This can be solved by using the binomial distribution to find the probability of rolling 0,1, 2, 3 or 4 6's in 25 rolls. Then that value of probability is subtracted from 1 to find the probability of 5 or more 6's.
[tex]P(zero\ sixes)=25C0\times(\frac{1}{6})^{0}\times(\frac{5}{6})^{25}=0.0105[/tex]
[tex]P(one\ six)=25C1\times\frac{1}{6}\times(\frac{5}{6})^{24}=0.0524[/tex]
P(2 sixes) = 0.1258
P(3 sixes) = 0.1929
[tex]P(4\ sixes)=25C4\times(\frac{1}{6})^{4}\times(\frac{5}{6})^{21}=0.2122[/tex]
P(0, 1, 2, 3 or 4 sixes) = 0.0105 + 0.0524 + 0.1258 + 0.1929 + 0.2122 = 0.5938
P(5 or more sixes) = 1.0000 - 0.5938 = 0.4062
