Question

Medical cases histories indicate that different illnesses may produce identical symptoms. Suppose a particular set of symptoms, "H" occurs only with probabilities 0.01,0.005 and 0.02, respectively. The probability of developing the symptoms H, given illness A,B and C are 0.90,0.95 and 0.75, respectively. Assuming that an ill person shows symptoms H, what is the probability that a person has illness A ? [4 Marks]

Answer 1

To find the probability that a person has illness A given that they show symptoms H, we can use Bayes' theorem. Bayes' theorem states that P(A|H) = (P(H|A) * P(A)) / P(H).

Step-by-step explanation:

To find the probability that a person has illness A given that they show symptoms H, we can use Bayes' theorem. Bayes' theorem states that P(A|H) = (P(H|A) * P(A)) / P(H), where P(A|H) is the probability of having illness A given symptoms H, P(H|A) is the probability of having symptoms H given illness A, P(A) is the probability of having illness A, and P(H) is the probability of having symptoms H.

Given the information provided, P(H|A) = 0.90, P(H|B) = 0.95, P(H|C) = 0.75, P(A) = 0.01, P(B) = 0.005, P(C) = 0.02.

To calculate P(H), we can use the law of total probability. P(H) = P(H|A) * P(A) + P(H|B) * P(B) + P(H|C) * P(C).

Plugging the values into the formula, we get P(H) = (0.90 * 0.01) + (0.95 * 0.005) + (0.75 * 0.02) = 0.02335.

Finally, we can calculate P(A|H) using Bayes' theorem. P(A|H) = (0.90 * 0.01) / 0.02335 ≈ 0.386 (or 38.6%).