So I have a very simple probability question here.
An Urn contains five coins of which three are fair, one is two-headed, and one is two-tailed. A coin is drawn at random and tossed twice.
a) What is the probability that 'head' appears both times?
b) If 'head' appears both times, what is the probability that the coin is two-headed?
c) Suppose that 'head' appears both times. Draw another coin from the urn (Which has four coins remaining). What is the probability that the second coin drawn from the urn is a fair one?
I already got parts a and b. It's part c I am having issues with. My idea was to use: \begin{align} \text{P(2nd fair) }&\text{= P(1st coin fair)*P(2nd coin fair)+ P(Two headed coin)*P(2nd coin fair)}\\ &= \frac{3}{5}*\frac{2}{4}+\frac{1}{5}*\frac{3}{4}\\ &=0.45 \end{align} I ignored tails since I Want two heads and that can only happen with a fair coin and the two headed coin. Is my reasoning right? I just need to clarify this. Thanks!
Denote the type of the first coin drawn by a symbol $s\in \{F,T,H\}$, where $F$ stands for fair and the other two for tails and heads, respectively. We can compute the conditional probability that the second coin is fair given the type $s$ as follows: $$ \mathbb P(\text{second is fair}\mid s=F)=\frac{2}{4},\quad \mathbb P(\text{second is fair}\mid s=H)=\mathbb P(\text{second is fair}\mid s=T)=\frac{3}{4}. $$ Now we can use Bayes' Theorem to finish. The probabilities of seeing two heads for each of the three coin types $F,T,H$ are $$ \frac{1}{4},\quad 0,\quad 1, $$ respectively. Combining these allows us to find the distribution of the type, $s$: $$ \mathbb P(s=F)=\frac{1}{5},\quad \mathbb P(s=T)=0,\quad \mathbb P(s=H)=\frac{4}{5}. $$ Consequently, $$ \mathbb P(\text{second is fair})=\frac{1}{5}\cdot \frac{2}{4}+0\cdot \frac{3}{4}+\frac{4}{5}\cdot \frac{3}{4}=\frac{7}{10}. $$