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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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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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BiostatisticsEpidemiology, Step1Questions 
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