of throws, follow these steps:

Total Possible Outcomes: Each die has 6 faces, so there are a total of $ 6 \times 6 = 36 $ possible outcomes when throwing two dice.

Outcomes for Sum of 6:

Possible pairs: $ (1,5), (2,4), (3,3), (4,2), (5,1) $.

Number of outcomes: 5

Probability of not getting a sum of 6 in one throw is:

$ \frac{36 - 5}{36} = \frac{31}{36} $

Probability of Not Getting Sum of 6 in 9 Throws:

$ \left(\frac{31}{36}\right)^9 $

Probability of Getting Sum of 6 at Least Once ($ P_1 $):

$ P_1 = 1 - \left(\frac{31}{36}\right)^9 $

Outcomes for Sum of 8:

Possible pairs: $ (2,6), (3,5), (4,4), (5,3), (6,2) $.

Number of outcomes: 5

Probability of not getting a sum of 8 in one throw is:

$ \frac{36 - 5}{36} = \frac{31}{36} $

Probability of Not Getting Sum of 8 in 9 Throws:

$ \left(\frac{31}{36}\right)^9 $

Probability of Getting Sum of 8 at Least Once ($ P_2 $):

$ P_2 = 1 - \left(\frac{31}{36}\right)^9 $

Since the expressions for $ P_1 $ and $ P_2 $ are identical:

$ P_1 : P_2 = \frac{1 - \left(\frac{31}{36}\right)^9}{1 - \left(\frac{31}{36}\right)^9} = 1:1 $

Thus, the probability ratio is $ 1:1 $.

Probability

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