Question:
A clinic specializes in testing for a disease \(D\). The result of the test can be either positive or negative.
A study revealed that if a person suffers from the disease \(D\), the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease \(D\), the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease \(D\) occurs in 30% of the individuals.
If a person tests positive for \(D\) in that clinic, the probability that he/she actually suffers from the disease \(D\) is _______. (Rounded off to two decimal places)
A clinic specializes in testing for a disease \(D\). The result of the test can be either positive or negative.
A study revealed that if a person suffers from the disease \(D\), the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease \(D\), the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease \(D\) occurs in 30% of the individuals.
If a person tests positive for \(D\) in that clinic, the probability that he/she actually suffers from the disease \(D\) is _______. (Rounded off to two decimal places)
A study revealed that if a person suffers from the disease \(D\), the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease \(D\), the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease \(D\) occurs in 30% of the individuals.
If a person tests positive for \(D\) in that clinic, the probability that he/she actually suffers from the disease \(D\) is _______. (Rounded off to two decimal places)
Show Hint
Use Bayes' theorem: \(P(D|\text{positive}) = \dfrac{P(\text{positive}|D)P(D)}{P(\text{positive}|D)P(D) + P(\text{positive}|D^c)P(D^c)}\).
Updated On: Jul 22, 2026
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Verified By Collegedunia
Correct Answer: 0.77
Solution and Explanation
Step 1: Write down the given probabilities.
Let \(D\) be the event that a person has the disease and \(D^c\) be the event that a person does not have the disease. Let \(P\) be the event that the test comes out positive.
From the question: \(P(D) = 0.30\), so \(P(D^c) = 0.70\).
The test is positive 80% of the time when the person has the disease: \(P(P \mid D) = 0.80\).
The test is positive 10% of the time when the person does not have the disease: \(P(P \mid D^c) = 0.10\).
Step 2: Set up Bayes' theorem.
We want \(P(D \mid P)\), the chance the person actually has the disease given a positive result. Bayes' theorem gives
\[ P(D \mid P) = \frac{P(P \mid D)\,P(D)}{P(P \mid D)\,P(D) + P(P \mid D^c)\,P(D^c)} \]
The denominator is the total probability of testing positive, made up of positives coming from sick people plus positives coming from healthy people.
Step 3: Plug in the numbers.
Numerator: \(P(P \mid D)\,P(D) = 0.80 \times 0.30 = 0.24\)
Denominator, first term: \(0.80 \times 0.30 = 0.24\)
Denominator, second term: \(P(P \mid D^c)\,P(D^c) = 0.10 \times 0.70 = 0.07\)
Total probability of a positive test: \(0.24 + 0.07 = 0.31\)
Step 4: Compute the final probability.
\[ P(D \mid P) = \frac{0.24}{0.31} = 0.7742\ldots \]
Rounded to two decimal places, this is \(0.77\).
Final Answer:
The probability that a person who tests positive actually has the disease is about 0.77. \[ \boxed{0.77} \]
Let \(D\) be the event that a person has the disease and \(D^c\) be the event that a person does not have the disease. Let \(P\) be the event that the test comes out positive.
From the question: \(P(D) = 0.30\), so \(P(D^c) = 0.70\).
The test is positive 80% of the time when the person has the disease: \(P(P \mid D) = 0.80\).
The test is positive 10% of the time when the person does not have the disease: \(P(P \mid D^c) = 0.10\).
Step 2: Set up Bayes' theorem.
We want \(P(D \mid P)\), the chance the person actually has the disease given a positive result. Bayes' theorem gives
\[ P(D \mid P) = \frac{P(P \mid D)\,P(D)}{P(P \mid D)\,P(D) + P(P \mid D^c)\,P(D^c)} \]
The denominator is the total probability of testing positive, made up of positives coming from sick people plus positives coming from healthy people.
Step 3: Plug in the numbers.
Numerator: \(P(P \mid D)\,P(D) = 0.80 \times 0.30 = 0.24\)
Denominator, first term: \(0.80 \times 0.30 = 0.24\)
Denominator, second term: \(P(P \mid D^c)\,P(D^c) = 0.10 \times 0.70 = 0.07\)
Total probability of a positive test: \(0.24 + 0.07 = 0.31\)
Step 4: Compute the final probability.
\[ P(D \mid P) = \frac{0.24}{0.31} = 0.7742\ldots \]
Rounded to two decimal places, this is \(0.77\).
Final Answer:
The probability that a person who tests positive actually has the disease is about 0.77. \[ \boxed{0.77} \]
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