Sentence examples for month probability of from inspiring English sources

We calculate each individual's per month probability of infection with either the DS or the DR strain (φDS, φDR) by first considering the probability of not being infected (∼φDS, ∼φDR).

In month 5, only two are at risk for death at the beginning of the interval, because one patient was censored in the previous month (probability of survival is 2/2 = 1).

In the absence of more suitable data, a 12 month probability of clearance of 87.5% was estimated both for those warts that are treated and those for which treatment is not sought.

For example, in the third month, four patients are at risk for death at the beginning of the month, with three alive at the end of the month (probability of survival is 3/4 = 0.75).

The 24-month probability of TLR for each treatment was weighted by sample size.

The 12-month probability of remaining seizure-free was 14 and 16%, respectively (P=0.74).

The 3-month probability of graft failure was 69% when the interval between the first HCT and second HCT was less than 3 months, compared with 23% when the interval was longer (P <.001).

HPV16/16-like HPV16/16-likelike groups showed heterogeneity (p<0.05) in terms of how CD4 counts, HIV VL, and HPV18/18-like HPV18/18-likeof clearance of each HPV infection.This new model predicts the 3-month probability of HPV infection clearance based on CD4 cell counts and other HIV-1-related clinical measurements.

The 3-month probability of occurrence of rash or hepatotoxicity did not differ between groups 1 and 2 (5.3% vs. 7.5%; Log-Rank test, p = 0.35).

The 24-month probability of occurrence of hepatotoxicity or rash was not different between women with a CD4 cell count >250 cells/mm3 and women with a CD4 cell count ≤250 cells/mm3 (8.3% vs. 9.9%, Log-rank test: p = 0.75).

The 24-month probability of occurrence of rash or hepatotoxicity was not different between women with a CD4 cell count >250 cells/mm3 and women with a CD4 cell count ≤250 cells/mm3 (8.3% vs. 9.9%, Log-rank test: p = 0.75).