+77 Mr. C rolls four dice. What is the probability of getting at most three 6's on top?
+118654 1-[P(4 sixes and 2 non sixes)+(5 sixes and one non six)+P(6 sixes)]
\(1-[\binom{6}{4}(\frac{1}{6})^4(\frac{5}{6})^2+\binom{6}{5}(\frac{1}{6})^5(\frac{5}{6})^1+\binom{6}{6}(\frac{1}{6})^6]\\ 1-[15(\frac{1}{6})^4(\frac{5}{6})^2+6(\frac{1}{6})^5(\frac{5}{6})^1+(\frac{1}{6})^6]\\ 1-[15(\frac{5^2}{6^6})+6(\frac{5}{6^6})+(\frac{1}{6^6})]\\ 1-\frac{1}{6^6}[15*25+30+1]\\ 1-\frac{406}{6^6}\\ \)
1-406/6^6 = 0.9912980109739369
+129840 "At most" means either no dice show a "6", 1 die shows a "6", 2 die show a "6" or 3 die show a "6"
No die shows a "6' = (5/6)^4 = 625/1296
I die shows a "6" = C(4,1) * (1/6)*(5/6)^3 = 125/324 = 500/1296
2 die show a "6" = C(4,2) * (1/6)^2*(5/6)^2 = 25 /216 = 150 / 1296
3 die show a "6" = C(4,3) *(1/6)^3 * (5/6) = 5/324 = 20/ 1296
P ( getting at most three 6's on top) = [ 625 + 500 + 150 + 20 ] / 1296 =
0.9992283950617284 ≈ 99.99% [almost a sure thing....!!!!]
+118654 I'll have to put my specs on the right side up next time. I answered for 6 dice. 
