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Medical researchers know that the probability of getting

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Medical researchers know that the probability of getting [#permalink] New post 15 May 2004, 16:42
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Medical researchers know that the probability of getting lung cancer if a person smokes is .34. The probability that a nonsmoker will get lung cancer is .03. It is also known that 11% of the population smokes. What is the probability that a person with lung cancer will have been a smoker?
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 [#permalink] New post 16 May 2004, 04:51
p(s) = person smokes = 0.11
p(n) = person does not smoke = 1-0.11 = 0.88
p(c1) = cancer if smoking = 0.34
p(c2) = cancer if not smoking = 0.03

no of people who get cancer = [ p(c1) * p(s) + p(c2) * p(n) ] * 100
no of people who smoke and get cancer = p(c1) * p(s) * 100

This is so because we already know that a person has cancer and we need to determine if he/she did smoke ( we are dealing with a much narrower range of the sample )

P = ( 0.11 * 0.34 ) / [ 0.11 * 0.34 + 0.88 * 0.03 ] = 0.5862 ( I used a calc)

What this says is that there is high probability( higher than just 0.34 ) that a person smoked if he/she had a lung cancer.

I hope I am right.
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Re: PS-38 [#permalink] New post 17 May 2004, 13:30
hallelujah1234 wrote:
Medical researchers know that the probability of getting lung cancer if a person smokes is .34. The probability that a nonsmoker will get lung cancer is .03. It is also known that 11% of the population smokes. What is the probability that a person with lung cancer will have been a smoker?


----------smoke---No smoke
Lung------.34*11--.03*89---6.41
Not-Lung-.66*11--.97*89--93.59
Total----------11-------89----100

Answer = .34*11/6.41 = .583
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 [#permalink] New post 17 May 2004, 13:49
hi ndidi04,

What are you trying to do here?

Anand. :shock:
  [#permalink] 17 May 2004, 13:49
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