8 Drawer Desk Conditional Probability Problem: Use Bayes' Theorem using the given statement

Question

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 a probability of 1/2 that this person didn't 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?

This answer makes use of a slightly different Event A while this answer takes advantage of $P(A \cap B)$. I want to use the alternative form $P(A) * P(B|A)$ instead. This answer touches on the other part of the disjoint equation in the denominator that I am trying to figure out. Here is what I have so far:

Let $A$ denote the event in which there is a letter in the 8th drawer.

Let $B$ denote the event in which the first 7 drawers are all empty.

Then $P(A|B) = \frac{P(A) * P(B|A)}{P(B)}$

$P(A) = \frac{1}{2} * \frac{1}{8}$ since there's a 1/2 chance that there's a letter in the drawers and only 1 of the 8 possible scenarios is there a letter in the 8th drawer.

$P(B) = P(B|A)*P(A) + P(B|A')*P(A')$ - This means the probability of having the first seven drawers empty is equal to:

• The probability that the first seven drawers are empty given that the letter is in the 8th drawer $P(B|A)$ times the probability that the letter is in the 8th drawer $P(A)$ PLUS
• The probability that the first seven drawers are empty given that the letter is NOT in the 8th drawer $P(B|A')$ times the probability that the letter is NOT in the 8th drawer $P(A')$

$P(B|A)$ = 1 since if it's in the 8th drawer, the other seven have to be empty since there's only one letter.

$P(A)$ was already calculated as $\frac{1}{2} * \frac{1}{8}$

$P(B|A')$ = unsure

$P(A')$ = $(\frac{1}{2} * \frac{7}{8}) + \frac{1}{2} * 1$ (I think this is true - 1/2 of the time we will be in the letter state and there will be 7 out of 8 states in which the letter is not in the 8th spot). The other 1/2 of the time when there is no letter, it will be 8/8 times that the letter is not in the 8th spot.

I am focused on filling in the gaps on this specific method. There are other elegant solutions that I understand, but this is how I've been solving other Baye's Theorem problems, so I want to solidify my understanding by pointing out where I went wrong or how I can count $P(B|A')$

Judging by the correct answers, if everything I wrote is correct, then $P(B|A')$ has to be 8/15, but what does 8/15 represent? Also, I wouldn't know what this is beforehand, so I'd like to know how to calculate this.

Answer 1

The letter is exclusively either:

• $A$: In the eighth drawer with probability $1/16$.
• $B'$: In one among the other drawers with probability $7/16$.
• $A'\cap B$: Not in any drawer with probability $1/2$ .

$$\begin{align}\mathsf P(A~\cap B~)&= 1/16 &\text{because }&A~\cap B~ = A\\\mathsf P(A'\cap B') &= 7/16 &&A'\cap B' = B'\\\mathsf P(A'\cap B~) &=1/2\\\mathsf P(A~\cap B')&=0\end{align}$$

Use this to calculate your probabilities. Such as: $$\begin{align}\mathsf P(A\mid B) &=\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\\&=\dfrac{1/16}{9/16}\end{align}$$

Answer 2

You don't explicitly say so, but I'm assuming that if there is a letter in one of the drawers, it's a priori equally likely to be in any of them.

Assume we perform $16$ trials. In $8$ of those trials, there's no letter. In the other $8$ trials, there's a letter in a different one of the $8$ drawers.

Given the information we now have, there are $9$ remaining cases and each of them is equally likely. So the probability that there's a letter in the remaining drawer is $\frac 19$.