A self-administered pregnancy test detects 80% of those who are pregnant but does not detect pregnancy in 20%. It is 92% accurate in indicating women who are not pregnant but indicates 8% of this group as being pregnant. Suppose it is known that 2% of the women in a neighborhood are pregnant. If a woman is chosen at random from those living in this neighborhood, and if the test indicates she is pregnant, what is the probability, rounded to four decimal places, that she really is?
A)Pr(pregnant | tested pregnant) = 0.1949
B)Pr(pregnant | tested pregnant) = 0.1695
C)Pr(pregnant | tested pregnant) = 0.2041
D)Pr(pregnant | tested pregnant) = 0.2000
E)Pr(pregnant | tested pregnant) = 0.0063

Question

A self-administered pregnancy test detects 80% of those who are pregnant but does not detect pregnancy in 20%. It is 92% accurate in indicating women who are not pregnant but indicates 8% of this group as being pregnant. Suppose it is known that 2% of the women in a neighborhood are pregnant. If a woman is chosen at random from those living in this neighborhood, and if the test indicates she is pregnant, what is the probability, rounded to four decimal places, that she really is? ​
A)Pr(pregnant | tested pregnant) = 0.1949
B)Pr(pregnant | tested pregnant) = 0.1695
C)Pr(pregnant | tested pregnant) = 0.2041
D)Pr(pregnant | tested pregnant) = 0.2000
E)Pr(pregnant | tested pregnant) = 0.0063

Quizplus · Accepted Answer

We are looking for the probability that a woman is actually pregnant given that the test indicates she is pregnant. This is a conditional probability, and we can use Bayes' Theorem to solve for it:

Pr(pregnant | tested pregnant) = Pr(tested pregnant | pregnant) * Pr(pregnant) / Pr(tested pregnant)

where
- Pr(pregnant | tested pregnant) is the probability that the woman is actually pregnant given that the test indicates she is pregnant (this is what we want to find)
- Pr(tested pregnant | pregnant) is the probability that the test indicates pregnant given that the woman is actually pregnant (this is the sensitivity of the test, which is 0.8)
- Pr(pregnant) is the overall probability of a woman being pregnant in the neighborhood (which is given as 0.02)
- Pr(tested pregnant) is the overall probability of the test indicating pregnant, regardless of whether the woman is actually pregnant or not

To find Pr(tested pregnant), we need to consider two cases:
- A woman is pregnant and the test indicates she is pregnant (true positive): this happens with probability Pr(tested pregnant | pregnant) * Pr(pregnant) = 0.8 * 0.02 = 0.016
- A woman is not pregnant but the test indicates she is pregnant (false positive): this happens with probability 0.08 * (1 - Pr(pregnant)) = 0.08 * 0.98 = 0.0784

Therefore, Pr(tested pregnant) = 0.016 + 0.0784 = 0.0944

Plugging this and the other given values into Bayes' Theorem, we get:

Pr(pregnant | tested pregnant) = 0.8 * 0.02 / 0.0944 = 0.1695

So the best choice is B.

A Self-Administered Pregnancy Test Detects 80% of Those Who Are