Exercise 2.13

Answers

(a)
Let B be the event that the test done by company B is successfull. Let A be the event that the test done by company A is successfull. Let D be the event that a random person has the disease.

P(B) = P(D)P(B|D) + P(Dc)P(B|Dc) = 0.01 ∗ 0 + 0.99 ∗ 1 = 0.99

P(A) = P(D)P(A|D) + P(Dc)P(A|Dc) = 0.01 ∗ 0.95 + 0.99 ∗ 0.95 = 0.95

Thus, P(B) > P(A).

(b)
Since the disease is so rare, most people don’t have it. Company B diagnoses them correctly every time. However, in the rare cases when a person has the disease, company B fails to diagnose them correctly. Company A however shows a very good probability of an accurate diagnoses for afflicted patients.
(c)
If the test conducted by company A has equal specifity and sensitivity, then it’s accuracy surpasses that of company B’s test if the specifity and the sensitivity are larger than 0.99. If company A manages to achieve a specifity of 1, then any positive sensitivity will result in a more accurate test. If company A achieves a sensitivity of 1, it still requires a specificity larger than 0.98, since positive cases are so rare.
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2021-12-05 00:00
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