This is a binomial distribution because there are only `2` possible outcomes (we get a `5` or we don't).

Now, `n = 3` for each part. Let `X =` number of fives appearing.

(a) Here, x = 0.

`P(X=0)` `=C_x^np^xq^[n-x]` `=C_0^3 (1/6)^0 (5/6)^3` `=125/216` `=0.5787 `

(b) Here, x = 1.

`P(X=1)` `=C_x^np^xq^[n-x]` `=C_1^3 (1/6)^1 (5/6)^2` `=75/216` `=0.34722 `

(c) Here, x = 3.

`P(X=3)=C_x^np^xq^[n-x]` `=C_3^3 (1/6)^3 (5/6)^0` `=1/216` `=4.6296times10^-3 `