Answer: $\frac{2}{3}$

## Explanation This is a classic Bayes' theorem problem. Let's break it down step by step: ### Given: - Coin 1: Double-headed (always lands on heads) - Coin 2: Normal unbiased coin (50% chance of heads) - Both coins are equally likely to be selected: P(coin 1) = P(coin 2) = 1/2 - We observe a head after the toss ### Applying Bayes' Theorem: We want to find P(coin 1 | head) $$\text{P(coin 1 | head)} = \frac{\text{P(coin 1)} \times \text{P(head | coin 1)}}{\text{P(coin 1)} \times \text{P(head | coin 1)} + \text{P(coin 2)} \times \text{P(head | coin 2)}}$$ ### Calculating probabilities: - P(coin 1) = 1/2 - P(head | coin 1) = 1 (since coin 1 is double-headed) - P(coin 2) = 1/2 - P(head | coin 2) = 1/2 (since coin 2 is unbiased) ### Substituting values: $$\text{P(coin 1 | head)} = \frac{(\frac{1}{2} \times 1)}{(\frac{1}{2} \times 1) + (\frac{1}{2} \times \frac{1}{2})} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$$ ### Intuitive understanding: When we see a head, it could come from either coin. However, coin 1 always produces heads, while coin 2 only produces heads half the time. Therefore, when we observe a head, it's more likely to have come from the double-headed coin than from the fair coin. The probability $\frac{2}{3}$ means that given we observed a head, there's a 2/3 chance it came from the double-headed coin and a 1/3 chance it came from the fair coin.

Author: Tanishq Prabhu