3.88 - A First Course in Probability

Question

In Example 5e, what is the conditional probability that the ith coin was selected given that the first n trials all result in heads? 

Step-by-Step Solution

Answer

$\text{ The conditional probability that the }{i}^{\text{th }}\text{ coin is selected }=\frac{{\left(\frac{i}{k}\right)}^n}{{\sum }_{j=0}^k{\left(\frac{j}{k}\right)}^n}\text{. }$

Step 1: Set up the problem using Bayes' theorem

In Example 5e, there are coins with probability \(i/k\) of heads for \(i = 0, 1, 2, \ldots, k\). A coin is randomly selected (each with probability \(1/(k+1)\)) and flipped \(n\) times, all resulting in heads. We want \(P(\text{coin } i \mid n \text{ heads})\).

Step 2: Apply Bayes' theorem

By Bayes' theorem:
\(P(\text{coin } i \mid n \text{ heads}) = \frac{P(n \text{ heads} \mid \text{coin } i) \cdot P(\text{coin } i)}{\sum_{j=0}^{k} P(n \text{ heads} \mid \text{coin } j) \cdot P(\text{coin } j)}\)
Since each coin is equally likely: \(P(\text{coin } i) = \frac{1}{k+1}\).
\(P(n \text{ heads} \mid \text{coin } i) = \left(\frac{i}{k}\right)^n\).

Step 3: Simplify

\(P(\text{coin } i \mid n \text{ heads}) = \frac{(i/k)^n}{\sum_{j=0}^{k}(j/k)^n} = \frac{i^n}{\sum_{j=0}^{k} j^n}\)

Step 4: Final answer

The conditional probability that the \(i\)-th coin was selected given that all \(n\) trials resulted in heads is:
\(P(\text{coin } i \mid n \text{ heads}) = \frac{i^n}{\sum_{j=0}^{k} j^n} = \frac{i^n}{0^n + 1^n + 2^n + \cdots + k^n}\)