An integer is chosen from the set {2k mid -9 | vedclass

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(D) The set is given by $S = \{2k \mid -9 \leq k \leq 10\}$.Since $k$ ranges from $-9$ to $10$,the number of values for $k$ is $10 - (-9) + 1 = 20$.Th

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$\frac{3}{20}$

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(D) The set is given by $S = \{2k \mid -9 \leq k \leq 10\}$.Since $k$ ranges from $-9$ to $10$,the number of values for $k$ is $10 - (-9) + 1 = 20$.Thus,the total number of elements in set $S$ is $20$.The elements are $\{-18, -16, -14, -12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$.An integer is divisible by both $4$ and $6$ if it is divisible by $	ext{lcm}(4, 6) = 12$.We look for elements in $S$ that are multiples of $12$.These are $\{-12, 0, 12\}$.There are $3$ such favourable outcomes.Therefore,the required probability is $\frac{3}{20}$.

An integer is chosen from the set $\{2k \mid -9 \leq k \leq 10\}$. The probability that the chosen integer is divisible by both $4$ and $6$ is

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