Final answer:
To find the probability that the larger number is a 5 given the smaller is a 3 when rolling two six-sided dice, we consider the reduced sample space {(3,4), (3,5), (3,6)} resulting in a probability of 1/3.
Step-by-step explanation:
When we roll two fair six-sided dice, determining the probability of a specific outcome requires considering all possible combinations. In this scenario, we're looking for the probability that the larger of the two numbers rolled is a 5, knowing that the smaller number is already a 3. This question reflects an understanding of conditional probability.
We can illustrate the sample space for rolling two dice. Since the dice are fair, each die has an equal chance of landing on any face, represented by the numbers {1, 2, 3, 4, 5, 6}. But because we are given that the smaller number is 3, the only possible combinations that we can have which are greater than 3 are {(3,4), (3,5), (3,6)}. This gives us 3 favorable outcomes.
Since we're only considering rolls where the smaller number is a 3, the sample space is reduced to those outcomes where at least one die shows a 3. If we consider the first die to be the one with the smaller number, there are three outcomes for the second die it may land on – 4, 5, or 6. Hence, the probability we're looking for is the number of outcomes where the second die is 5 given it is one of the three possible numbers that are larger than 3. Therefore, the event that the dice show 5 given that one die is locked at 3 would be 1/3.
Because this was a conditional probability problem, where we had information about one of the outcomes (the smaller number being a 3), we could narrow down our sample space to only those outcomes that aligned with the given conditions hence finding the sought probability.