A test designed to check for sickle-cell anemia has an 85% chance of detection in that, if a person has the disease, there is an 85% chance that the test will be positive. And, if the person does not have the disease, there is an 85% chance that the test will be negative. We also know that the proportion of people in the U.S. that has sickle-cell anemia is 0.00023.
Suppose an individual is drawn at random from the population of the U.S., and that the individual tests positive for sickle-cell anemia using the test described above. What is the probability that the individual has the disease?
Answer 0.15 0.0000766 0.85 0.0013 0.2499425
To find the probability that the individual has the disease given a positive test result, we can use Bayes' theorem:
P(Disease|Positive) = (P(Positive|Disease) * P(Disease)) / P(Positive)
Where:
P(Disease|Positive) is the probability that the individual has the disease given a positive test result.
P(Positive|Disease) is the probability of testing positive given that the individual has the disease.
P(Disease) is the proportion of people in the U.S. that has sickle-cell anemia.
P(Positive) is the probability of testing positive.
Given that the test has an 85% chance of detection, we can calculate P(Positive|Disease) as 0.85.
P(Disease) is given as 0.00023, which is the proportion of people in the U.S. that has sickle-cell anemia.
To calculate P(Positive), we need to consider both the probability of testing positive given the presence of the disease (0.85 * 0.00023) and the probability of testing positive given the absence of the disease (which is the false positive rate). As we know that the false positive rate is 15%, the probability of testing positive in the absence of the disease is 0.15 * (1 - 0.00023).
Now we can substitute the values into Bayes' theorem:
P(Disease|Positive) = (0.85 * 0.00023) / (0.85 * 0.00023 + 0.15 * (1 - 0.00023))
Calculating further:
P(Disease|Positive) = 0.0001955 / (0.0001955 + 0.15 * 0.99977)
P(Disease|Positive) = 0.0001955 / 0.0001955 + 0.1499655
P(Disease|Positive) ≈ 0.0001955 / 0.150161
P(Disease|Positive) ≈ 0.0013
Therefore, the probability that the individual has the disease given a positive test result is approximately 0.0013.
To find the probability that the individual has the disease given a positive test result, we can use Bayes' theorem.
Let's define the following probabilities:
P(D) = Probability of having the disease = 0.00023 (given)
P(~D) = Probability of not having the disease = 1 - P(D) = 1 - 0.00023 = 0.99977 (complement of P(D))
P(Pos|D) = Probability of testing positive given having the disease = 0.85 (given)
P(Neg|~D) = Probability of testing negative given not having the disease = 0.85 (given)
We need to determine P(D|Pos), which is the probability of having the disease given a positive test result.
Applying Bayes' theorem:
P(D|Pos) = (P(Pos|D) * P(D)) / P(Pos)
P(Pos) can be calculated using the law of total probability:
P(Pos) = (P(Pos|D) * P(D)) + (P(Pos|~D) * P(~D))
Substituting the given values:
P(Pos) = (0.85 * 0.00023) + (0.15 * 0.99977)
P(Pos) = 0.0001955 + 0.1499655
P(Pos) = 0.150161
Now, we can calculate P(D|Pos):
P(D|Pos) = (P(Pos|D) * P(D)) / P(Pos)
P(D|Pos) = (0.85 * 0.00023) / 0.150161
P(D|Pos) = 0.0001955 / 0.150161
P(D|Pos) ≈ 0.0013
Therefore, the probability that the individual has the disease given a positive test result is approximately 0.0013.