Lupus is an autoimmune disease, and it is believed that 2% of the population suf

Question

Lupus is an autoimmune disease, and it is believed that 2% of the population suffer from this disease. The test is 98% accurate if a person actually has the disease. The test is 74% accurate if a person does not have the disease.

If in individual has tested negative for lupus, what is the probablity that this person does not have lupus?(In other words, what is the probability of a true negative result?)Give your result to four decimal places. please include a probability tree to make information clear and include formula. thank you.

Explanation / Answer

Here we are given that 2% of the population suffer from this disease. Therefore

P( Lupus ) = 0.02

Also we are given that: test is 98% accurate if a person actually has the disease

Therefore P( positive | Lupus ) = 0.98 therefore P( negative | Lupus ) = 0.02

And also test is 74% accurate if a person does not have the disease, therefore

P( negative | No Lupus ) = 0.74 Therefore we get: P( positive | No Lupus ) = 1 - 0.74 = 0.26

By law of total probability, the probability that the test is negative is computed as:

P( negative ) = P( negative | Lupus ) P( Lupus ) + P( negative | No Lupus ) P( No Lupus)

P( negative ) = 0.02*0.02 + 0.74*(1 - 0.02) = 0.7256

Now given that the test is negative, probablity that this person does not have lupus is computed as: ( Using Bayes theorem as )

P( No Lupus | negative) = P( negative | No Lupus ) P( No Lupus) / P( negative )

P( No Lupus | negative) = 0.74*(1 - 0.02) / 0.7256 = 0.9994

Therefore 0.9994 is the required probability here.

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