A desk has eight drawers. There is a probability of 1/2 that someone placed a letter in one of the desk's eight drawers and probability of 1/2 that this person did not place a letter in any of the desk's eight drawers. You open the first 7 drawers and find that they are all empty. What is the probability that the 8th drawer has a letter in it?
I am having difficulty determining $P(B)$, the probability of 7 Empty Drawers. The first half of the solution to this sub-problem is clear:
$P(B) = P(B|NotInDesk)(P(NotInDesk)) + P(B|InDesk)P(InDesk)$
$ = 1(\frac{1}{2}) + P(B|InDesk)(P(InDesk))$
But the second part eludes me. The solution states that the second probability should be $\frac{1}{2} (\frac{1}{8})$, but when I attempted to solve it I thought it made more sense that it ought to be $\frac{1}{2} (\frac{7}{8})^7$.
The reasoning is that we know that the letter isn't in the first 7 drawers, and that happens 7 times as independent events, so the probability of it not being in the drawer given we know its in the desk and in the 8th drawer should be 7/8 raised to the power of seven.
The only other logical explanation I can think of is that the probability of getting 7 empty drawers given the letter is in the desk is equivalent to probability of getting it in the 8th drawer given that it is in the desk, but would that make the probability $\frac{1}{8} (\frac{1}{2})$ or would it make it 100%, based on the prior knowledge that the desk has it AND that the first seven drawers have already turned up empty?
First, note that $P(X|Y)P(Y) = P(X\cap Y)$.
So, $P(B|InDesk)(P(InDesk)) = P(B \cap InDesk)$. There is a $\color{blue}{0.5}$ probability that a letter is placed in one of the drawers, and you're correct there.
As you have noticed, your answer is wrong because of the $(7/8)^7$. "The reasoning is that we know that the letter isn't in the first 7 drawers, and that happens 7 times as independent events" is wrong, since they are not independent. We know the letter is not in the first drawer, and this has a $7/8$ probability. The probability that it is not in the second box is not $7/8$, since you already know it's not in the first box. It should be $6/7$, since $7$ boxes remain. Similarly, the probability that it's not in the third is $5/6$ and so on till the event box which has probability $1/2$. All terms cancel except $1$ in the numerator and $8$ in denominator, so you get $\color{blue}{1/8}$. Now, multiply the blue things.
To explain the fault in your reasoning more intuitively, suppose there were $3$ boxes instead of $8$. According to you, the probability should be $4/9$. Clearly, the correct probability is $1/3$.
The easiest way would be to see that each box is equiprobable, hence you directly get $1/8$, and multiply it by $1/2$ for the probability of the man deciding to keep the letter in some drawer.
You can also view my first argument like this. Suppose we know that the letter is in one of the boxes (we'll multiply by $\color{blue}{0.5}$ to account for it). So, "it is in the first box" and "not in the first box" are complementary events. So, $$\underbrace{\frac 78}_{\text{not in 1}}+\underbrace{\frac 18}_{\text{in 1}} = 1$$ Note that the "not in 1" can be written as the union of two disjoint events "not in 2 and 1" and "in 2 but not in 1". $$\underbrace{\frac 78}_{\text{not in 1}}\left(\underbrace{\frac 67}_{\text{not in 2}}+\underbrace{\frac17}_{\text{in 2}}\right)+\underbrace{\frac 18}_{\text{in 1}} = 1$$ Similarly, doing this for "not in 2" and so on, we get $$\underbrace{\frac 78}_{\text{not in 1}}\left(\underbrace{\frac 67}_{\text{not in 2}}\left(\underbrace{\frac 56}_{\text{not in 3}}\left(\underbrace{\frac 45}_{\text{not in 4}}\left(\underbrace{\frac 34}_{\text{not in 5}}\left(\underbrace{\frac 23}_{\text{not in 6}}\left(\underbrace{\frac 12}_{\text{not in 7}}+\underbrace{\frac12}_{\text{in 7}}\right)+\underbrace{\frac13}_{\text{in 6}}\right)+\underbrace{\frac14}_{\text{in 5}}\right)+\underbrace{\frac15}_{\text{in 4}}\right)+\underbrace{\frac16}_{\text{in 3}}\right)+\underbrace{\frac17}_{\text{in 2}}\right)+\underbrace{\frac 18}_{\text{in 1}} = 1$$This representation infact gives you the probability of all possible events. To find the probability of "in 8", just multiply out the probabilities of "not in 1" till "not in 7". (Or just observe that in 1 and in 8 are equiprobable, so their probabilities are equal.)