| im not 100% sure but since it's at least 2 eggs, you find the probability of 1 egg and 0 egg being broken then u subtract that from 1.
0 egg- .8^6 = .262~
1 egg- (.8^5) * (.2^1) * 6C1 (combination) = .393~
1 - .262 - .393 = .345
So theres around a 34.5% chance that at least 2 eggs are broken if 6 eggs are chosen at random. But like i said I dunno if i did it right...but anyways is that even a real sat problem? |