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  • In a population of 10,000 individuals, a rare disease affects 1 in every 1000 people. A diagnostic test for the disease has been developed, with a sensitivity of 98% and a specificity of 95%. If a

In a population of 10,000 individuals, a rare disease affects 1 in every 1000 people. A diagnostic test for the disease has been developed, with a sensitivity of 98% and a specificity of 95%. If a randomly selected individual tests positive for the disease, what is the probability that they actually have the disease if they belong to a high-risk group with a disease prevalence of 5%?

 

Option: 1

0.10%


Option: 2

0.49%


Option: 3

1.96%


Option: 4

9.80%


Answers (1)

best_answer

To find the probability that an individual actually has the disease given a positive test result, we need to use Bayes' theorem. Let's define the events:

A: Person has the disease

B: Person tests positive for the disease

We are given:
\mathrm{P(A)=1 / 1000} (prevalence of the disease in the general population)
\mathrm{P(B \mid A)=0.98} (sensitivity of the test)
\mathrm{P(B \mid not A)=0.05} (1-specificity of the test )
\mathrm{P(A \mid high - risk \, group )=5 \%} ( disease prevalence in the high - risk group )
We want to find

\mathrm{\mathrm{P}(\mathrm{A} \mid \mathrm{B})},the probability that the individual actually has the disease given a positive test result.
Using Bayes' theorem:
 

\mathrm{P(A \mid B)=P(B \mid A) \times P(A \mid \text { high-risk group }) / P(B)}
To calculate \mathrm{P(B)}, we can use the law of total probability:

\mathrm{P(B)=P(B \mid A) \times P(A)+P(B \mid \text { not } A) \times P(\text { not } A)}

Given:
\begin{aligned} & P(A)=1 / 1000 \\ & P(B \mid A)=0.98 \\ & P(B \mid \text { not } A)=0.05 \\ & P(A \mid \text { high-risk group })=5 \% \end{aligned}
Substituting the values into the equation, we can calculate \mathrm{P(A \mid B)}  to be approximately \mathrm{0.49 \%}.

Therefore, the probability that an individual actually has the disease given a positive test result in the high-risk group is 0.49%.

 

Posted by

himanshu.meshram

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