Let's analyze and solve the problem step by step.

Problem Breakdown

1. Coins Description:

   • First Coin: Double-headed (HH).
   • Second Coin: Double-tailed (TT).
   • Third Coin: Fair coin (HT).

2. Process:

   • Coins are mixed, and one is chosen randomly.
   • The chosen coin is placed on the ground, and it is observed that the upper side is head.

3. Question: What is the probability that the other side of the chosen coin is tail?

Solution

We solve using Bayes' Theorem and conditional probability:

Step 1: Probability of Choosing Each Coin

• There are three coins, so the probability of choosing any specific coin is $P(\text{Coin}) = \frac{1}{3}$.

Step 2: Probability of Observing a Head for Each Coin

• For the HH coin, both sides are heads, so $P(\text{Head}|\text{HH}) = 1$.
• For the TT coin, both sides are tails, so $P(\text{Head}|\text{TT}) = 0$.
• For the HT coin (fair coin), one side is head, so $P(\text{Head}|\text{HT}) = \frac{1}{2}$.

Step 3: Total Probability of Observing a Head

Using the law of total probability: $P(\text{Head}) = P(\text{Head}|\text{HH})P(\text{HH}) + P(\text{Head}|\text{TT})P(\text{TT}) + P(\text{Head}|\text{HT})P(\text{HT})$ Substitute values: $P(\text{Head}) = (1)\left(\frac{1}{3}\right) + (0)\left(\frac{1}{3}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{3}\right)$ $P(\text{Head}) = \frac{1}{3} + 0 + \frac{1}{6} = \frac{1}{2}$

Step 4: Probability the Coin is HT Given a Head is Observed

Using Bayes' Theorem: $P(\text{HT}|\text{Head}) = \frac{P(\text{Head}|\text{HT})P(\text{HT})}{P(\text{Head})}$ Substitute values: $P(\text{HT}|\text{Head}) = \frac{\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)}{\frac{1}{2}}$ $P(\text{HT}|\text{Head}) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$

Step 5: Probability of the Other Side Being Tail

If the coin is HT, and a head is observed, the other side is definitely tail. Thus: $P(\text{Other side is Tail}) = P(\text{HT}|\text{Head}) = \frac{1}{3}$

Final Answer:

The probability that the other side of the coin is tail is $\frac{1}{3}$.

Would you like a detailed breakdown of any part of this solution? Here are some follow-up questions to explore further:

1. How does Bayes' Theorem work in other scenarios involving conditional probability?
2. Can we extend this problem to a set of coins with more than three types?
3. How would the result change if the probabilities of selecting each coin were not equal?
4. What if the observation was "tail" instead of "head"?
5. Can you generalize this problem to dice or other random objects?

Tip: Always double-check conditional probabilities by summing over all possible outcomes to ensure consistency.