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#1
06-05-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,259 Times in 881 Posts Reputation: 3269
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
 The above post was thanked by: docoftheworld (06-06-2011), dr_lizard (09-01-2011), pass7 (06-05-2011), struggle (06-06-2011)

#2
06-05-2011
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I'd say the answer is D
 The above post was thanked by: bebix (06-05-2011)
#3
06-05-2011
 USMLE Forums Scout Steps History: Not yet Posts: 21 Threads: 1 Thanked 16 Times in 13 Posts Reputation: 26

I think is a)
 The above post was thanked by: bebix (06-05-2011)
#4
06-05-2011
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i think is D.....
 The above post was thanked by: bebix (06-05-2011)
#5
06-05-2011
 USMLE Forums Guru Steps History: 1+CK+CS+3 Posts: 406 Threads: 30 Thanked 140 Times in 108 Posts Reputation: 150

I dont know

????? D
 The above post was thanked by: bebix (06-05-2011)
#6
06-06-2011
 USMLE Forums Veteran Steps History: 1 + CS Posts: 204 Threads: 56 Thanked 41 Times in 36 Posts Reputation: 51

I think C

could you email me the answer as well
 The above post was thanked by: bebix (06-06-2011)
#7
06-06-2011
 USMLE Forums Guru Steps History: Not yet Posts: 324 Threads: 70 Thanked 236 Times in 143 Posts Reputation: 246

is it E?
I really dont remember how to calculate the probablity
 The above post was thanked by: bebix (06-06-2011)
#8
06-06-2011
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 124 Threads: 5 Thanked 204 Times in 70 Posts Reputation: 214
Ans.

I choose D
 The above post was thanked by: bebix (06-06-2011)
#9
06-06-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,259 Times in 881 Posts Reputation: 3269
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

 The above post was thanked by: Amenah (06-06-2011), dr_lizard (09-01-2011), pass7 (06-06-2011), radkur (10-16-2014), rulz (10-06-2011), struggle (06-06-2011)

 Tags Biostatistics-Epidemiology, Step-1-Questions

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