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(D) Let the two selected numbers be $x$ and $y$ where $x < y$. The total number of ways to select two numbers from $10$ is $^{10}C_2 = \frac{10 	imes

Vedclass · Accepted Answer

$\frac{19}{45}$

Vedclass · Answer

(D) Let the two selected numbers be $x$ and $y$ where $x Let $A$ be the event that the minimum is divisible by $3$. The possible values for the minimum are $3, 6, 9$.If min is $3$,$y \in \{4, 5, 6, 7, 8, 9, 10\}$ ($7$ pairs).If min is $6$,$y \in \{7, 8, 9, 10\}$ ($4$ pairs).If min is $9$,$y = 10$ ($1$ pair).Total for $A = 7 + 4 + 1 = 12$.Let $B$ be the event that the maximum is divisible by $4$. The possible values for the maximum are $4, 8$.If max is $4$,$x \in \{1, 2, 3\}$ ($3$ pairs).If max is $8$,$x \in \{1, 2, 3, 4, 5, 6, 7\}$ ($7$ pairs).Total for $B = 3 + 7 = 10$.Intersection $A \cap B$: Pairs where min is divisible by $3$ $AND$ max is divisible by $4$.Pairs: $(3, 4), (3, 8), (6, 8)$. Total = $3$.Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{12}{45} + \frac{10}{45} - \frac{3}{45} = \frac{19}{45}$.

Two numbers are selected randomly from the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ without replacement one by one. The probability that the minimum of the two numbers is divisible by $3$ or the maximum of the two numbers is divisible by $4$ is:

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