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10% of the population in a town is HIV+. A new diagnostic kit for HIV detection is available; this kit correctly identifies HIV+ individuals 95% of the time, and HIV− individuals 89% of the time. A particular patient is tested using this kit and is found to be positive. The probability that the individual is actually positive is _______
Correct answer is between '0.48,0.49'. Can you explain this answer?
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10% of the population in a town is HIV+. A new diagnostic kit for HIV ...
-negative individuals 98% of the time.
a) What is the probability that a randomly selected individual from the town has HIV and tests positive for it using the new kit?
b) What is the probability that a randomly selected individual from the town does not have HIV but tests positive for it using the new kit?
c) If an individual tests positive for HIV using the new kit, what is the probability that they actually have HIV?
a) Let P(HIV) be the probability that an individual from the town has HIV, which is 10% or 0.1. Let P(positive|HIV) be the probability that an individual with HIV tests positive for it using the new kit, which is 95% or 0.95. Then the probability that a randomly selected individual from the town has HIV and tests positive for it using the new kit is:
P(HIV and positive) = P(HIV) * P(positive|HIV) = 0.1 * 0.95 = 0.095 or 9.5%
b) Let P(negative|no HIV) be the probability that an individual without HIV tests positive for it using the new kit, which is 2% or 0.02. Then the probability that a randomly selected individual from the town does not have HIV but tests positive for it using the new kit is:
P(no HIV and positive) = P(no HIV) * P(positive|no HIV) = (1 - P(HIV)) * P(negative|no HIV) = 0.9 * 0.02 = 0.018 or 1.8%
c) Let P(HIV|positive) be the probability that an individual who tests positive for HIV using the new kit actually has HIV. This is the probability we want to find. We can use Bayes' theorem to calculate it:
P(HIV|positive) = P(positive|HIV) * P(HIV) / P(positive)
where P(positive) is the probability of testing positive for HIV regardless of whether the individual has HIV or not. We can calculate this by adding up the probabilities of the two cases:
P(positive) = P(HIV and positive) + P(no HIV and positive) = 0.095 + 0.018 = 0.113 or 11.3%
Then we can substitute the values we calculated in parts (a) and (b) to get:
P(HIV|positive) = 0.95 * 0.1 / 0.113 = 0.841 or 84.1%
Therefore, if an individual tests positive for HIV using the new kit, the probability that they actually have HIV is about 84%.
a) What is the probability that a randomly selected individual from the town has HIV and tests positive for it using the new kit?
b) What is the probability that a randomly selected individual from the town does not have HIV but tests positive for it using the new kit?
c) If an individual tests positive for HIV using the new kit, what is the probability that they actually have HIV?
a) Let P(HIV) be the probability that an individual from the town has HIV, which is 10% or 0.1. Let P(positive|HIV) be the probability that an individual with HIV tests positive for it using the new kit, which is 95% or 0.95. Then the probability that a randomly selected individual from the town has HIV and tests positive for it using the new kit is:
P(HIV and positive) = P(HIV) * P(positive|HIV) = 0.1 * 0.95 = 0.095 or 9.5%
b) Let P(negative|no HIV) be the probability that an individual without HIV tests positive for it using the new kit, which is 2% or 0.02. Then the probability that a randomly selected individual from the town does not have HIV but tests positive for it using the new kit is:
P(no HIV and positive) = P(no HIV) * P(positive|no HIV) = (1 - P(HIV)) * P(negative|no HIV) = 0.9 * 0.02 = 0.018 or 1.8%
c) Let P(HIV|positive) be the probability that an individual who tests positive for HIV using the new kit actually has HIV. This is the probability we want to find. We can use Bayes' theorem to calculate it:
P(HIV|positive) = P(positive|HIV) * P(HIV) / P(positive)
where P(positive) is the probability of testing positive for HIV regardless of whether the individual has HIV or not. We can calculate this by adding up the probabilities of the two cases:
P(positive) = P(HIV and positive) + P(no HIV and positive) = 0.095 + 0.018 = 0.113 or 11.3%
Then we can substitute the values we calculated in parts (a) and (b) to get:
P(HIV|positive) = 0.95 * 0.1 / 0.113 = 0.841 or 84.1%
Therefore, if an individual tests positive for HIV using the new kit, the probability that they actually have HIV is about 84%.
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Community Answer
10% of the population in a town is HIV+. A new diagnostic kit for HIV ...
HIV+ population in a town=10%
HIV- population in a town=90%
kit correctly identify HIV+=95%
kit incorrectly identify HIV-=(100-89)% =11%
by Baye's theorem,
the probability of that individual is actually positive=(0.1x0.95) /(0.1x0.95+0.11x0.9)= 0.489
HIV- population in a town=90%
kit correctly identify HIV+=95%
kit incorrectly identify HIV-=(100-89)% =11%
by Baye's theorem,
the probability of that individual is actually positive=(0.1x0.95) /(0.1x0.95+0.11x0.9)= 0.489
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10% of the population in a town is HIV+. A new diagnostic kit for HIV detection is available; thiskit correctly identifies HIV+ individuals 95% of the time, and HIV− individuals 89% of the time. Aparticular patient is tested using this kit and is found to be positive. The probability that theindividual is actually positive is _______Correct answer is between '0.48,0.49'. Can you explain this answer?
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