I have some confusion regarding one of the examples from probability and statistics course. The translation of the examples goes something as:
The chef knows how to prepare 4 different meals. He starts his week by preparing any meal with equal probability.(he prepares one meal per day just to be clear). The next day he prepares a meal from previous day with probability of 0.4 or he prepares one of the remaining meals which all are equally likely. What is the probability that the choice of meals will be b,b,a,c,c?
So I have the solution which goes as: $$A1 - \text{he prepared meal b}$$ $$A2 - \text{he prepared meal b}$$ $$A3 - \text{he prepared meal a}$$ $$A4 - \text{he prepared meal c}$$ $$A5 - \text{he prepared meal c}$$ then, $$P(A1) = 0.25, $$$$P(A2|A1) = 0.4$$ $$P(A3|A1\cap A2) = 0.2$$$$P(A4|A1\cap A2\cap A3)=0.2$$ $$P(A5|A1\cap A2\cap A3\cap A4)=0.4$$ The problem is, I have no idea how to derive the result with multiple events. I know how the formula for multiplication rule goes and I know how we got $P(A2|A1)$, but I can't solve other results. I tried some things like to apply formula but I can't get exact numbers. Can anyone explain how to get those?