Just saw Joe's post -- I am way too slow.
probability of expectation is the product of combinations times the probability of rolling only one 1 or 5
P(E)=3!/(1!2!)*(2/6)*(4/6)*(4/6)=3*32/216=96/216=4/9
Rap
oolongteasup wrote:
P(none of 1 or5) is 3C0*(2/6)^0* (4/6)^3 = 64/216
P(exactly one of 1 or 5) is 3C1*(2/6)^1*(4/6)^2=96/216
P(exactly two of 1 or 5) is 3C2*(2/6^2*(4/6)^1=48/216
P(exactly three of 1 or 5) is 3C3*(2/6)^3*(4/6)0=8/216
ergo 144/216 , funny how the probabilities add to 1
All that work and you still got it wrong.
my mistake
8 ways to roll all 1's and 5's only 2 of which are all1's or 5's
so 150/216 is correct
