Question

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probability 0.99, but falsely detects a nonexistent bug with probability .02. Give numerical answers to each of the following questions:

(i) If the de-bugging program claims to have found a bug in program A, what is the probability that the bug is actually present?

(ii) Suppose that Program A is tested twice, the two tests conducted independently. If the de-bugging program claims that bugs are present on both the first and second tests, what is the probability that a bug is actually present?

(iii) Suppose that Program A is tested three times, the three tests conducted independently. If the de-bugging program claims that bugs are present on all three tests, what is the probability that a bug is actually present?

Answer 1

Introduction:

Denote B as the event that a bug is actually present, so that Bꞌ denotes the event that no bug is actually present. Denote T+ as the event that the de-bugging program detects the presence of a bug, that is, it tests positive, and T– as the event that the de-bugging program does not detect the presence of a bug, that is, it tests negative.

(i)

The given probabilities are as follows:

P (B) = 0.01 (= 1%) (Bug is present).

P (Bꞌ) = 0.99 (= 1 – 0.01) (Therefore, bug is absent).

P (T+ | B) = 0.99 (Tests positive when there is actual bug).

P (T+ | Bꞌ) = 0.02 (Tests positive when there is no actual bug).

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 (Aꞌ | B) + P (A | B) = 1.

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

P (T– | B) = 0.01 (= 1 – 0.99) (Tests negative when there is actual bug).

P (T– | Bꞌ) = 0.98 (= 1 – 0.02) (Tests negative when there is no actual bug).

P (T+ ⋂ B) = 0.0099 (= 0.99 * 0.01).

P (T– ⋂ B) = 0.0001 (= 0.01 * 0.01).

P (T+ ⋂ Bꞌ) = 0.0198 (= 0.02 * 0.99).

P (T– ⋂ Bꞌ) = 0.9702 (= 0.98 * 0.99).

The probability that there is a bug present, given that the test is positive, is as follows:

P (B | T+) = P (T+ ⋂ B) / P (T+).

Now, P (T+) = P (T+ ⋂ B) + P (T+ ⋂ Bꞌ).

= 0.0099/ (0.0099 + 0.0198)

= 0.0099/0.0297

≈ 0.3333.

Thus, if the de-bugging program claims to have found a bug in program A, the probability that the bug is actually present is 0.3333.

(ii)

Since Program A is tested twice and independently, if the de-bugging program claims that bugs are present on both the first and second tests, the situation described in Part (i) will be replicated twice; then the probability that the bug is actually present is:

[P (B | T+) ∙ P (B | T+)]

= [P (B | T+)]²

= (0.3333)²

≈ 0.1111.

Thus, if the de-bugging program claims that bugs are present on both the first and second tests, the probability that the bug is actually present is 0.1111.

(iii)

Again, if Program A is tested thrice, independently, and the result is positive for all three tests, the probability of the presence of a bug is:

[P (B | T+) ∙ P (B | T+) ∙ P (B | T+)]

= [P (B | T+)]³

= (0.3333)³

≈ 0.0370.

Thus, if the de-bugging program claims that bugs are present on all three tests, the probability that the bug is actually present is 0.0370.