Let p be the probability of success that is getting 5 =

1

6

⇒q=1−p=1−

1

6

=

5

6

Let X has a binomial distribution with n=7,p=

1

6

,q=

5

6

∴ P(getting 5 exactly twice) = P(X = 2) =7C2q5p2
=

7×6

2×1

×(

5

6

)5(

1

6

)2
=

7

12

(

5

6

)5