Compute the probability that a person who tests positive actually has the disease.
A certain disease has an incidence rate of 0.3%. If the false negative rate is 4% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease.
The false positive rate is 1%, so the probability that somebody who tests positive doesn't have the disease is 1% or 0.01. Therefore the probability that somebody who tests positive has the disease is 99% or 0.99.
Answer: 99%
To compute the probability that a person who tests positive actually has the disease, we need to consider the incidence rate, false negative rate, and false positive rate.
1. Incidence rate: The incidence rate of the disease is given as 0.3%. This means that out of 1000 people, 3 individuals have the disease.
2. False negative rate: The false negative rate is given as 4%. This means that if someone has the disease, there is a 4% chance that they will test negative.
3. False positive rate: The false positive rate is given as 1%. This means that out of 1000 people without the disease, 10 individuals will test positive.
Now, let's calculate the probability that a person who tests positive actually has the disease.
Out of the 1000 people, 3 individuals have the disease (0.3% incidence rate). The false negative rate is 4%, so 96% of the people with the disease will test positive.
Therefore, out of the 3 individuals with the disease, 96% or approximately 2.88 individuals will test positive.
Out of the 1000 people without the disease, 1% or approximately 10 individuals will test positive due to the false positive rate.
So, the total number of individuals who test positive is 2.88 + 10 = 12.88.
Now, to calculate the probability that a person who tests positive actually has the disease, we can use Bayes' theorem.
The probability of having the disease given that a person tests positive can be calculated as: (probability of having the disease * probability of testing positive given the disease) divided by the probability of testing positive.
Probability of having the disease = 3/1000 = 0.003
Probability of testing positive given the disease = 0.96 (96%),
Probability of testing positive = 12.88/1000 = 0.01288
Therefore, the probability that a person who tests positive actually has the disease is:
(0.003 * 0.96) / 0.01288 ≈ 0.224 or 22.4%
Answer: The probability that a person who tests positive actually has the disease is approximately 22.4%.
To compute the probability that a person who tests positive actually has the disease, we need to use Bayes' theorem. Bayes' theorem is a mathematical formula that calculates the probability of an event based on prior knowledge of conditions that might be related to the event.
The formula for Bayes' theorem is:
P(A|B) = (P(B|A) * P(A)) / P(B)
In this case:
A represents the event "having the disease"
B represents the event "testing positive"
Now, let's break down the values we have:
P(A) = incidence rate of the disease = 0.3% = 0.003
P(B|A) = false negative rate = 4% = 0.04
P(B) = probability of testing positive = ?
To calculate P(B), we need to consider both the true positive and false positive rates. The true positive rate is the complement of the false negative rate, which means:
True positive rate = 1 - False negative rate = 1 - 0.04 = 0.96
The false positive rate is given as 1% or 0.01.
So, P(B) = (P(B|A) * P(A)) + (P(B|not A) * P(not A))
= (0.96 * 0.003) + (0.01 * (1 - 0.003))
Now, we can substitute the values into Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(A|B) = (0.96 * 0.003) / [(0.96 * 0.003) + (0.01 * (1 - 0.003))]
Calculating this expression yields the probability that a person who tests positive actually has the disease.
Please substitute the actual values into the equation to get the final result.