times <- 931
random <- 10000
prob <- 1/10
mu <- random*prob
a_sq <- (random*prob)*(1-prob)
a <- sqrt(a_sq)
answer <- (times-mu)/a
pnorm(answer)*100
## [1] 1.072411
Probability that among 10,000 random digits the digit 3 appears is 1.072411 %