Question

2. Suppose next that we have even less knowledge of our patient, and we are only given the accuracy of the blood test and prevalence of the disease in our population. We are told that the blood test is 96% percent reliable, this means that the test will yield an accurate positive result in 96% of the cases where the disease is actually present. Gestational diabetes affects 7 percent of the population in our patient’s age group, and that our test has a false positive rate of 10% percent. Use your knowledge of Bayes’ Theorem and Conditional Probabilities to compute the following quantities based on the information given only in part 2:

1. If 100,000 people take the blood test, how many people would you expect to test positive and actually have gestational diabetes?
2. What is the probability of having the disease given that you test positive?
3. If 100,000 people take the blood test, how many people would you expect to test negative despite actually having gestational diabetes?
4. What is the probability of having the disease given that you tested negative?

Answer 1

Let A be the event that the patient has Gestational diabetes.
Since, Gestational diabetes affects 7 percent of the population : P(A) = 0.07

Let B be the event that the patient tested positive
Since the blood test is 96% reliable given the disease is actually present : P(B|A) = 0.96

From conditional probability we know that P(A∩B) = P(A) * P(B|A)
P(A∩B) = 0.96*0.07 = 0.0672

If 100,000 people take the blood test, the number of people who would test positive and actually have gestational diabetes is
100,000 * P(A∩B)
= 6,720

Using law of total probability
P(B) = P(A₁) * P(B|A₁) + P(A₂) * P(B|A₂)

=0.96* 0.07 + 0.10 * (1-0.07)
=0.0672 + 0.093
P(B) =0.1602

Therefore, 0.1602 is the total probability that the patient tests positive.

what is the probability of having the disease given that you test Positive? To find : P(A|B)

From Bayes' theorem we know that
P(B) * P(A|B) = P(A)* P(B|A)
Substituting the above values we have

P(A|B) = (0.96 * 0.07)/0.1602
= 0.419 = 41.9%

Let P(B^c|A) be the probability that the patient tests negative given they have gestational diabetes.
Since the test test is 96% reliable when they actually have the disease, there is a 4% chance they test negative even when they have the disease. Note that B^c is also read as B complement.

Therefore P(B^c|A) =0.04
P(A∩B^c) = P(A) * P(B^c|A) = 0.04* 0.07 = 28/10000
If 100,000 people take the blood test, 100,000 * P(A∩B^c) = 280 people would test negative, despite actually having gestational diabetes.

P(A) * P(B^c|A) = P(B^c) * P(A|B^c)

P(B^c) = 0.04*0.07 + 0.90*0.93 ( 0.90*0.93 is the chance of false negative in population who do not have the disease)

Total probability of getting a negative result on the test is
P(B^c) = 0.8398

What is the probability of having the disease given that you tested negative. To find : P(A|B^c)

We know that P(A) * P(B^c|A) = 0.0028 = 28/10000
Therefore P(A|B^c) = 0.0028/ 0.8398
= 0.0033 = 0.33%