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(C) For $p_1$: Two persons throw a die. Total outcomes = $6^2 = 36$. Favorable outcomes (both same) are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$,so

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$p_1 > p_2$

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(C) For $p_1$: Two persons throw a die. Total outcomes = $6^2 = 36$. Favorable outcomes (both same) are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$,so $6$ outcomes. Thus,$p_1 = \frac{6}{36} = \frac{1}{6} \approx 0.1667$.For $p_2$: Four persons throw a die. Total outcomes = $6^4 = 1296$. We want exactly three to be equal. First,choose the value that appears three times: $6$ ways. Choose the three persons who get this value: $\binom{4}{3} = 4$ ways. Choose the value for the fourth person (must be different from the first): $5$ ways. Total favorable outcomes = $6 	imes 4 	imes 5 = 120$.Thus,$p_2 = \frac{120}{1296} = \frac{5}{54} \approx 0.0926$.Comparing the two,$p_1 = \frac{9}{54}$ and $p_2 = \frac{5}{54}$,so $p_1 > p_2$.

Two persons each make a single throw with a die. The probability they get equal value is $p_1$. Four persons each make a single throw and the probability of exactly three being equal is $p_2$,then:

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