Suppose that we toss a fair coin. If a head $H$ comes up, we roll a fair dice; otherwise we choose a random number in the interval $(0,10)$. Let $X$ be the number observed. If the value observed was equal to $2$, give the probability that we got a $H$.
Here is my solution:
$F_X(x)= P[X\leq x] = P[H \mbox { and } X\leq x \mbox{ or } T \mbox{ and } X\leq x]$ =$$1/2 (P[X\leq x | H] + P[X\leq x |T]).$$
With this equation i obbtained the following distribution function $F_X(x)$:
$$0, x<0\\x/20, 0\leq x<1\\ 1/12 +x/20, 1\leq x <2\\2/12+x/20,2\leq x < 3\\ 3/12+x/20, 3\leq x <4 \\ 4/12 + x/20, 4\leq x <5 \\5/12 + x/20 , 5\leq x < 6\\ 6/12 + x/20, 6\leq x < 10\\ 1, 10\leq x$$ Then,
$P[H|X=2] = P[H].P[X=2|H]/P[X=2] = \frac{1/2\times1/6}{1/12} = 1$
Is this result correct? Because it is saying that the only way to obttain a 2 is by rolling the dice, which i find much counter-intuitive, since it is still possible to obtain $2$ by choosing it randombly.
Provided that the random number chosen when the result of the coin is tails is a continuous random variable (i.e., can take on any real value between $0$ and $10$), then the probability of that being $2$ is indeed $0$. That doesn't mean it doesn't happen; it means that in the probability space, it happens over a set of measure zero.
Heuristically, we might observe that the probability distribution between $0$ and $10$ is constant (uniform), so the probability that the random variable is $2$ is equal to the probability that it is any other value in that interval. Since there are an uncountably infinite number of such values, that probability must be zero.