Understanding the Joint Probability Distribution

The question asks for a conditional probability based on a given full joint distribution table for three Boolean variables: Toothache, Cavity, and Catch. A full joint distribution table provides the probability for every possible combination of states of the variables in the domain. In this case, with three Boolean variables, there are $2^3 = 8$ possible states, and the table provides the probability for each of these states.

The table is structured as follows:

Let's denote the variables as follows: T for Toothache, C for Cavity, and H for Catch. The table entries represent the joint probabilities of the form $P(T, H, C)$. For example, the entry 0.108 corresponds to $P(\text{Toothache} \land \text{Catch} \land \text{Cavity})$.

We can list all the joint probabilities from the table:

• $P(T, H, C) = 0.108$
• $P(T, \neg H, C) = 0.012$
• $P(\neg T, H, C) = 0.016$
• $P(\neg T, \neg H, C) = 0.064$
• $P(T, H, \neg C) = 0.072$
• $P(T, \neg H, \neg C) = 0.008$
• $P(\neg T, H, \neg C) = 0.144$
• $P(\neg T, \neg H, \neg C) = 0.576$

The sum of all these probabilities is $0.108 + 0.012 + 0.016 + 0.064 + 0.072 + 0.008 + 0.144 + 0.576 = 1.000$, confirming it is a valid joint distribution.

Calculating the Conditional Probability

The question asks for the probability of Cavity given that either Toothache or Catch is true. This can be written as $P(\text{Cavity} | \text{toothache} \lor \text{catch})$.

The formula for conditional probability is $P(A|B) = \frac{P(A \land B)}{P(B)}$.

In this case, event A is Cavity (C) and event B is toothache $\lor$ catch (T $\lor$ H).

So, we need to calculate $P(C | T \lor H) = \frac{P(C \land (T \lor H))}{P(T \lor H)}$.

Step 1: Calculate the probability of the condition $P(\text{toothache} \lor \text{catch})$

The event "toothache $\lor$ catch" (T $\lor$ H) is true if Toothache is true, or Catch is true, or both are true. We can find the probability of this event by summing the probabilities of all states where T is true OR H is true.

States where T is true:

• (T, H, C): 0.108
• (T, $\neg$H, C): 0.012
• (T, H, $\neg$C): 0.072
• (T, $\neg$H, $\neg$C): 0.008

States where H is true (excluding those already listed where T is also true):

• ($\neg$T, H, C): 0.016
• ($\neg$T, H, $\neg$C): 0.144

Summing these probabilities gives $P(T \lor H)$: $P(T \lor H) = P(T, H, C) + P(T, \neg H, C) + P(T, H, \neg C) + P(T, \neg H, \neg C) + P(\neg T, H, C) + P(\neg T, H, \neg C)$ $P(T \lor H) = 0.108 + 0.012 + 0.072 + 0.008 + 0.016 + 0.144 = 0.360$.

Alternatively, we can calculate $P(T \lor H)$ using the complement rule: $P(T \lor H) = 1 - P(\neg (T \lor H)) = 1 - P(\neg T \land \neg H)$.

$P(\neg T \land \neg H)$ corresponds to the states where Toothache is false AND Catch is false. From the table, these are:

• ($\neg$T, $\neg$H, C): 0.064
• ($\neg$T, $\neg$H, $\neg$C): 0.576

$P(\neg T \land \neg H) = 0.064 + 0.576 = 0.640$.

So, $P(T \lor H) = 1 - 0.640 = 0.360$. Both methods yield the same result for the denominator.

Step 2: Calculate the probability of Cavity AND the condition $P(\text{Cavity} \land (\text{toothache} \lor \text{catch}))$

This is the probability of states where Cavity (C) is true AND (Toothache (T) is true OR Catch (H) is true). We look for states where C is true and at least one of T or H is true.

Let's look at the states where Cavity (C) is true (the first row of the table):

• (T, H, C): C is true, T is true, H is true. T $\lor$ H is true. This state is included. Prob = 0.108.
• (T, $\neg$H, C): C is true, T is true, H is false. T $\lor$ H is true. This state is included. Prob = 0.012.
• ($\neg$T, H, C): C is true, T is false, H is true. T $\lor$ H is true. This state is included. Prob = 0.016.
• ($\neg$T, $\neg$H, C): C is true, T is false, H is false. T $\lor$ H is false. This state is NOT included. Prob = 0.064.

Summing the probabilities for the included states:

$P(C \land (T \lor H)) = P(T, H, C) + P(T, \neg H, C) + P(\neg T, H, C)$ $P(C \land (T \lor H)) = 0.108 + 0.012 + 0.016 = 0.136$.

Step 3: Calculate the conditional probability

Now, we divide the numerator by the denominator:

$\qquad P(\text{Cavity} | \text{toothache} \lor \text{catch}) = \frac{P(C \land (T \lor H))}{P(T \lor H)} = \frac{0.136}{0.360}$

Let's simplify the fraction:

$\qquad \frac{0.136}{0.360} = \frac{136}{360}$

Both 136 and 360 are divisible by 8:

$\qquad \frac{136 \div 8}{360 \div 8} = \frac{17}{45}$

To express this as a decimal, we perform the division:

$\qquad \frac{17}{45} \approx 0.3777...$

Comparing with Options

The calculated probability is approximately 0.3778. Let's look at the given options:

• 1. 0.6000
• 2. 0.5384
• 3. 0.8000
• 4. 0.4615

Our calculated value does not match any of the provided options exactly. However, following the standard method of calculating conditional probability from a joint distribution table using the formula $P(A|B) = P(A \land B) / P(B)$, the result obtained is approximately 0.3778.

Revision Table: Probability Calculation

Additional Information: Conditional Probability and Joint Distributions

Conditional probability is a fundamental concept in probability theory, allowing us to update our belief about an event based on the occurrence of another event. The formula $P(A|B) = P(A \land B) / P(B)$ quantifies this updated belief.

A full joint distribution table is the cornerstone for answering any probabilistic query within the domain it covers. From the joint distribution, we can calculate:

• Marginal probabilities: Summing over the unwanted variables. For example, $P(\text{Cavity}) = P(\text{C} \land \text{T} \land \text{H}) + P(\text{C} \land \text{T} \land \neg \text{H}) + P(\text{C} \land \neg \text{T} \land \text{H}) + P(\text{C} \land \neg \text{T} \land \neg \text{H}) = 0.108 + 0.012 + 0.016 + 0.064 = 0.200$.
• Joint probabilities of subsets of variables: Summing over the variables not included in the subset. For example, $P(\text{Toothache} \land \text{Catch}) = P(T, H, C) + P(T, H, \neg C) = 0.108 + 0.072 = 0.180$.
• Conditional probabilities: Using the formula $P(A|B) = P(A \land B) / P(B)$, where $P(A \land B)$ and $P(B)$ are calculated by summing the relevant entries from the joint distribution table.

In this problem, the condition is a disjunction (OR), which means the event occurs if any part of the disjunction is true. Calculating the probability of a disjunction $P(B_1 \lor B_2 \lor ...)$ involves summing the probabilities of all atomic states (entries in the joint table) where at least one of the conditions $B_i$ is met.