Consider a diagnostic test scenario where a patient is either sick or not sick. The probability of a patient being sick is 0.2 and not sick is 0.8. If a patient is sick, the test yields a positive result with probability 0.9 and a negative result with probability 0.1. If a patient is not sick, the test yields a positive result with probability 0.15 and a negative result with probability of 0.85.
Answer the following:
(a) Draw a tree diagram representing this scenario.
(b) Calculate the probability that a patient is sick and tests positive, and the probability that a patient is not sick and tests positive.
(c) Determine the overall probability that a patient tests positive and, using this, compute the conditional probability that a patient is sick given a positive test result. Comment on the reliability of the test.
Question
Draw a tree diagram representing the following scenario: A patient is sick with probability 0.2 and not sick with probability 0.8. If sick, the test yields a positive result with probability 0.9 and a negative result with probability 0.1. If not sick, the test yields a positive result with probability 0.15 and a negative result with probability 0.85. Label each branch with the appropriate probabilities.
Hint
Hint
Hint
Begin by creating two branches for ‘Sick’ and ‘Not Sick’, then add branches for the test results, ensuring each branch is labeled with its probability.
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Answer
Answer
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Explanation
Explanation
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Complete Answer
The tree diagram should have a first split into ‘Sick (0.2)’ and ‘Not Sick (0.8)’. Under ‘Sick’, add branches ‘Positive (0.9)’ and ‘Negative (0.1)’. Under ‘Not Sick’, add branches ‘Positive (0.15)’ and ‘Negative (0.85)’.
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Answer Explanation
Answer Explanation
Question
Calculate the following branch probabilities:
(i) The probability that a patient is sick and tests positive.
(ii) The probability that a patient is not sick and tests positive.
Hint
Hint
Hint
Multiply the probabilities along each branch (e.g., P(Sick and Positive) = P(Sick) × P(Positive|Sick)).
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Answer
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Explanation
Explanation
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Complete Answer
P(Sick and Positive) = 0.2 × 0.9 = 0.18
P(Not Sick and Positive) = 0.8 × 0.15 = 0.12
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Answer Explanation
Answer Explanation
Question
Using the branch probabilities from part (b):
(i) Determine the overall probability that a patient tests positive.
(ii) Compute the conditional probability that a patient is sick given a positive test result.
(iii) Briefly comment on the reliability of the test.
Hint
Hint
Hint
For (i) add the probabilities from both branches. For (ii), use Bayes’ theorem: P(Sick|Positive) = P(Sick and Positive)/P(Positive).
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Answer
Answer
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Explanation
Explanation
- Formula Booklet
Complete Answer
Overall P(Positive) = 0.18 + 0.12 = 0.30
P(Sick|Positive) = 0.18 / 0.30 = 0.6
Interpretation: A 60% chance that a patient is sick given a positive test suggests moderate reliability, indicating a significant rate of false positives.
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