Dice is thrown 6 times. What's the probability of numbers 5 and 6 showing up at least once.
This ought to be basic probability calculus but i just can't get my head around this. Some kind of attempt(ish) below. THe answer ought to be 0.418 or 41.8%.
The Attempt at a Solution
Now, i know the amount of different permutations is 6^6. Suppose A = [a number 5 appears] and B = [a number 6 appears].
I think i'd need to do this as
P(A n B) = 1 - P(A^c) - P(B^c) + P(A^c n B^c) where ^c denotes a complement.
Probability of A being false (5 not showing) = (5/6)^6 and it's the same for B. I just can't figure out what P(A^c n B^c) is... It's "something" divided by 6^6 but that's as far as i get... Any help would be appreciated.