A fair 8-sided dice labeled 1 to 8 is tossed three times. What is the probability the value of the dice shown is 3 at least once?	Choose the correct choice.
A) 57/512	B) 7/512	C) 55/512	D) 9/512	E) 11/512
This is a fair dice with 8 fair sides, so the probability of the dice landing on any sides is equal.   The question ask if the value of the dice is shown 3 at least once   So the calculation of the first possible outcomes is 1/8*1/8*1/8 (all of the dice thrown shown the value of 3)   This is follow by the second possible outcomes if the dice thrown show the value of 3 2 times ( 1/8*1/8*7/8) *7 here referring to any number besides 3   The third possible outcomes is that the dice show the value 3 one time ( 1/8*7/8*7/8)   So the answer is 1/512+7/512+49/512= 57/512   57/512nswer: 57/512
<result>A</result>
A fair 8-sided dice labeled 1 to 8 is tossed three times. What is the probability the value of the dice shown is 3 at least once?