## Solution :

Probability of getting score 9 in a single throw

= \(4\over 36\) = \(1\over 9\)

Probability of getting score 9 exactly in double throw

= \(^3C_2\) \(\times\) \(({1\over 9})^2\) \(\times\) \(8\over 9\) = \(8\over 243\)

Probability of getting score 9 in a single throw

= \(4\over 36\) = \(1\over 9\)

Probability of getting score 9 exactly in double throw

= \(^3C_2\) \(\times\) \(({1\over 9})^2\) \(\times\) \(8\over 9\) = \(8\over 243\)