You can
put this solution on YOUR website!Your numbers have taken care of the independent nature of the event but
not the mutually exclusive nature of the event.
For example getting one "1" on the 8 dice is (1/6)^1(5/6)^7 takes care
of getting the "1" on the 1st dice; but getting the one on the
2nd dice is a mututally exclusive event and has the same probability,
and the same for a "1" on the 3rd dice etc.
So the probability of getting one "1" is 8C1(1/6)(5/6)^7 = 8(1/6)(5/6)^7
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Similarly the probability of getting two "1's" is 8C2(1/6)^2(5/6)^6=
28(1/6)^2(5/6)^6
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If you continue this to 8C8(1/6)^8(5/6)^0 and add up the terms they will
add up to one.
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Cheers,
Stan H.