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(A) The sample space $S$ for throwing two dice has $6 	imes 6 = 36$ outcomes. $p_k = P(E_k) = \frac{|E_k|}{36}$. For $k=1$: $E_1 = \{(1, 1)\}$, so $p

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The sample space $S$ for throwing two dice has $6 	imes 6 = 36$ outcomes. $p_k = P(E_k) = \frac{|E_k|}{36}$. For $k=1$: $E_1 = \{(1, 1)\}$, so $p_1 = \frac{1}{36}$. For $k=2$: $E_2 = \{(1, 2), (2, 1)\}$, so $p_2 = \frac{2}{36}$. For $k=4$: $E_4 = \{(1, 4), (4, 1), (2, 2)\}$, so $p_4 = \frac{3}{36}$. For $k=6$: $E_6 = \{(1, 6), (6, 1), (2, 3), (3, 2)\}$, so $p_6 = \frac{4}{36}$. For $k=30$: $E_{30} = \{(5, 6), (6, 5)\}$, so $p_{30} = \frac{2}{36}$. For $k=11$: $E_{11} = \emptyset$, so $p_{11} = 0$. For $k=36$: $E_{36} = \{(6, 6)\}$, so $p_{36} = \frac{1}{36}$. Comparing the values: $p_1 = \frac{1}{36}$, $p_{30} = \frac{2}{36}$, $p_4 = \frac{3}{36}$, $p_6 = \frac{4}{36}$. Thus, $p_1 < p_{30} < p_4 < p_6$ is correct.

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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