I am confused about a conditional probability question from One Thousand Exercises in Probability by Geoffery Grimmett & David Stirzaker:
A man possesses five coins, two of which are double-headed, one is double-tailed, and two are
normal. He shuts his eyes, picks a coin at random, and tosses it. What is the probability that the lower face of the coin is a head?He opens his eyes and sees that the coin is showing heads; what is the probability that the lower face is a head?
He shuts his eyes again, and tosses the coin again. What is the probability that the lower face is a head?
I'm ok with the first two questions. For the third question, I think he asks for the probability that the lower face is a head on the second toss given the upper face is a head on the first toss. Let $D_H$ be the event that the coin is double-headed, $D_T$ the event that the coin is double-tailed, $N$ the event the coin is normal, $H_i$ the event the lower face is head on the $i$th toss, and finally $U_i$ the event the upper face is head on the $i$th toss.
Now we need to find $\mathbb{P}(H_2 \mid U_1)$. However, what we know from the second question is $$\mathbb{P}(D_H \mid U_1)= \frac{2}{3}, \quad \mathbb{P}(D_T \mid U_1) =0, \quad \mathbb{P}(N\mid U_1)=\frac{1}{3},$$
and I failed to deduce what $\mathbb{P}(H_2 \mid U_1) = \frac{\mathbb{P}(H_2 \; \cap\; U_1)}{\mathbb{P}(U_1)}$ is from what we know.
Best Answer
There are three types of coins $HH$, $HT$, and $TT$, with probabilities of $2/5$, $2/5$ and $1/5$ respectively. Let $L$, $U$ be the events that the lower/upper face is a head.
$$P(L) = \frac{2}{5}\times1+\frac{2}{5}\times\frac{1}{2}+\frac{1}{5}\times0=\frac{3}{5}$$ (Or, simpler, there are six heads out of $10$ faces.)
$$P(U)=P(L)=\frac{3}{5}$$ $$\therefore P(L|U)=\frac{P(L\cap U)}{P(U)}=\frac{2/5}{3/5}=\frac{2}{3}$$
$$P(HH|U_1)=\frac{P(HH\cap U_1)}{P(U_1)}=\frac{2/5}{3/5}=\frac{2}{3}$$ $$\therefore P(HT|U_1)=\frac{1}{3}$$ \begin{align}\therefore P(L_2|U_1)&=P(L_2|HH)P(HH|U_1)+P(L_2|HT)P(HT|U_1)+P(L_2|TT)P(TT|U_1)\\ &=\frac{2}{3}\times1+\frac{1}{3}\times\frac{1}{2}+0=\frac{5}{6}\end{align}