Question

Medical case histories indicate that different illnesses may produce identical symptoms. Suppose a particular set of symptoms, which we will denote as event \(H,\) occurs only when any one of three illnesses \(-A, B\), or \(C\) - occurs. (For the sake of simplicity, we will assume that illnesses \(A\), \(B\), and \(C\) are mutually exclusive.) Studies show these probabilities of getting the three illnesses: $$\begin{array}{l}P(A)=.01 \\\P(B)=.005 \\\P(C)=.02\end{array}$$ The probabilities of developing the symptoms \(H\), given a specific illness, are $$\begin{array}{l}P(H \mid A)=.90 \\\P(H \mid B)=.95 \\\P(H \mid C)=.75\end{array}$$ Assuming that an ill person shows the symptoms \(H\), what is the probability that the person has illness \(A\) ?

Short answer

Answer: The probability that a person has illness A, given that they exhibit the symptoms H, is approximately 31.33%.

Step-by-step solution

Understand Bayes' theorem

Bayes' theorem is a fundamental principle in probability theory that provides a method for updating probabilities based on new information. The theorem is as follows: $$P(A_i \mid H) = \frac{P(H \mid A_i) P(A_i)}{P(H)}$$ In this exercise, we want to find \(P(A \mid H)\), which is the probability of having illness A given that a person exhibits the symptoms H.

Calculate the probability of H

The denominator, \(P(H)\), is the probability that a person exhibits the symptoms H. We can calculate this using the Law of Total Probability: $$P(H) = P(H \mid A)P(A) + P(H \mid B)P(B) + P(H \mid C)P(C)$$ The probabilities of developing the symptoms H given a specific illness (A, B, or C) are provided in the problem statement: $$P(H) = (0.90)(0.01) + (0.95)(0.005) + (0.75)(0.02)$$ Now, calculating the probability of H: $$P(H) = 0.009 + 0.00475 + 0.015 = 0.02875$$

Apply Bayes' theorem to find \(P(A \mid H)\)

Now that we have the probability of H, we can apply Bayes' theorem to find the desired probability \(P(A \mid H)\): $$P(A \mid H) = \frac{P(H \mid A) P(A)}{P(H)}$$ Plugging in the given values and the calculated probability of H: $$P(A \mid H) = \frac{(0.90)(0.01)}{0.02875}$$ Calculating the probability: $$P(A \mid H) \approx 0.3133$$

Interpret the result

The probability that a person has illness A, given that they exhibit the symptoms H, is approximately 31.33%.

Key concepts

Law of Total Probability

The Law of Total Probability is a key concept that helps us find the probability of an event by considering all possible ways that this event can occur. This law is particularly useful when we have several mutually exclusive events that cover all possibilities. In the context of the exercise, we are given three mutually exclusive illnesses, labeled as A, B, and C. Each illness can cause the symptoms denoted by H. The law tells us:

• Calculate the total probability of the symptom, P(H), by adding up the probabilities of the symptom given each illness, weighted by the probability of each illness itself.
• This is expressed in the formula: \[ P(H) = P(H \mid A)P(A) + P(H \mid B)P(B) + P(H \mid C)P(C) \]
• Each term in the formula represents one possible way H can occur, considering only the contribution from each illness.

Applying this law allows us to find the probability of any general symptoms by considering all underlying causes.

conditional probability

Conditional probability is a measure of the probability of an event occurring, given that another event has already occurred. In the exercise, we analyze the probability that a person has illness A given that they have symptom H. Represented as \(P(A \mid H)\), this tells us how likely one illness is, knowing that the symptom is present.

• The condition we know beforehand is that the person shows symptoms H.
• This information helps us "update" our belief about the likelihood of illness A using Bayes' theorem.
• Instead of considering the overall probability of having illness A, we refine it in light of the knowledge that H is true.

Using conditional probability allows us to make smarter predictions and decisions based on available information.

mutually exclusive events

Mutually exclusive events are scenarios where two or more events cannot occur simultaneously. In probability terms, if one event happens, the others cannot happen at the same time. In our example, the illnesses A, B, and C are mutually exclusive.

• This means if one has illness A, they cannot simultaneously have illness B or C.
• The concept simplifies calculations using the Law of Total Probability since each possible outcome (one illness occurring) does not overlap with others.
• Mutual exclusivity is crucial for calculating probabilities accurately when considering how an event like displaying symptoms (H) can occur.

Understanding mutually exclusive events helps simplify complex probability problems by clearly separating different possibilities.