To solve this problem, we can use Bayes' theorem, which allows us to update the probability estimate for an event based on new evidence.

Problem Recap

• Alice knows the term paper is in drawer $j$ with probability $P$.
• If she searches drawer $i$, the probability that she finds the paper (if it's there) is $d$.
• The search in drawer $i$ was unsuccessful.

We are required to find the conditional probability that the term paper is in drawer $j$, given that the search in drawer $i$ was unsuccessful.

Solution

Notation

• $P(A)$: Probability that the paper is in drawer $j$.
• $P(B|A)$: Probability that Alice does not find the paper in drawer $i$ given that the paper is in drawer $j$.
• $P(B|\neg A)$: Probability that Alice does not find the paper in drawer $i$ given that the paper is not in drawer $j$.
• $P(\neg A) = 1 - P(A)$: Probability that the paper is not in drawer $j$.

We are interested in finding $P(A|B)$, the probability that the paper is in drawer $j$ given that the search in drawer $i$ was unsuccessful.

Step 1: Determine the probability of an unsuccessful search

1. If the paper is in drawer $j$, and since Alice is searching drawer $i$ (not necessarily $j$), the probability that she does not find it is $1$, because she is searching in the wrong drawer. Therefore: $P(B|A) = 1$

2. If the paper is not in drawer $j$, the paper is either in drawer $i$ or some other drawer:

   • If the paper is in drawer $i$, then the probability she doesn't find it is $1 - d$.
   • If the paper is in another drawer (not $j$ or $i$), the probability that she doesn't find it is $1$ (because she is looking in the wrong place).

   Therefore: $P(B|\neg A) = \frac{1}{n-1} \cdot (1 - d) + \frac{n-2}{n-1} \cdot 1$ where $n$ is the total number of drawers.

Step 2: Apply Bayes' Theorem

Bayes' theorem states: $P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$ Where $P(B)$ (total probability of not finding the paper in drawer $i$) can be computed as: $P(B) = P(B|A) \cdot P(A) + P(B|\neg A) \cdot P(\neg A)$ Substituting the known quantities: $P(B) = 1 \cdot P + \left[\frac{1}{n-1} \cdot (1-d) + \frac{n-2}{n-1}\right] \cdot (1-P)$ $P(B) = P + \left[\frac{1}{n-1} \cdot (1-d) + \frac{n-2}{n-1}\right] \cdot (1-P)$

Thus, the conditional probability that the paper is in drawer $j$ given the unsuccessful search is: $P(A|B) = \frac{P}{P + \left[\frac{1}{n-1} \cdot (1-d) + \frac{n-2}{n-1}\right] \cdot (1-P)}$

This equation provides the updated probability that Alice's paper is in drawer $j$ given that she searched drawer $i$ and did not find it.

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Do you want details on any of the steps? Here are some related questions:

1. What happens if $P$ is close to 1?
2. How does the value of $d$ affect the probability $P(A|B)$?
3. What is the impact of the total number of drawers $n$ on $P(A|B)$?
4. How does the scenario change if Alice finds the paper in drawer $i$?
5. How can we generalize this for multiple unsuccessful searches?

Tip: Bayes' theorem is a powerful tool for updating probabilities based on new information. Understanding the relationship between prior and conditional probabilities is key to solving many real-world problems involving uncertainty.