An algebra problem by Sunil Gupta

We know that P(no 6s) + P(at least one 6) = 1, since the sum of all possible outcomes (in terms of getting sixes) must be 1.

Thus, P(at least one 6) = 1 - P(no 6s)

The probability of not rolling a 6 on one die is 5/6. And since the outcome of each die is independent from the others, the probability of getting no 6s is (5/6)^3 = 125/216

Therefore, A = P(at least one 6) = 1 - (125/216) = 91/216

216A = 91

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