Answer:
[tex]\frac{1}{6}[/tex]
Step-by-step explanation:
The sum is larger than 9, it means we have: 10, 11, and 12
P(12) = [tex]\frac{1}{6} *\frac{1}{6}[/tex] = [tex]\frac{1}{36}[/tex]
P(11) = 2* [tex]\frac{1}{6} *\frac{1}{6}[/tex] = [tex]\frac{1}{18}[/tex]
P(10) = 3* [tex]\frac{1}{6} *\frac{1}{6}[/tex] = [tex]\frac{1}{12}[/tex]
So the probability of getting a number larger than 9 for the first time on the third roll is:
P(12) + P(11) +P(10 )
= [tex]\frac{1}{36}[/tex] + [tex]\frac{1}{18}[/tex] + [tex]\frac{1}{12}[/tex] = [tex]\frac{1}{6}[/tex]