Two Bayes Problems (1)

My friends asked me two Bayes questions a while ago. Due to the rapid spread of COVID-19 in US, I’ve been at home for five days. So I decide to put my solutions here for references. The first question is as follows.

There are five balls in a bag. Each ball is colored either black or white with equal chance. Now suppose that someone have already drawn the ball 4 times with replacements, among which he/she got 1 white ball and 3 black balls as a result. Find the probability distribution of the number of white balls in that bag.

This problem is a straightforward application of Bayes’ Rule. The only tricky part is to formulate the problem statement with math language.

Let $X$ be the number of white balls and let event $A = [1 white ball and 3 black balls among 4 trials]$ . The prior distribution of $X$ is simply the binomial distribution $B (5, \frac{1}{2})$ . Actually, we are interested in finding the posterior probability $P (X = x | A)$ . Note that

P (A | X = x) = {\begin{cases} b (1; 5, \frac{x}{5}), & x = 1, 2, 3 or 4 \\ 0, & x = 0 or 5 \end{cases},

where $b (x; n, p)$ is the pmf at $x$ for a $B (n, p)$ r.v.

We use Bayes’s Rule as follows:

\begin{aligned} P (X = x | A) & = \frac{P (X = x, A)}{P (A)} = \frac{P (A | X = x) P (X = x)}{\sum_{i} P (A | X = i) P (X = i)} \\ = \frac{(\binom{4}{3}) \cdot {(\frac{x}{5})}^{1} \cdot {(\frac{5 - x}{5})}^{3} \cdot (\binom{5}{x}) \cdot {(\frac{1}{2})}^{5}}{\sum_{i = 1}^{4} (\binom{4}{3}) \cdot {(\frac{i}{5})}^{1} \cdot {(\frac{5 - i}{5})}^{3} \cdot (\binom{5}{i}) \cdot {(\frac{1}{2})}^{5}}, x = 1, . . ., 4. \end{aligned}

I used R to compute the above quantity:

> x <- rep(0,4)
> for (i in 1:4){
+     x[i] <- 4 * (i/5) * ((5-i)/5)^3 * choose(5, i) / 2^5
+ }
> y <- x/sum(x)
> y
[1] 0.28571429 0.48214286 0.21428571 0.01785714

Thus, we have

P (X = x | A) = {\begin{cases} 0.28571429, & x = 1 \\ 0.48214286, & x = 2 \\ 0.21428571, & x = 3 \\ 0.01785714, & x = 4 \\ 0, & x = 0 or 5 \end{cases} .

When we are asked to give a point estimate of the number of white balls, we typically use the posterior mode, which is also known as the MAP (maximum a posteriori) estimation. In this case, it is $2$ .

Two Bayes Problems (1) Ibuki