There are three coins. One is a two-headed coin, 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 was the two-headed coin?

Hint: In this question it is given that there are three coins and among them one is a two-headed coin, 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, we have to find what is the probability that it was the two-headed coin. So to find the solution we have to know that if $$E_{1}$$, $$E_{2}$$, $$E_{3}$$ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin and A be the event that the coin shows heads, then we have to find the conditional probability, $$P\left( E_{1}|A\right) $$.
Where, $$P\left( E_{1}|A\right) $$= P(coin is two-headed, given that it shows head)

Complete step-by-step solution:
Let us consider, $$E_{1}$$, $$E_{2}$$, $$E_{3}$$ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.
Also let A be the event that the coin shows heads.
So from the three events $$E_{1}$$, $$E_{2}$$, $$E_{3}$$ the probability of selecting one event is $$\dfrac{1}{3}$$,
i.e, $$P\left( E_{1}\right) =P\left( E_{2}\right) =P\left( E_{3}\right) =\dfrac{1}{3}$$
Now as we know that a two-headed coin always shows head.
$$\therefore P\left( A|E_{1}\right) =P\left( \text{coin showing head, given that it is a two-headed coin} \text{} \right) =1$$

Now for another coin, it is a biased coin that comes up heads 75%,
So we can write,
$$\therefore P\left( A|E_{2}\right) $$
$$=P\left( \text{coin showing head, given that it is a biased coin} \text{} \right) $$
$$=\dfrac{75}{100} =\dfrac{3}{4}$$

Since the third coin is unbiased, therefore it has one side head and another side is tail. So the probability of getting a head is $$\dfrac{1}{2}$$.
$$\therefore P\left( A|E_{3}\right) $$
$$=P\left( \text{coin showing head, given that the coin is unbiased} \text{} \right) $$
$$=\dfrac{1}{2} $$

So by baye’s theorem, we can write, the probability that the coin is two-headed, given that it shows heads, is
$$P\left( E_{1}|A\right) =\dfrac{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) }{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) +P\left( E_{2}\right) \cdot P\left( A|E_{2}\right) +P\left( E_{3}\right) \cdot P\left( A|E_{3}\right) }$$
$$=\dfrac{\dfrac{1}{3} \times 1}{\dfrac{1}{3} \times 1+\dfrac{1}{3} \times \dfrac{3}{4} +\dfrac{1}{3} \times \dfrac{1}{2} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{1}{3} +\dfrac{1}{4} +\dfrac{1}{6} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{4+3+2}{12} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{8}{12} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{3}{4} }$$
$$=\dfrac{1}{3} \times \dfrac{4}{3}$$
$$=\dfrac{4}{9}$$

Note: Bayes’ theorem is a way to figure out conditional probability. Conditional probability is the probability of an event happening, given that it has some relationship to one or more other events. So this theorem states that, if $$E_{1},\ E_{2},\ E_{3},\cdots ,E_{n}$$ be the n number of events and A be the another event (condition) then the conditional probability of any event $$E_{i}$$ given A is,
$$P\left( E_{i}|A\right) =\dfrac{P\left( E_{i}\right) \cdot P\left( A|E_{i}\right) }{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) +P\left( E_{2}\right) \cdot P\left( A|E_{2}\right) +\ldots +P\left( E_{i}\right) \cdot P\left( A|E_{i}\right) +\ldots +P\left( E_{n}\right) \cdot P\left( A|E_{n}\right) }$$.