Probability of rolling sixes in 50 rolls of a fair die A fair die is rolled 50 times. Find the probability of observing:

a) exactly 10 sixes
   b) no more than 10 sixes 
   c) at least 10 sixes

I know how to do 

a) $\frac{50!}{10!(40!)}$x$(\frac{1}{6})^{10}$x$(1-\frac{1}{6})^{50-10}$ 
  =0.1155

Please help me out to do b) & c)  I have tried same formula above and changing power to 9, 11 etc...
But Can't get right answer.!
Appreciate your help!
 A: Hint: You will get no more than 10 sixes if you get no sixes or one six or two sixes or three sixes or ... or nine sixes.  Since these possibilities are mutually exclusive, you can add the individual probabilities to get the total probability.
Once you've solved (b), think how you can use that answer to solve (c).
A: Hint: Your formula can be generalized to exactly $k$ sixes with
$$
\binom{50}{k}\left(\frac16\right)^k\left(\frac56\right)^{50-k}\tag{1}
$$
where
$$
\binom{n}{k}=\frac{n!}{k!\,(n-k)!}\tag{2}
$$
A: Hint: If you get no six, you don't get exactly one six, or exactly two, or exactly three, and so on.
Similarly, if you get exactly one six, you don't get nonte, nor exactly two, and so on.
If you have several possibilities which exclude each other (i.e. if one of then occurs, none of the other has occurred), then the probability of getting any of those possibilities is just the sum of the probabilities of the individual probabilities.
"No more than 10" is "None, or one, or …, or 8, or 9, or 10".
"At least 10" is "more than 9", which is the opposite of "not more than none".
Since you already know how to calculate the probability of "Exactly $k$ sixes", those hints should enable you to solve the other questions.
