What is the probability of getting at least one six in a single throw of three unbiased dice ?  Select.
A) 31/216	B) 91/216	C) 125/216	D) 81/216	E) 81/359
Explanation :   Three cases arise:   Case 1: When only one dice shows up a six   This dice can be any of the 1st, 2nd or 3rd dice. Find the probability for these three independent events and add them up to get the total probability   Probability that only 1st dice shows up a six: (probability that first dice shows up a 6) and (probability that second dice shows up other than 6) and (probability that third dice shows up other than 6)   =(1/6)*(5/6)*(5/6)   =25/216   similarly probability that 2nd dice shows up a six: (5/6)*(1/6)*(5/6) = 25/216   And, probability that 3rd dice shows up a six: (5/6)*(5/6)*(1/6) = 25/216   So probability that only one dice shows up a six: (25/216)+(25/216) +(25/216) = 75/216   Case 2:When two dice show up a six   Total number of ways of selecting a pair of dice that show up a six from a set of 3 dice are: 3C2=3   Find the probability of getting six on a pair of dice and multiply it by total number of such possible pairs   Probability of getting a six on a pair of dice = (1/6)*(1/6)*(5/6) = 5/216   So, total probability = 3*(5/216) = 15/216   Case 3: When all dice show up a six   In this case total probability is just (1/6)*(1/6)*(1/6) = 1/216   So total probability of getting at least one six = (75/216) + (15/216) + (1/216) = 91/216   Answer : 91/216
<result>B</result>
What is the probability of getting at least one six in a single throw of three unbiased dice ?