We are asked to find the probability that the coin is fake, given that two heads were tossed.
This is a problem of conditional probability. We can use Bayes' Theorem to solve it. Bayes' Theorem is given by:
\[
P({Fake} \mid {Two heads}) = \frac{P({Two heads} \mid {Fake}) P({Fake})}{P({Two heads})}
\]
We will calculate each term in this formula:
1. Prior probability of choosing the fake coin:
Since there is 1 fake coin out of 5 coins, the probability of selecting the fake coin is:
\[
P({Fake}) = \frac{1}{5}
\]
2. Likelihood of getting two heads given the fake coin:
If the fake coin is chosen, the probability of getting heads on each toss is 1. Therefore, the probability of getting two heads is:
\[
P({Two heads} \mid {Fake}) = 1 \times 1 = 1
\]
3. Likelihood of getting two heads given a regular coin:
If a regular coin is chosen, the probability of getting heads on each toss is 0.5. Therefore, the probability of getting two heads is:
\[
P({Two heads} \mid {Regular}) = 0.5 \times 0.5 = 0.25
\]
4. Prior probability of choosing a regular coin:
There are 4 regular coins out of 5, so the probability of selecting a regular coin is:
\[
P({Regular}) = \frac{4}{5}
\]
5. Total probability of getting two heads:
The total probability of getting two heads is given by the law of total probability:
\[
P({Two heads}) = P({Two heads} \mid {Fake}) P({Fake}) + P({Two heads} \mid {Regular}) P({Regular})
\]
Substituting the values:
\[
P({Two heads}) = 1 \times \frac{1}{5} + 0.25 \times \frac{4}{5} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}
\]
6. Bayes' Theorem:
Now, applying Bayes' Theorem:
\[
P({Fake} \mid {Two heads}) = \frac{P({Two heads} \mid {Fake}) P({Fake})}{P({Two heads})} = \frac{1 \times \frac{1}{5}}{\frac{2}{5}} = \frac{1}{2}
\]
Thus, the probability that the coin you have chosen is the fake coin is \( \boxed{0.50} \).