N <- seq(from=100,to=2000,by=1)
P <- choose(N-100, 50) / choose(N, 60)
## Normalization of Ps:
sum(P)
P <- P / sum(P)
## Plot it:
plot(N, P, type = "l",las=1)

P[which(N) = 386]

# Set seed for reproducible example
set.seed(42)

# Create a dataframe of random numbers
mydf <- data.frame(X=rnorm(45),Y=rnorm(45))

# For example set the value of 5th element to 386 as in your question
mydf$X[5] <- 386

# Print the fifth value of Y
print(mydf$Y[5])

# Use to find X where it is value and print equivalent value of Y
print(mydf$Y[which(mydf$X==386)])

P[which(N == 386)]