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(A) Given that $E_k = \{(a, b) \in S : ab = k\}$ for $k \geq 1$ and $p_k = P(E_k)$.Since there are $6 	imes 6 = 36$ total outcomes in the sample spac

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$p_1 < p_{30} < p_4 < p_6$

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(A) Given that $E_k = \{(a, b) \in S : ab = k\}$ for $k \geq 1$ and $p_k = P(E_k)$.Since there are $6 	imes 6 = 36$ total outcomes in the sample space $S$,the probability $p_k$ is given by $\frac{|E_k|}{36}$.For $k=1$: $E_1 = \{(1, 1)\}$,so $|E_1| = 1$ and $p_1 = \frac{1}{36}$.For $k=2$: $E_2 = \{(1, 2), (2, 1)\}$,so $|E_2| = 2$ and $p_2 = \frac{2}{36}$.For $k=4$: $E_4 = \{(1, 4), (4, 1), (2, 2)\}$,so $|E_4| = 3$ and $p_4 = \frac{3}{36}$.For $k=6$: $E_6 = \{(1, 6), (6, 1), (2, 3), (3, 2)\}$,so $|E_6| = 4$ and $p_6 = \frac{4}{36}$.For $k=30$: $E_{30} = \{(5, 6), (6, 5)\}$,so $|E_{30}| = 2$ and $p_{30} = \frac{2}{36}$.Comparing the values: $p_1 = \frac{1}{36}$,$p_{30} = \frac{2}{36}$,$p_4 = \frac{3}{36}$,$p_6 = \frac{4}{36}$.Thus,$p_1 Therefore,the correct option is $A$.

Let $S$ be the sample space of the random experiment of throwing simultaneously two unbiased dice with six faces (numbered $1$ to $6$) and let $E_k = \{(a, b) \in S : ab = k\}$ for $k \geq 1$. If $p_k = P(E_k)$ for $k \geq 1$,then which of the following is correct?

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