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(A) The sample space $S$ consists of $6 	imes 6 = 36$ outcomes.$E_1 = \{(1, 1)\} \implies p_1 = \frac{1}{36}$.$E_4 = \{(1, 4), (2, 2), (4, 1)\} \impl

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$p_1 < p_{10} < p_4$

Vedclass · Answer

(A) The sample space $S$ consists of $6 	imes 6 = 36$ outcomes.$E_1 = \{(1, 1)\} \implies p_1 = \frac{1}{36}$.$E_4 = \{(1, 4), (2, 2), (4, 1)\} \implies p_4 = \frac{3}{36} = \frac{1}{12}$.$E_{10} = \{(2, 5), (5, 2)\} \implies p_{10} = \frac{2}{36} = \frac{1}{18}$.Comparing the values: $p_1 = \frac{1}{36} \approx 0.0277$, $p_{10} = \frac{2}{36} \approx 0.0555$, $p_4 = \frac{3}{36} \approx 0.0833$.Thus, $p_1 < p_{10} < p_4$ is correct.

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

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