Alice and Bob both need to buy a bicycle. The bike store has a stock of four green, three yellow, and two red bikes. Alice randomly picks one of the bikes and buys it. Immediately after, Bob does the same. ... Let A be the event that Alice bought a green bike, and B be the event that Bob bought a green bike. a. (5 points) What is P(A)? What is P(A|B)? Solution: We have P(A) = 4/9 (4 green bikes out of 9), and P(A|B) = 3/8 (since we know that Bob has a green bike, Alice can have one of 3 green bikes out of the remaining 8).

Since Bob buys his bike after Alice how can it affect the number of green bikes Alice has to choose and the total number of bikes? Alice goes first so how can Bob's subsequent choice affect the probability of what she did in the past? Or is this just a typo, i.e. it means to ask P(B|A) which I agree would be 3/8.

The answer is correct: If we know Bob bought a green bike, then there are only three other green bikes that Alice could have bought out of the eight bikes Bob didn't buy. Thus the probability that Alice bought a green bike given we know Bob bought a green bike is $$3/8$$.

It is not that Bob's subsequent choice is affecting what Alice did in the past. It is that knowing Bob's choice gives us information about what Alice did in the past. The fact that Bob also happened to buy a green bike, when choosing randomly, makes it less likely that Alice had (because, symmetrically, when Alice buys a green bike, it is less likely that Bob would buy a green bike randomly).

Note that $$P(B|A)=P(A|B)$$ here. The order Alice and Bob buy bikes does not matter.

A simpler example to illustrate the point: Suppose that the bike store had one green bike and one red bike. Then:

• if you know Bob bought a green bike, it tells you Anne could not have bought a green bike: $$P(A|B)=0$$;
• if you know Alice bought a green bike, it tells you Bob did not buy a green bike: $$P(B|A)=0$$.
• The simpler example clinches it. Thanks! A good illustration of the power of choosing the simplest case possible to get a clear understanding. Oct 7, 2018 at 0:13
• Similarly in the simple example, P(A) = P(B). This isn't intuitively obvious to me, it seems that P(B) would depend on what A did but if I use what this text calls the probability law I do indeed get P(B) = P(A). Oct 7, 2018 at 1:27

The other parts of the solution consistently re-use the probability $$P(A\mid B),$$ and the explanation clearly says we "know" that Bob's bike is green, which does not suit a calculation of $$P(B\mid A)$$ at all. I think the writer of this solution meant exactly what they wrote.

So it seems there is no way to "correct" the solution, since it is not wrong. It might, however, help some people if the answer were more thoroughly explained.

You may be imagining a scenario in which we watch Alice and Bob as they leave the store and take note of which bike each bought. If we see Alice leave with a green bike, we can say at that point that Bob has a $$3/8$$ chance of buying another green bike next. That's analogous to the conditional probability $$P(B\mid A).$$

But suppose we were not watching at the time, but just heard about the store's inventory and the fact that Alice and then Bob randomly bought bikes, exactly as described in the question. Then we go visit Bob and see that he has a green bike. We plan to visit Alice next. What's the probability that her bike also will be green?

Whatever bike Bob bought, we know Alice must have bought one of the others. At this point, as far as we know (which is what conditional probability is analogous to in most cases), the bike Alice bought is equally likely to have been any one of those eight bikes as any other; all we can say that distinguishes the chances of any of the original nine bikes from another is that there is a zero chance that Alice bought the same bike Bob did. And now that we know Bob's bike is green, we know that three of the other eight bikes were green.

$$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$. Let $$Y$$ be the event that Alice buys the yellow bike, $$R$$ be the event Alice buys the red bike. Then $$P(B) = P(B|A)P(A) + P(B|R)P(R) + P(B|Y)P(Y) = \frac{3}{8}\frac{4}{9} + \frac{4}{8}\frac{2}{9} + \frac{4}{8}\frac{3}{9} = \frac{4}{9}$$. So $$P(A|B) = (\frac{3}{8}\frac{4}{9}) / \frac{4}{9} = \frac{3}{8}$$