Flip 3 biased coins One about flipping coins. Coin A has 0.6 probability of being heads, Coin B has a 0.3 probability of being heads and Coin C has a 0.1 probability of heads. Given that a coin is flipped and lands on heads, what is the probability that when it is flipped again it will also be heads.
I'm not sure how I can apply Bayes' rule here.
I did a Monte Carole simulation the answer come out to be around 0.46, I am not sure how to give a analytical answer. 
 A: As a way to get intuition for such problems:  suppose we toss each coin $10$ times.  Then we expect to get $10$ Heads... $6$ Heads from $A$, $3$ from $B$, and $1$ from $C$.  Accordingly if our prior was that our mystery coin could be any of the three equally, then seeing Heads now makes us rethink.  Now Bayes' tells us that there is a $.6$ probability the coin was $A$, $.3$ that it was $B$, and $.1$ that it was $C$.  
Note:  the fact that these numbers coincide with the weightings is a consequence of the fact that the weightings add to $1$, it is not generally true.
Thus the probability that tossing the coin again yields another $H$ is:  $$.6^2+.3^2+.1^2=.46$$
Which, unsurprisingly, lines up with the result of your Monte Carlo simulation.
A: Making use of tree diagram,
\begin{align*}
  P(H|H) &= \frac{P(HH|A)+P(HH|B)+P(HH|C)}{P(H|A)+P(H|B)+P(H|C)} \\
         &= \frac{\frac{1}{3} \times 0.6^{2}+
                  \frac{1}{3} \times 0.3^{2}+
                  \frac{1}{3} \times 0.1^{2}}
                 {\frac{1}{3} \times 0.6+
                  \frac{1}{3} \times 0.3+
                  \frac{1}{3} \times 0.1} \\
         &= 0.46
\end{align*}

