Let’s focus on Question 14 from the provided image. Here’s the problem and how to solve it:

Problem (Q14):

• There are 3 identical filing cabinets, and each contains two folders, each with one bill.

  • First cabinet: 1 folder containing a telephone bill only.
  • Second cabinet: 1 folder with an electricity bill.
  • Third cabinet: 1 folder with a telephone bill and another with an electricity bill.

• If a cabinet is selected randomly and one folder from that cabinet is opened, revealing an electricity bill, what is the probability that a telephone bill is in the other folder?

Solution:

This is a Bayesian probability problem where we need to compute the conditional probability.

We are given that:

• An electricity bill is found after opening one folder.

We need to determine:

• P(telephone bill in the other folder | electricity bill was found).

Step-by-step solution:

Step 1: Analyze the sample space

There are 3 cabinets with the following distributions:

1. Cabinet 1:

   • Folder 1: Telephone bill
   • Folder 2: (Empty)

2. Cabinet 2:

   • Folder 1: Electricity bill
   • Folder 2: (Empty)

3. Cabinet 3:

   • Folder 1: Telephone bill
   • Folder 2: Electricity bill

Step 2: Which cases involve seeing an electricity bill?

• Cabinet 2:
  There is only one folder here with an electricity bill, so this cabinet has no chance of containing a telephone bill if we draw it.

• Cabinet 3:
  If we draw the electricity bill from one folder, the other folder has the telephone bill.

So, the favorable case for finding a telephone bill in the other folder is only when we are dealing with Cabinet 3.

Step 3: Use Bayes' Theorem

We need to calculate the conditional probability:
P(Cabinet 3 | Electricity bill found)

By Bayes' theorem:

$P(Cabinet 3 \mid \text{Electricity bill found}) = \frac{P(\text{Electricity bill found} \mid \text{Cabinet 3}) \cdot P(\text{Cabinet 3})}{P(\text{Electricity bill found})}$

• P(Electricity bill found | Cabinet 3): $\frac{1}{2}$ (Since there are 2 bills, and we could have picked either)
• P(Cabinet 3): $\frac{1}{3}$ (Each cabinet is equally likely to be chosen)
• P(Electricity bill found):
  $P(\text{Electricity bill found}) = \left( \frac{0}{2} \cdot \frac{1}{3} \right) + \left( 1 \cdot \frac{1}{3} \right) + \left( \frac{1}{2} \cdot \frac{1}{3} \right) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$

Step 4: Calculate the probability

$P(Cabinet 3 \mid \text{Electricity bill found}) = \frac{\frac{1}{2} \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{1}{3}$

Final Answer:

The probability that a telephone bill is in the other folder, given that an electricity bill was found, is:

$\boxed{\frac{1}{3}}$

Would you like a further breakdown or explanation of any step?

5 Related Questions:

1. How does Bayes’ theorem work in conditional probability problems?
2. How would the answer change if there were more cabinets with different bill combinations?
3. Can you explain the concept of prior and posterior probabilities in Bayes' theorem?
4. What happens if a new cabinet is added with both bills in both folders? How would it change the probabilities?
5. How do real-world scenarios use similar reasoning, like in spam email filtering?

Tip:

When working with conditional probability, always clearly identify the prior probabilities and the favorable cases to avoid confusion.