Probability Case Study question-10

Probability Case Study: Medical Diagnosis Using Bayes’ Theorem

Background: A medical clinic conducts screenings for a rare disease called “Xenovirus.” The disease is rare, with an estimated prevalence in the general population of 0.1%. The clinic has developed a diagnostic test for Xenovirus, but like many medical tests, it is not perfect. The test’s sensitivity (true positive rate) is 95%, and its specificity (true negative rate) is 90%.(Probability Case Study question)

Question 1:

Suppose a patient tests positive for Xenovirus using the diagnostic test. What is the probability that the patient actually has the disease?

Solution:

To solve this problem, we can use Bayes’ Theorem. Bayes’ Theorem is a statistical formula that calculates the probability of an event based on prior knowledge of conditions that might be related to the event. In this case, we want to calculate the probability of having the disease given a positive test result.

Step 1: Define the Events:

Let A be the event that a patient has Xenovirus (disease).
Let A’ be the event that a patient has not Xenovirus (disease).
Let B be the event that a patient tests positive on the diagnostic test.

Step 2: Gather Information:

The prevalence of Xenovirus in the general population, P(A), is 0.1% or 0.001.
The sensitivity of the diagnostic test, P(B|A), is 95% or 0.95.
The specificity of the diagnostic test, P(B’|A’), is 90% or 0.90.

Step 3: Calculate the Probability: We want to find P(A|B), the probability that a patient has the disease given a positive test result. According to Bayes’ Theorem:

$P(A/B) = \dfrac{P(A).P(B/A)}{P(A).P(B/A)+P(A')P(B/A')}$

Now, we can plug in the values:

P(A) = 0.001 (prevalence of the disease).

P(B/A)=0.95 (sensitivity of the test).

P(A’) for not having the disease,

P(A’) = 1 – P(A) = 1 – 0.001 = 0.999

P(B/A’) is the probability of testing positive given not having the disease. This is the complement of specificity, which

is 1−0.90=0.10

Now, calculate:

$P(A/B) = \dfrac{P(A).P(B/A)}{P(A).P(B/A)+P(A')P(B/A')}$

$\Rightarrow P(A/B) = \dfrac{0.001\times 0.95}{0.001\times 0.95+0.999\times 0.10}$

$\Rightarrow P(A/B) = \dfrac{0.00095}{0.00095+0.0999}$

$\Rightarrow P(A/B) = \dfrac{0.00095}{0.10085}$

$\Rightarrow P(A/B) = \dfrac{95}{10085}$

⇒ P(A/B) = 0.00942

So, the probability that a patient actually has Xenovirus (the disease) given a positive test result is approximately 0.00942, or about 0.942%. This is a relatively low probability, despite the high sensitivity of the test, because the disease is rare in the general population, and false positives can occur.

Question 2:

Suppose a patient tests negative for Xenovirus using the diagnostic test. What is the probability that the patient is truly free from the disease?

Solution:

To answer this question, we need to calculate the probability that a patient does not have the disease (Xenovirus) given a negative test result.

Step 1: Define the Events:

Let A be the event that a patient has Xenovirus (disease).
Let A’ be the event that a patient has not Xenovirus (disease).
Let B be the event that a patient tests negative on the diagnostic test.

Step 2: Gather Information:

The prevalence of Xenovirus in the general population, P(A), is 0.1% or 0.001.
The sensitivity of the diagnostic test, P(B/A), is 95% or 0.95.
The specificity of the diagnostic test, P(B’/A’), is 90% or 0.90.

Step 3: Calculate the Probability: We want to find P(A’/B’), the probability that a patient does not have the disease given a negative test result. According to Bayes’ Theorem:

$P(A'/B) = \dfrac{P(A').P(B/A')}{P(A').P(B/A')+P(A)P(B/A)}$

Now, we can plug in the values:

P(A)=0.001 (prevalence of the disease).

P(A’) For not having the disease, which is

P(A’) = 1−P(A)=1−0.001=0.999.

P(B/A’) is the probability of testing negative given not having the disease. This is the complement of specificity, which is 1−0.90=0.10

P(B/A) is the probability of testing negative given having the disease. Since the test is not perfect, it’s the complement of sensitivity, which is 1−0.95=0.05

Now, calculate:

$P(A'/B) = \dfrac{P(A').P(B/A')}{P(A').P(B/A')+P(A)P(B/A)}$

$P(A'/B) = \dfrac{0.999\times 0.10}{0.999\times 0.10 + 0.001\times 0.05}$

$\Rightarrow P(A'/B) = \dfrac{0.0999}{0.0999 + 0.00005}$

$\Rightarrow P(A'/B) = \dfrac{0.0999}{0.09995}$

⇒ P(A’/B) = 0.9995

So, the probability that a patient does not have Xenovirus (the disease) given a negative test result is approximately 0.9995, or about 99.95%. This is a high probability, indicating that a negative test result is a strong indicator of being free from the disease.

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Probability Case Study question-10

Probability Case Study: Medical Diagnosis Using Bayes’ Theorem

Question 1:

Solution:

Question 2:

Solution:

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