A coin is flipped when dice hits 6 (conditional probability) Prof gave us homework on conditional probability that is due on the day of the lecture on conditional probability. Yeah, this has been a bad week and I've no idea what I'm doing.

Q: 3 dice are rolled, then, a coin is flipped as many times as the number 6 is obtained.
a) find the probability of getting less than 2 heads.
b) knowing this experiment results in less than 2 heads, what is the conditional probability that exactly 2 sixes were obtained?

I don't even know where to start...

Attempt:
a)

The only way I figure is:
Rolling one six, probability of head < 2 = $ \displaystyle \frac{3}{6^3} \cdot \frac{2}{2}$
Rolling two sixes: $\displaystyle \frac{3}{6^3} \cdot \frac{2}{2^2}$
Rolling three sixes: $\displaystyle \frac{1}{6^3} \cdot \frac{4}{2^3}$
Then adding those.

b)

$\displaystyle \frac{\frac{1}{6^3} \cdot \frac{4}{2^3}}{\frac{1}{6^3}}$
 A: a)   The computation takes a while. It may be useful to draw a tree in order not to lose track of the possibilities.  Initially, we toss $3$ dice. We get $0$, $1$, $2$, or $3$ $6$'s.  Then, depending on the outcome, we toss a certain number of coins.  So from the "start" position, there are $4$ branches, corresponding to the number of $6$'s obtained.   
If $3$ dice are rolled, then the probability of $k$ $6$'s is equal to
$$\binom{3}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{3-k}.$$
In particular, the probability of $0$ $6$'s is $\frac{125}{216}$, the probability of exactly $1$ $6$ is $\frac{75}{216}$, the probability of exactly $2$ $6$'s is $\frac{15}{216}$, and the probability of $3$ $6$'s is $\frac{1}{216}$.
If we have $0$ $6$'s, we toss $0$ coins, so for sure (probability $1$) we get fewer than $2$ heads.  If we get $1$ $6$, then we toss $1$ coin, and again for sure we get fewer than $2$ heads.  If we get $2$ $6$'s, we toss $2$ coins, and the probability of fewer than $2$ heads is easily computed to be $3/4$.  If we get $3$ $6$'s, we toss $3$ coins. By symmetry the probability of fewer than $2$ heads is then $1/2$.  It follows that the overall probability of fewer than $2$ heads is
$$\frac{125}{216}\cdot 1 + \frac{75}{216}\cdot 1 +\frac{15}{216}\cdot \frac{3}{4}+\frac{1}{216}\cdot\frac{1}{2}.$$
This may simplify to $\dfrac{847}{864}$.  But check the arithmetic!
b)  Let $E$ be the event that exactly $2$ sixes were obtained, and let $L$ be the event we got fewer than $2$ heads. We want $P(E|L)$.
There are various formulas that we could now use.  We go for the simplest one, though using it is a bit less mechanical than the full Bayes formula.  Note that
$$P(E|L)P(L)=P(E \cap L).$$
This formula comes straight from the the definition of conditional probability.
From the above formula, we see that we will be essentially finished once we find $P(E\cap L)$ and $P(L)$.  
The probability $P(L)$ that we get fewer than $2$ heads has been computed in part (a). So we only need to find $P(E \cap L)$.
In the solution of part (a), we already saw that
$$P(E)=\binom{3}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^1=\frac{15}{216}=\frac{5}{72}.$$
Given that we got exactly $2$ $6$'s, we got to flip the coin twice, and thus the probability of fewer than $2$ heads is $3/4$, as shown by @Yuval Filmus.  It follows that
$$P(E\cap L)=P(L|E)P(E)=\left(\frac{3}{4}\right)\left(\frac{5}{72}\right)=\frac{5}{96}.$$
Now put the pieces together. I think that $P(E|L)=\dfrac{45}{847}$.
Do check the arithmetic!  
A: Your thinking is correct, though you're not calculating the probabilities right. As an example, if you roll two sixes, then you're tossing two coins; the probability of rolling two heads is $1/2 \cdot 1/2 = 1/4$, so the probability of rolling less than two heads is $3/4$.
As for the second part, look up Bayes' rule.
