Bayes' Theorem: Finding the probability of choosing a trick coin given the result is a tail

Question

A trick coin with two tails is put in a jar with three normal coins. One coin is selected at random and tossed. If the result is a tail, what is the probability that the trick coin was selected?

My Attempt:

Let $A$ be the event of obtaining a tail and let $B$ be the event that the coin is fair.

So we have $P(A) = 0.5$ and $P(B) = 0.75$. Therefore, $P(\bar{A}) = 0.5$ and $P(\bar{B}) = 0.25$.

Now, I know that we need to use the Bayes' Theorem here to find $P(A|\bar{B})$. The problem is I am not sure how to find $P(\bar{B}|A)$ that is used in the equation. Can someone please explain to me how to find that.

Thanks

Answer 1

Let $T$ be the event that we had a tail, $F$ the event that the coin is fair and $K$ the event that the coin is the trick coin.

$$P(K|T)=\frac{P(K\cap T)}{P(T)}=\frac{P(K\cap T)}{P(T\cap F)+P(T\cap K)}=\frac{P(T|K)P(K)}{P(T|F)P(F)+P(T|K)P(K)}$$

where we have $P(T|K)=1, P(T|F)=\frac{1}{2}$ and $P(K)=\frac{1}{4}, P(F)=\frac{3}{4}$. Replacing those values, we get:

$$P(K|T)=\frac{1\times \frac{1}{4}}{\frac{1}{2}\times \frac{3}{4}+1\times \frac{1}{4}}=\frac{\frac{1}{4}}{\frac{5}{8}}=\frac{2}{5}$$

Answer 2

I know you have to use Bayes' Theorem but still it's easier to look at the tail set since we know we get a tail and consider the favorables cases for the trick coin $C=2$ the number of all tails $A=5$ so $P=\frac C A=\frac 2 5$.