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(D) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$. For a sum of $6$,the favorable outcomes are $(1, 5), (2, 4), (3, 3), (

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$1 : 1$

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(D) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$. For a sum of $6$,the favorable outcomes are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$,so there are $5$ outcomes. The probability of getting a sum of $6$ in one trial is $p_1 = \frac{5}{36}$. The probability of not getting a sum of $6$ in one trial is $q_1 = 1 - \frac{5}{36} = \frac{31}{36}$. The probability of getting a sum of $6$ at least once in $9$ trials is $P_1 = 1 - (q_1)^9 = 1 - \left(\frac{31}{36}ight)^9$. For a sum of $8$,the favorable outcomes are $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$,so there are $5$ outcomes. The probability of getting a sum of $8$ in one trial is $p_2 = \frac{5}{36}$. The probability of not getting a sum of $8$ in one trial is $q_2 = 1 - \frac{5}{36} = \frac{31}{36}$. The probability of getting a sum of $8$ at least once in $9$ trials is $P_2 = 1 - (q_2)^9 = 1 - \left(\frac{31}{36}ight)^9$. Since $P_1 = P_2$,the ratio $P_1 : P_2 = 1 : 1$.

In a random experiment,two dice are thrown and the sum of the numbers appeared on them is recorded. This experiment is repeated $9$ times. If the probability that a sum of $6$ appears at least once is $P_1$ and a sum of $8$ appears at least once is $P_2$,then $P_1 : P_2 =$

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