Here is a standard sampling-without-replacement problem. Suppose we have \(N\) balls in a box, \(B\) that are black and the remaining \(W = N - B\) balls are white.

You sample \(n\) balls from the box and \(b\) turn out to be black. What have you learned about the number \(B\) that are black in the box?

This is a Bayes’ rule problem. We’ll illustrate it for the case where \(N = 50\).

- (Prior) First the number of black balls in the box \(B\) could be any value from 0 to 50. We place a uniform prior on the values 0, 1, 2, …, 50.

```
library(TeachBayes)
bayes_df <- data.frame(B=0:50, Prior=rep(1/51, 51))
```

- (Data) We take a sample of \(n = 10\) balls without replacement and observe that the number of black balls is \(b = 3\). The likelihood is the probability of this outcome, expressed as a function of \(B\). I use the special function
`dsampling()`

.

```
sample_b <- 3
pop_N <- 50
sample_n <- 10
bayes_df$Likelihood <- dsampling(sample_b, pop_N,
bayes_df$B, sample_n)
```

- Last we turn the Bayesian crank using the
`bayesian_crank()`

function and obtain the posterior probabilities for \(B\).

`bayes_df <- bayesian_crank(bayes_df)`

I compare the prior and posterior probabilities for \(B\) graphically.

`prior_post_plot(bayes_df)`

Here is a 90 percent probability interval for \(B\):

```
library(dplyr)
discint(select(bayes_df, B, Posterior), 0.90)
```

```
## $prob
## [1] 0.9125933
##
## $set
## [1] 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
```