$ at a time. The probability that the smaller of the two numbers is less than $4$ is

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