A fair die is cast for times calculate the probability of obtaining at least two sixes

You can solve this using the binomial probability formula.

The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.

Then, we can set the equation as follows:

P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) 

when x=2  (4 2)(1/6)^2(5/6)^4-2 = 0.1157

when x=3  (4 3)(1/6)^3(5/6)^4-3 = 0.0154

when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008

Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)

P(2 sixes) = (1/6)2(5/6)2

Add these up.  I got 31/64

Check my work, and compute the ratio to obtain the decimal value, then convert to % by the usual process.