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Activity Discussion Math Probability

• # Probability

Posted by on June 23, 2023 at 6:55 pm

In a certain population, a rare disease affects 1% of individuals. There is a diagnostic test for this disease, which correctly identifies the disease in 95% of cases when a person has it and incorrectly identifies the disease in 2% of cases when a person does not have it. If a person tests positive for the disease, what is the probability that they actually have the disease?

replied 6 months, 2 weeks ago 2 Members · 1 Reply
• ### Kunal

Member
January 30, 2024 at 5:57 pm
0

To find the probability that a person actually has the disease given that they tested positive, we can use Bayes’ theorem. Let’s define the following probabilities:

P(D) = Probability of having the disease = 0.01 (1% of individuals)

P(Pos|D) = Probability of testing positive given that a person has the disease = 0.95 (95% accuracy)

P(Pos|Â¬D) = Probability of testing positive given that a person does not have the disease = 0.02 (2% false positive rate)

We want to calculate P(D|Pos), which is the probability of having the disease given that the person tested positive. According to Bayes’ theorem:

P(D|Pos) = (P(Pos|D) * P(D)) / P(Pos)

To calculate P(Pos), the probability of testing positive, we need to consider both scenarios: the person having the disease and testing positive, and the person not having the disease but testing positive.

P(Pos) = P(Pos|D) * P(D) + P(Pos|Â¬D) * P(Â¬D)

P(Â¬D) = Probability of not having the disease = 1 – P(D) = 0.99 (99% of individuals)

Now we can substitute these values into the equation:

P(D|Pos) = (P(Pos|D) * P(D)) / (P(Pos|D) * P(D) + P(Pos|Â¬D) * P(Â¬D))

P(D|Pos) = (0.95 * 0.01) / (0.95 * 0.01 + 0.02 * 0.99)

Simplifying the equation:

P(D|Pos) = 0.0095 / (0.0095 + 0.0198)

P(D|Pos) â‰ˆ 0.324

Therefore, the probability that a person actually has the disease given that they tested positive is approximately 0.324 or 32.4%.

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