Biostatistics #12 - Probability of urinary tract infection - USMLE Forums
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Old 06-05-2011
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Stats Biostatistics #12 - Probability of urinary tract infection

Five percent of pregnant women have evidence of urinary tract infection when they are first seen for prenatal care. Four percent of those who are not found to be infected at the first prenatal visit develop an infection between that time and delivery. The probability that a woman will have a urinary tract infection during pregnancy is thus:

a. 0.04 x 0.95 = 0.038
b. 0.05
c. 0.04 + 0.05 = 0.09
d. 0.05 + [0.04 x 0.95] = 0.088
e. 0.04 x 0.05 = 0.002
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I'd say the answer is D
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I think is a)
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i think is D.....
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I dont know

????? D
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I think C

could you email me the answer as well
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is it E?
I really dont remember how to calculate the probablity
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I choose D
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Correct Answer correct answer

a. 0.04 x 0.95 = 0.038
b. 0.05
c. 0.04 + 0.05 = 0.09
d. 0.05 + [0.04 x 0.95] = 0.088
e. 0.04 x 0.05 = 0.002


So:
- 5% (0.05) = UTI when they are first seen for prenatal care
- The remaining 95% (0.95) = 4% UTI between that time and delivery

Then:
5% plus 4% of the 95% = 0.05 + [0.04 x 0.95] = 0.088

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Biostatistics-Epidemiology, Step-1-Questions

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