Silly question regarding probability?

Question

A probability exercise which I believe that it is written in slightly dodgy manner as I have trouble finding a solution for it. The way I go about the solution is to have the problem split into three expected values - {(5,3)-default, (6,2), (4,4)}, but I have no idea how to finish it. Could it be that I am going totaly wrong about it?

Here is the exrcise I have:

Answer 1

As discussed in the comments, it is not clear what it means to select a cake at random. There are two competing ideas:

A. ("Random Cake") Choose a cake at random (so probability $\frac 58$ of chocolate and probability $\frac 38$ of banana).

B. ("Random Box") Choose one of the two boxes at random (so probability $\frac 12$ of either chocolate or banana).

Not hard to do both cases simultaneously

Let's go over what happened with the annoying first customer.

Case I. The cake was a chocolate cake. (prob = $\frac 58$ or $\frac 12$).

Ia. It is returned to the chocolate box (prob = $\frac 12$)

Ib. It is placed in the banana box (prob= $\frac 12$)

Case II. The cake was a banana cake. (prob = $\frac 38$ or $\frac 12$).

IIa. It is returned to the banana box (prob = $\frac 12$)

IIb. It is placed in the chocolate box (prob= $\frac 12$)

The reasonable customer now arrives. the only scenario in which they might get a banana cake is IIb (which we are in with probability $\frac 38\times \frac 12$ or $\frac 12\times \frac 12$ ). If we are in that case, the chocolate box contains the five real chocolate cakes and the one imposter, so there is just a $\frac 16$ probability of a mistake. Putting it all together we get $$\frac 38\times \frac 12\times \frac 16=\frac 3{96}\;\;or\;\;\frac 12\times \frac 12\times \frac 16=\frac 1{24}$$.

Answer 2

The experiment has three stages:

1. Pick a cake at random
2. Return the cake to a randomly chosen box
3. Pick a cake at random from the chocolate box

It's unclear whether in stage (1) the box is chosen at random, or the cake is chosen at random, but you can proceed under either assumption.

There's only one combination of events where the second customer gets a banana cake: In stage (1) a banana cake is selected (prob either $\frac12$ or $\frac38$); and in stage (2) the banana cake gets returned to the chocolate box (prob $\frac12$ assuming independence from the stage 1 event); and in stage (3) that selfsame banana cake is selected for the second customer (prob $\frac16$ conditional on the previous stage 1 and stage 2 events). Now multiply the probabilities at the individual stages to get the requested probability.