x <- c(A = 100, B = 900) # the group B distribution of positive hits (A) and the rest (B) from a sample of 1000 produced by the model

p <- c(0.0078, 1-0.0078) # probability of naturally occuring event in the population of 136k, 0.78% for positive hits, resulting in the already mentioned 8 per 1000 hit rate

chisq.test(x, p = p)

X-squared = 1098.4, df = 1, p-value < 2.2e-16