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  • In a city, 80% of the people have a particular genetic marker. A genetic test has been developed to identify the presence of this marker, with a sensitivity of 90% and a specificity of 95%. If a ra

In a city, 80% of the people have a particular genetic marker. A genetic test has been developed to identify the presence of this marker, with a sensitivity of 90% and a specificity of 95%. If a randomly selected person tests positive for the genetic marker, and an independent second test with the same sensitivity and specificity is conducted, what is the probability that both tests correctly identify the presence of the genetic marker?

 

Option: 1

92.4%


Option: 2

85.5%


Option: 3

76.2%


Option: 4

69.6%


Answers (1)

best_answer

Let's define the events:

A: Person has the genetic marker

B: Person tests positive for the genetic marker in the first test

C: Person tests positive for the genetic marker in the second test

We are given:

\mathrm{P(A)=0.80} (prevalence of the genetic marker in the city)
\mathrm{P(B \mid A)=0.90} (sensitivity of the first genetic test)
\mathrm{P(C \mid A)=0.90} (sensitivity of the second genetic test)
\mathrm{P(B \mid\, not \, A)=0.05} (1 - specificity of the first genetic test )
\mathrm{P(C \mid not A)=0.05} (1- specificity of the second genetic test )

We want to find \mathrm{P(B \cap C)}, the probability that both tests correctly identify the presence of genetic marker given that a person tests positive for it.
Using the formula for conditional probability:

\mathrm{P(B \cap C)=P(B \mid A) \times P(C \mid A) \times P(A)}

Given:
\begin{aligned} & P(A)=0.80 \\ & P(B \mid A)=0.90 \\ & P(C \mid A)=0.90 \end{aligned}
Substituting the values into the equation, we can calculate \mathrm{P\left ( B\cap C \right )} to be approximately 0.724 or 92.4%.

Therefore, the probability that both tests correctly identify the presence of the genetic marker given that a person tests positive for it is 92.4%.

 

Posted by

Divya Prakash Singh

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