To find the probability of a toothache given evidence of a cavity, P(toothache | cavity), we need to use the definition of conditional probability:

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

In this case, A is 'toothache' and B is 'cavity'. So, we need to calculate P(toothache ∩ cavity) and P(cavity) from the provided joint distribution table.

Understanding the Joint Distribution Table

The table shows the joint probabilities for all combinations of the three Boolean variables: Toothache, Cavity, and Catch. Let's represent Toothache as T, Cavity as C, and Catch as Ca.

The table values represent P(T, C, Ca), P(T, C, ¬Ca), P(¬T, C, Ca), P(¬T, C, ¬Ca), and so on, for all 8 combinations.

Calculating P(toothache ∩ cavity)

The probability P(toothache ∩ cavity) is the joint probability that both Toothache and Cavity are true. This corresponds to the first row of the table where 'cavity' is true, summing the probabilities where 'toothache' is also true (the first two columns within the 'cavity' row).

$$P(\text{toothache} \cap \text{cavity}) = P(\text{toothache}, \text{cavity}, \text{catch}) + P(\text{toothache}, \text{cavity}, \neg \text{catch})$$

From the table:

• P(toothache, cavity, catch) = 0.108
• P(toothache, cavity, ¬catch) = 0.012

So,

$$P(\text{toothache} \cap \text{cavity}) = 0.108 + 0.012 = 0.120$$

Calculating P(cavity)

The probability P(cavity) is the marginal probability that Cavity is true. This involves summing all probabilities in the joint distribution where 'cavity' is true, regardless of the state of Toothache and Catch. This corresponds to summing all values in the first row of the table.

$$P(\text{cavity}) = P(\text{toothache}, \text{cavity}, \text{catch}) + P(\text{toothache}, \text{cavity}, \neg \text{catch}) + P(\neg \text{toothache}, \text{cavity}, \text{catch}) + P(\neg \text{toothache}, \text{cavity}, \neg \text{catch})$$

From the table:

• P(toothache, cavity, catch) = 0.108
• P(toothache, cavity, ¬catch) = 0.012
• P(¬toothache, cavity, catch) = 0.072
• P(¬toothache, cavity, ¬catch) = 0.008

So,

$$P(\text{cavity}) = 0.108 + 0.012 + 0.072 + 0.008 = 0.200$$

Calculating P(toothache | cavity)

Now we can apply the conditional probability formula:

$$P(\text{toothache} | \text{cavity}) = \frac{P(\text{toothache} \cap \text{cavity})}{P(\text{cavity})}$$

$$P(\text{toothache} | \text{cavity}) = \frac{0.120}{0.200}$$

$$P(\text{toothache} | \text{cavity}) = 0.600$$

Thus, the probability of a toothache, given evidence of a cavity, is 0.600.

Revision Table: Key Probabilities

Additional Information: Conditional Probability and Joint Distributions

Conditional probability is a fundamental concept in probability theory that measures the probability of an event occurring given that another event has already occurred. It is expressed as P(A | B), the probability of event A given event B.

A joint probability distribution table, like the one provided, lists the probabilities of all possible outcomes of a set of random variables occurring together. For discrete variables, the sum of all entries in the joint distribution table must equal 1.

To find the probability of a single variable or a subset of variables (marginal probability) from a joint distribution, you sum the probabilities over the variables you are not interested in. For example, to find P(cavity), we summed over all possible states of Toothache and Catch when Cavity was true.

Conditional probabilities are often calculated using joint probabilities, as shown in this problem. They are crucial for tasks like inference in probabilistic graphical models, such as Bayesian networks, where we want to update our belief about a variable based on observed evidence for other variables.