Biostatistics #12 - Probability of urinary tract infection - USMLE Forums
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#1
06-05-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,357 Threads: 194 Thanked 3,268 Times in 881 Posts Reputation: 3278
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
 USMLE Forums Newbie Steps History: Step 1 Only Posts: 8 Threads: 0 Thanked 4 Times in 4 Posts Reputation: 14

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 141 Times in 109 Posts Reputation: 151

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 42 Times in 37 Posts Reputation: 52

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 239 Times in 144 Posts Reputation: 249

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 3,268 Times in 881 Posts Reputation: 3278

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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