Bayes' Theorem: Videos & Practice Problems

The concept of conditional probability is essential in understanding how the occurrence of one event can influence the likelihood of another event. Specifically, when we want to determine the probability of event B occurring given that event A has already occurred, we utilize the conditional probability formula:

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

However, in many scenarios, directly calculating the probabilities of A and B can be challenging. This is where Bayes' theorem becomes a valuable tool. Bayes' theorem can be expressed as:

$$P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A|B) \cdot P(B) + P(A|B') \cdot P(B')}$$

In this context, event A is the occurrence of a specific outcome (e.g., drawing a red marble), while event B represents the condition we are interested in (e.g., the marble coming from the left bag). The complement of B, denoted as B', indicates the alternative scenario (the marble coming from the right bag).

To illustrate this, consider a game where marbles are drawn from two bags. The left bag contains 2 red and 4 blue marbles, while the right bag has 1 red and 5 blue marbles. Observing that 3 out of every 4 marbles drawn come from the left bag, we can establish the following probabilities:

• $$P(B) = \frac{3}{4}$$ (probability of drawing from the left bag)
• $$P(B') = \frac{1}{4}$$ (probability of drawing from the right bag)
• $$P(A|B) = \frac{2}{6}$$ (probability of drawing a red marble from the left bag)
• $$P(A|B') = \frac{1}{6}$$ (probability of drawing a red marble from the right bag)

Using Bayes' theorem, we can calculate the probability that the marble came from the left bag given that it is red:

$$P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A|B) \cdot P(B) + P(A|B') \cdot P(B')}$$

Substituting the known values:

$$P(B|A) = \frac{\left(\frac{2}{6}\right) \cdot \left(\frac{3}{4}\right)}{\left(\frac{2}{6} \cdot \frac{3}{4}\right) + \left(\frac{1}{6} \cdot \frac{1}{4}\right)}$$

Calculating the numerator gives:

$$\frac{6}{24}$$

For the denominator, we compute:

$$\frac{6}{24} + \frac{1}{24} = \frac{7}{24}$$

Thus, the final probability simplifies to:

$$P(B|A) = \frac{6/24}{7/24} = \frac{6}{7}$$

This result indicates that the probability that the marble came from the left bag, given that it is red, is $\frac{6}{7}$. Understanding and applying Bayes' theorem in this manner allows for effective problem-solving in scenarios involving conditional probabilities.

Bayes' Theorem is a powerful tool for calculating conditional probabilities, particularly in scenarios involving medical testing. In this example, we explore how to determine the probability of having the flu given a positive test result.

At the peak of flu season, it is known that 10% of the population in a town contracts the flu. This probability can be denoted as:

$$P(B) = 0.1$$

where event B represents being sick with the flu. Consequently, the probability of not being sick, or the complement of event B, is:

$$P(B') = 1 - P(B) = 0.9$$

Next, we consider the test's accuracy. The probability of testing positive given that a person is sick (event A given event B) is:

$$P(A|B) = 0.95$$

Conversely, the probability of testing positive given that a person is not sick is:

$$P(A|B') = 0.01$$

To find the probability of being sick given a positive test result, we apply Bayes' Theorem, which is expressed as:

$$P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A|B) \cdot P(B) + P(A|B') \cdot P(B')}$$

Substituting the known values into the formula, we calculate the numerator:

$$P(A|B) \cdot P(B) = 0.95 \cdot 0.1 = 0.095$$

For the denominator, we compute:

$$P(A|B) \cdot P(B) + P(A|B') \cdot P(B') = 0.095 + (0.01 \cdot 0.9) = 0.095 + 0.009 = 0.104$$

Now, we can find the conditional probability:

$$P(B|A) = \frac{0.095}{0.104} \approx 0.913$$

This result indicates that the probability of having the flu given a positive test result is approximately 91.3%. Thus, if someone tests positive, there is a high likelihood that they are indeed sick with the flu, demonstrating the effectiveness of the test in this context.