# Probability of number 6 occurence between 8 and 12 when die is tossed 60 times

I tried solving this question:

Toss a die 60 times, what is the probability that the number of sixes is a) greater or equal 12, b) strictly between 8 and 12

My solution for part a was P(grt or equal 12) = 1 - P(less than 12) = 1 - (11 * 1/6)/60 but i don't think it's correct. I don't have any idea about part b.

(a) $$\sum_{k=12}^{60}\binom{60}{k}\left(\frac{1}{6}\right)^{k}\left(\frac{5}{6}\right)^{60-k}=1-\sum_{k=0}^{11}\binom{60}{k}\left(\frac{1}{6}\right)^{k}\left(\frac{5}{6}\right)^{60-k}.$$
(b) $$\sum_{k=9}^{11}\binom{60}{k}\left(\frac{1}{6}\right)^{k}\left(\frac{5}{6}\right)^{60-k}.$$
Explanation: suppose that you are looking for the probability of the number of sixes being exactly $k$ ($k$ integer, $0\leq k\leq60$). Then, the event of a six, which has probability $1/6$, must occur exactly $k$ times, whereas the complementary event, whose probability is $5/6$, must occur $60-k$ times. Finally, the number of ways in which you can choose which tosses out of the 60 tosses are to result in $6$ is $\binom{60}{k}$.
The answer for $(a)$ is approximately 0.292056, by the work performed in the above answer. The answer for $(b)$ is approximately 0.395897 by the work performed in the above answer.