Two persons each make a single throw with a d | vedclass

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(C) For ${p_1}$: Two persons throw a die. Total outcomes $= 6 	imes 6 = 36$. Favourable outcomes (equal values) are $(1,1), (2,2), (3,3), (4,4), (5,5

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${p_1} > {p_2}$

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(C) For ${p_1}$: Two persons throw a die. Total outcomes $= 6 	imes 6 = 36$. Favourable outcomes (equal values) are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$,which are $6$ outcomes. So,${p_1} = \frac{6}{36} = \frac{1}{6}$.For ${p_2}$: Four persons throw a die. Total outcomes $= 6^4 = 1296$. We need the probability that exactly three persons get the same value. Step $1$: Choose the value that appears three times: ${}^6C_1 = 6$ ways.Step $2$: Choose the three persons out of four who get this value: ${}^4C_3 = 4$ ways.Step $3$: Choose the value for the fourth person (must be different from the first value): ${}^5C_1 = 5$ ways.Favourable outcomes $= 6 	imes 4 	imes 5 = 120$.So,${p_2} = \frac{120}{1296} = \frac{5}{54}$.Comparing ${p_1} = \frac{1}{6} = \frac{9}{54}$ and ${p_2} = \frac{5}{54}$,we get ${p_1} > {p_2}$.

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 three being equal is ${p_2}$,then:

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