3年的跨度(1095 days)下一个班30人中至少有两个人一天生日的概率：

1-choose(1095,30)*factorial(30)/(1095^30)
=0.3302673

In case birthday is not equally distributed,  R code is as following:

#----------------------Start----------------------
N <- 365*3
m <- 30

# birthday is equally distributed
pbirth_equal <- function(x) {
  return(1.0/N)
}

# birthday is normally distributed
pbirth_norm <- function(x) {
  mu <- (N+1)/2
  #Note:
  #   1. In order to avoid too much truncate error: delta need to be less than N/6
  #   2. In order to maintain positive sign of sumPi in the following calculation, delta need to be greater than (m-1)/sqrt(2*pi)
  delta <- N/6  
  px <- (x-mu)/delta
  pbirth <- 1/sqrt(2*pi)*exp(-px*px/2)
  return(pbirth/delta)
}

Pbirthday <- 0
for (i in 1:N) {  
  Pi <- pbirth_norm(i)
  sumPi <- Pi
  
  for (j in 1:(m-1)) {
    sumPi <- sumPi * (1-j*Pi)
  }
  Pbirthday <- Pbirthday + sumPi
}

(1-Pbirthday)    # probability output
#----------------------End----------------------

Basically you need to implement your own birthday piror function. If the birthday is normall distributed with mean: (N+1)/2 and standard deviation: N/6, the probability will be 0.478.

Thank you very much!!!