A man forgets his umbrella to a store with a probability of $\frac{1}{4}$. What is the probability for the umbrella being left in the store $i$ ? The following solved problem is in our course material but I'm having a trouble understanding a part of it:
An absent-minded man forgets his umbrella to a store with a probability of $\frac{1}{4}$. One day, after visiting four stores, he realizes he had forgotten his umbrella. What is the probability for the umbrella being left in the store $i$ ? 
$ A_i $= " the umbrella has been left in the store $i$ ", $i=1,2,3,4$
$ B $= " the umbrella has been forgotten to the store "
Now
$P(A_1)=\frac{1}{4}=\frac{64}{256}, \quad P(A_2)=\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{16}=\frac{48}{256}, $
$P(A_3)=(\frac{3}{4})^2\cdot\frac{1}{4}=\frac{9}{64}=\frac{36}{256}, \quad P(A_4)=(\frac{3}{4})^3\cdot\frac{1}{4}=\frac{27}{256}.$
Since $A_i \subset B$ for every $i$, $P(B|A_i)=1$ for every $i$. Now
$\sum^4_{k=1}P(A_k)=\frac{64}{256}+\frac{48}{256}+\frac{36}{256}+\frac{27}{256}=\frac{175}{256}$
and with Bayes' theorem we get 
$P(A_i|B)=\frac{P(A_i)P(B|A_i)}{\sum^4_{k=1}P(A_k)P(B|A_k)}=\frac{P(A_i)}{\sum^4_{k=1}P(A_k)}=\frac{256}{175}P(A_i)$ 
for every $i=1,2,3,4$.
The probability of the umbrella being left in the store 1 is $P(A_1)=\frac{1}{4}=\frac{64}{256}$ but where do we get the $\frac{3}{4}$ in $P(A_2)=\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{16}=\frac{48}{256}$? And why do we multiply the $\frac{3}{4}$ with the probability of the umbrella being forgotten and also put it to the power in $P(A_3)$ and $P(A_4)$?
I know that there is a similar question here but I'm confused by the solutions.
 A: The probability of not leaving it in the first store multiplied by the probability of not leaving it in the second store is (3/4)•(3/4), which then multiplied by the probability of leaving it in the 3rd store (1/4) gives the 3rd term.
A: The event $A_2$ happening means that the person got to the second store without forgetting his umbrella at the first.
Let's mark with $C_i$ the event "the person would have forgotten his umbrella at store $i$, provided he had his umbrella with him", than the statement of the question means $C_1=C_2=C_3=C_4$.
So now we have $A_2 = A_1^\complement \cap C_2$. Also, by the definition of $C_i$ it's clear that $A_1$ and $C_2$ are independent, and thus $P(A_2)=P(A_1^\complement \cap C_2)=P(A_1^\complement)P(C_2)=(1-\frac{1}{4})\cdot\frac{1}{4}$.
By similiar arguments we can get $P(A_3)=P((A_1 \cup A_2)^\complement\cap C_3)$ and $P(A_4)$.
A: E.g. $$P(A_3)=P(A_1^{\complement}\cap A_2^{\complement}\cap A_3)=P(A_1^{\complement})P(A_2^{\complement}\mid A_1^{\complement})P(A_3\mid A_1^{\complement}\cap A_2^{\complement})=\frac34\frac34\frac14$$
