There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it is a two headed coin.

Explanation

To solve the problem, we will use Bayes' theorem. Let's denote the events as follows:

- Let \( C_1 \) be the event that we picked the two-headed coin.
- Let \( C_2 \) be the event that we picked the biased coin (75% heads).
- Let \( C_3 \) be the event that we picked the unbiased coin (50% heads).
- Let \( H \) be the event that we get heads when we toss the chosen coin.

We need to find the probability that the coin chosen was the two-headed coin given that we observed heads, which can be expressed as \( P(C_1 | H) \).

### Step 1: Calculate the prior probabilities
Since one of the three coins is chosen at random, the prior probabilities for each coin are:
\[ P(C_1) = P(C_2) = P(C_3) = \frac{1}{3} \]

### Step 2: Calculate the likelihoods
Next, we calculate the probability of getting heads given each coin:
- For the two-headed coin \( C_1 \):
\[ P(H | C_1) = 1 \]
- For the biased coin \( C_2 \):
\[ P(H | C_2) = 0.75 = \frac{3}{4} \]
- For the unbiased coin \( C_3 \):
\[ P(H | C_3) = 0.5 = \frac{1}{2} \]

### Step 3: Calculate the total probability of getting heads
Using the law of total probability, we can find \( P(H) \):
\[ P(H) = P(H | C_1) P(C_1) + P(H | C_2) P(C_2) + P(H | C_3) P(C_3) \]
Substituting the values we have:
\[ P(H) = 1 \cdot \frac{1}{3} + \frac{3}{4} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} \]
Calculating each term:
\[ P(H) = \frac{1}{3} + \frac{3}{12} + \frac{2}{12} \]
\[ P(H) = \frac{1}{3} + \frac{5}{12} \]
To add these fractions, we convert \( \frac{1}{3} \) to have a common denominator of 12:
\[ \frac{1}{3} = \frac{4}{12} \]
Thus,
\[ P(H) = \frac{4}{12} + \frac{5}{12} = \frac{9}{12} = \frac{3}{4} \]

### Step 4: Apply Bayes' theorem
Now we can use Bayes' theorem to find \( P(C_1 | H) \):
\[ P(C_1 | H) = \frac{P(H | C_1) P(C_1)}{P(H)} \]
Substituting the values:
\[ P(C_1 | H) = \frac{1 \cdot \frac{1}{3}}{\frac{3}{4}} = \frac{\frac{1}{3}}{\frac{3}{4}} = \frac{1}{3} \cdot \frac{4}{3} = \frac{4}{9} \]

### Final Answer
Thus, the probability that the coin chosen is the two-headed coin given that we observed heads is:
\[ \boxed{\frac{4}{9}} \]