Question

A common test for tuberculosis (TB) is a skin test, where protein is injected into a person’s arm and the person tests positive for TB if a rash develops. The probability of the person testing negative when they have TB is 1% (false negative). In some cases, a rash can develop even if the person does not have TB (false positive). Assume the probability of a false positive is 5%. Only 4 in 10,000 people have TB.

(a) Let T + mean “tests positive”, T − mean “tests negative”, and D mean “has the disease TB”. Use proper probabilistic notation to write three probabilities that are directly given in the information above.

(b) Draw a tree diagram (based upon the conditional information that is given) to show all possibilities. Fill in all probabilities along the tree.

(c) What is the probability that a randomly selected person will test positive for TB? Show all work, and use proper notation. You may leave your answer numerically unsimplified.

(d) What is the probability that a randomly selected person who tests positive actually has TB? Show all work, and use proper notation. You may leave your answer numerically unsimplified.

Answer 1

(a)

Denote Dꞌ as the complement of the event D, that is, Dꞌ denotes “does not have the disease TB”.

The three given probabilities are as follows:

P (D) = 0.0004 (= 4/10,000).

P (Dꞌ) = 0.9996 (= 1 – 0.0004).

P (T+ | Dꞌ) = 0.05.

P (T– | D) = 0.01.

(b)

Conditional probability:

For any 2 events A and B, the probability of the occurrence of A when B has already occurred, is the conditional probability: P (A | B) = P (A ⋂ B) / P (B). As a result, the intersection probability can be written as: P (A ⋂ B) = P (B) ∙ P (A | B).

Further recall that, P (A | B) + P (A | Bꞌ) = 1.

Using the above relationships and the given probabilities, the following can be obtained:

P (T– | Dꞌ) = 0.95 (= 1 – 0.05).

P (T+ | D) = 0.99 (= 1 – 0.01).

P (T+ ⋂ D) = 0.000396 (= 0.99 * 0.0004).

P (T– ⋂ D) = 0.000004 (= 0.01 * 0.0004).

P (T+ ⋂ Dꞌ) = 0.04998 (= 0.05 * 0.9996).

P (T– ⋂ Dꞌ) = 0.94962 (= 0.95 * 0.9996).

The tree diagram is drawn given below. Note that, the probabilities calculated by us (except that of Dꞌ) are written inside the oval areas.

(c)

The probability that a randomly selected person will test positive is as follows:

P (T+) = P (T+ ⋂ D) + P (T– ⋂ D)

= 0.000396 + 0.04998

= 0.050376.

The required probability is 0.050376.

(d)

The probability that a randomly selected person who tests positive actually has TB is as follows:

P (D | T+) = P (T+ ⋂ D)/P (T+)

= 0.000396/0.050376

= 0.007861.

The required probability is 0.007861.