A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic gonadotropin (hCG) hormone. Suppose that in a given population, 6% of fetuses have a chromosome abnormality. The test correctly produces a positive result for a fetus with a chromosome abnormality 88% of the time and correctly produces a negative result for a normal fetus 81% of the time. What proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality
To find the proportion of women who get a positive test result and are actually carrying a fetus with a chromosome abnormality, we can use conditional probability.
Let's denote the following:
A: The event that a fetus has a chromosome abnormality.
B: The event that the blood test result is positive.
We are given the following probabilities:
P(A) = 0.06 (6% of fetuses have a chromosome abnormality)
P(B|A) = 0.88 (the test produces a positive result for a fetus with a chromosome abnormality)
P(not A) = 1 - P(A) = 1 - 0.06 = 0.94 (the complement of A, i.e., the fetus being normal)
P(not B|not A) = 0.81 (the test produces a negative result for a normal fetus)
To find the proportion of women who get a positive test result and are actually carrying a fetus with a chromosome abnormality, we need to calculate the conditional probability:
P(A|B) = (P(A) * P(B|A)) / P(B)
Applying Bayes' theorem, we have:
P(B) = P(A) * P(B|A) + P(not A) * P(B|not A)
P(B) = 0.06 * 0.88 + 0.94 * (1 - 0.81)
P(B) = 0.0528 + 0.1736
P(B) = 0.2264
Substituting this value back into the formula for conditional probability:
P(A|B) = (0.06 * 0.88) / 0.2264
P(A|B) = 0.0528 / 0.2264
P(A|B) ≈ 0.233 (rounded to three decimal places)
Therefore, approximately 23.3% of women who get a positive test result are actually carrying a fetus with a chromosome abnormality.