Suppose that we have $2$ coins, a fair one (probability of $H$ or $T$ = $0.5$), lets call this $C_1$ and a biased one (probability of $H = 0.6,$ $T = 0.4$), lets call this $C_2$.
Then if we pick one of the coins at random and look at the result of a coin toss for which we get a $H$ result, why can we say that the probability of the coin being $C_1 = 0.4$ and being $C_2 = 0.6$ respectively.
Best Answer
Use Bayes' Theorem
$$\begin{align*} P(\text{Coin 1 | } H) &=\frac{P(\text{Coin 1} \cap H)}{P(H)}\\\\ &=\frac{0.5\cdot0.5}{(0.5\cdot0.5)+(0.5\cdot0.6)}\\\\ &\approx0.455 \end{align*}$$
and so $$P(\text{Coin 2 | } H)=1-P(\text{Coin 1 | } H)\approx0.545$$
It does not appear that the probability that it was coin one given that you obtained heads it $0.4$.