Hello, Our question is: A pair of six-sided fair dice is thrown. Find the probability that the sum is 10 or greater if it is given that a 5 appears on at least one of the dice. I see two methods to approach this; yielding different answers. We ignore the given dice and consider only the other dice. We need the other dice to be either a 5 or a 6, out of the possible choices from 1 to 6. This would yield 1/3 as the answer. If the dice are unique, then we obtain the following options when a five is included: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5) and (6,5). This creates 11 possible outcomes even though the sum may not be unique. In this scenario, given a 5 is present, there is a 3/11 probability of obtaining a sum greater than or equal to 10. I see this as if you are rolling 2 dice. The probability of rolling a sum of 3 is 2/36 where they arise from (1,2) and (2,1). Both arguments seem logical. Please tell us how you see it, and indicate the flaw in the logic of the other one, or may be both. Thank you for your help. Sincerely, Wallace Yang (student) and Mr. McMaster Hi Wallace, They are both logical and both yield correct answers, but correct answers to two different questions. One question is Suppose that I throw a red die and a blue die, and I tell you that 5 appears on the red die, what is the probability of getting a sum of at least 10? For this problem your first solution is correct. In this case you can ignore the given (blue) die and consider only the other (red) die to get a probability of 1/3. For the problem you have stated A pair of six-sided fair dice is thrown. Find the probability that the sum is 10 or greater if it is given that a 5 appears on at least one of the dice. your second argument is correct. In this case you need to consider all 36 possible outcomes, identify which have at least one 5, and then compute the proportion of these that have a sum of 10 or more. This is exactly what you did. These two problems are related to the following apparent paradox: Suppose that I throw the two dice where I can see them but you can't, and I tell you (honestly) that at least one 5 appears. Then you have a probability of 3/11 of having a sum of at least 10. If I had told you instead that a 5 appears on the red die, this probability climbs to 1/3, and similarly, if I were to tell you instead that a 5 appears on the blue die, this probability would also be 1/3. Now the problem is this: If it is true that at least one five appears, then it must be true that five appears on the red die or five appears on the blue die. So, instead of telling you "There is at least one five" (leading you to conclude that the probability is 3/11), I could always tell you one of "there is a 5 on the red dice" or "there is a 5 on the blue dice", (leading you to conclude you are in one of two situations, each of which yields to a probability of 1/3). Cheers, Penny and Claude Go to Math Central