There are three coins. One is two -heated coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was two-headed coin?

Answer

To solve the problem, we will use Bayes' theorem to find the probability that the coin chosen was the two-headed coin given that the result of the toss was heads.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( E_1 \): Event that the chosen coin is the two-headed coin.
- Let \( E_2 \): Event that the chosen coin is the biased coin with a 75% chance of heads.
- Let \( E_3 \): Event that the chosen coin is the biased coin with a 60% chance of heads (40% chance of tails).
- Let \( A \): Event that the result of the toss is heads.

2. **Determine Prior Probabilities**:
Since one of the three coins is chosen at random:
\[ P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \]

3. **Determine Conditional Probabilities**:
- For the two-headed coin:
\[ P(A | E_1) = 1 \quad (\text{since it always shows heads}) \]
- For the biased coin (75% heads):
\[ P(A | E_2) = 0.75 \]
- For the biased coin (60% heads):
\[ P(A | E_3) = 0.60 \]

4. **Calculate Total Probability of A**:
Using the law of total probability:
\[ P(A) = P(A | E_1)P(E_1) + P(A | E_2)P(E_2) + P(A | E_3)P(E_3) \]
Substituting the values:
\[ P(A) = (1 \cdot \frac{1}{3}) + (0.75 \cdot \frac{1}{3}) + (0.60 \cdot \frac{1}{3}) \]
\[ P(A) = \frac{1}{3} + \frac{0.75}{3} + \frac{0.60}{3} = \frac{1 + 0.75 + 0.60}{3} = \frac{2.35}{3} = \frac{47}{60} \]

5. **Apply Bayes' Theorem**:
We want to find \( P(E_1 | A) \):
\[ P(E_1 | A) = \frac{P(A | E_1) P(E_1)}{P(A)} \]
Substituting the known values:
\[ P(E_1 | A) = \frac{1 \cdot \frac{1}{3}}{\frac{47}{60}} = \frac{\frac{1}{3}}{\frac{47}{60}} = \frac{60}{141} = \frac{20}{47} \]

### Final Answer:
The probability that the chosen coin was the two-headed coin given that it showed heads is:
\[ \boxed{\frac{20}{47}} \]