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Let's break this down step by step to find the probability that the result of a coin flip is "Heads."

### Problem Summary:
- We have an infinite collection of biased coins indexed by the positive integers \(i = 1, 2, 3, \dots \).
- The probability of selecting coin \(i\) is \(2^{-i}\).
- If coin \(i\) is selected, the probability of flipping heads is \(3^{-i}\).
- We need to find the total probability that the result of the flip is "Heads."

### Total Probability:
The total probability of getting "Heads" can be found using the **Law of Total Probability**, which sums up the probability of getting heads from each individual coin \(i\), weighted by the probability of selecting that coin.

The probability that the result is "Heads" is:

\[
P(\text{Heads}) = \sum_{i=1}^{\infty} \left( P(\text{select coin } i) \times P(\text{Heads from coin } i) \right)
\]

Given:
- \(P(\text{select coin } i) = 2^{-i}\)
- \(P(\text{Heads from coin } i) = 3^{-i}\)

Thus:

\[
P(\text{Heads}) = \sum_{i=1}^{\infty} 2^{-i} \cdot 3^{-i} = \sum_{i=1}^{\infty} (2 \cdot 3)^{-i} = \sum_{i=1}^{\infty} 6^{-i}
\]

### Use the Geometric Series Formula:
We are given the geometric sum formula:

\[
\sum_{i=1}^{\infty} \alpha^i = \frac{\alpha}{1 - \alpha}, \quad \text{where} \ |\alpha| < 1
\]

In our case, \(\alpha = \frac{1}{6}\), so:

\[
P(\text{Heads}) = \frac{\frac{1}{6}}{1 - \frac{1}{6}} = \frac{\frac{1}{6}}{\frac{5}{6}} = \frac{1}{5}
\]

Thus, the probability that the result of the coin flip is **Heads** is:

\[
P(\text{Heads}) = \frac{1}{5}
\]

Would you like more details on any part of the solution or have any other questions?

### Related Questions:
1. How would the probability change if the selection probability for coin \(i\) was \(2^{-(i+1)}\)?
2. What if the probability of flipping heads for coin \(i\) was instead \(4^{-i}\)?
3. Can the geometric sum formula be applied to any similar problem, and if so, how?
4. How would the result change if the number of coins was finite (e.g., only the first 10 coins)?
5. What is the significance of the condition \(|\alpha| < 1\) in the geometric series formula?

### Tip:
When solving problems with infinite sums, always look for patterns (like geometric series) to simplify the computation.

We have an infinite collection of biased coins, indexed by the positive integers. Coin i has probability 2^-i of being selected. A flip of coin i results in Heads with probability 3^-i. We select a coin and flip it. What is the probability that the result is Heads?

Learn how to calculate the probability of getting heads from an infinite collection of biased coins, using the law of total probability and the geometric series sum formula. This problem is perfect for advanced students studying probability theory and geometric sums.

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