Final answer:
The question is about the probability of drawing at most two red balls from three urns without replacement. Calculating this probability requires considering all outcomes of drawing up to two red balls and working out their respective probabilities. The problem involves combinatorics and changes in probability after each draw without replacement.
Step-by-step explanation:
The student is asking about the probability of drawing at most two red balls from a collection of three urns without replacement. This calculation involves the concepts of probability and combinatorics. In this scenario, we need to consider all the possible ways we can draw up to two red balls from the urns and then calculate the total probability of these events.
The problem resembles the setup of a tree diagram, where each branch represents a possible outcome of a draw. Considering the first urn with 5 blue balls (B) and 5 red balls (R), the second and third urns each with 6 blue balls and 4 red balls, we want to calculate all possible outcomes of drawing at most 2 red balls from these urns.
Without replacement means once you draw a ball, you do not put it back. Therefore, the probability changes on the second draw, because there's one less ball in total. We should calculate these probabilities step-by-step to find out the probability of having at most 2 red balls drawn.
An example to illustrate this is if you draw a red ball on the first draw, then the probability of drawing another red ball decreases because there's one less red ball available. Conversely, if you draw a blue ball, the probability of then drawing a red ball remains the same. We should add up all possible combinations of drawing red balls (RR, RB, BR) and their respective probabilities to find the answer.