Fair Die Rolled 6 times: prob. of no more than three 2's?

This is my last question. I'll check answer in 2-hours from now, if someone kind enough to respond.

A fair die is rolled 6 times. What is the probability of no more than three twos?

I chose 0.3812. can't remember how i got there, but it worked for a second.

In my textbook, it shows this example, which I think is the same question. Why isn't the choice for the same answer given, or how is it that it's different? Plus, I do not get where the 15 comes from?

"Find the probability of obtaining exactly two 5s on six rolls of a fair die. Let 5 be "success". Then n = 6, p = 1/6, q = 5/6, and x = 2.

P(2) = C (6,2) (1/6)^2 (5/6)^4 = 15 (1/36) (625/1296) = 3125/15,552 = .201

My other choices were: 0.9649, 0.6774, 0.9913. I cannot arrive at any one of them!Please help.

\(\displaystyle \L\sum\limits_{k = 0}^3 {\left( {\begin{array}{c}
6 \\
k \\
\end{array}} \right)\left( {\frac{1}{6}} \right)^k \left( {\frac{5}{6}} \right)^{6 - k} } .\)

i know. throw the calculator away. i keep getting -2.999964635 :-(

I get 0.991298010973937.

thank you pka. must be the operator