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A test has been developed to detect a particular type of arthritis in individuals over 50 years old. From a national survey it is known that approximately 10% of the individuals in this age group suffer from this form of arthritis. The proposed test was given to individuals with confirmed arthritics disease, and a correct test result was obtained in 85% of the cases. When the test was administered to individuals of the same age group who were known to be free of the disease, 4% were reported to have the disease. What is the probability that an individual has this disease given that the test indicates its presence

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Answer:

The probability that an individual has this disease given that the test indicates its presence is 0.1

Step-by-step explanation:

Let A be the event which indicates Having disease

Let B be the event in which test indicates that disease presence

Now we are given that 10% of the individuals in this age group suffer from this form of arthritis

So, P(A)=0.1

We are also given that the proposed test was given to individuals with confirmed arthritics disease, and a correct test result was obtained in 85% of the cases.

P(B|A)=0.85

Now we are given that. When the test was administered to individuals of the same age group who were known to be free of the disease, 4% were reported to have the disease

So,P(B|A')=0.04

We are supposed to find  the probability that an individual has this disease given that the test indicates its presence i.e. P(A|B)

So we will use Bayes theorem

[tex]P(A|B)=\frac{P(B|A) P(A)}{\sum P(B|A) P(A)}[/tex]

[tex]P(A|B)=\frac{0.85 \times 0.1}{0.85 \times 0.1+0.85 \times 0.9}=0.1[/tex]

Hence the probability that an individual has this disease given that the test indicates its presence is 0.1