You roll a fair six-sided die. If the die shows an odd number then you toss a fair coin that number of times. If the die shows an even number you do not toss the coin at all. What is the probability that you see at least two Tails?
I kind of know where to start - you find the probability of getting 0 and 1 tails and take that away from 1. Don't know what to do next
$T: \text{tossing tails}$
$$\begin{aligned} P({T\ge2}) &= P(\texttt{roll 3}\texttt{ AND } T\in\{2,3\})+ P(\texttt{roll 5} \texttt{ AND }T\in\{2,3,4,5\}) \\ &=P(\texttt{roll 3})\cdot P(T\in\{2,3\}|\texttt{Flip coin thrice})+ P(\texttt{roll 5})\cdot P(T\in\{2,3,4,5\}|\texttt{Flip coin five times})\\ &=\frac{1}{6}\cdot P(T\in\{2,3\}|\texttt{Flip coin 3 times})+ \frac{1}{6}\cdot P(T\in\{2,3,4,5\}|\texttt{Flip coin 5 times})\\ &=\frac{1}{6}\cdot \frac{a}{8} + \frac{1}{6}\cdot \frac{b}{32} \end{aligned}$$
We'll leave $a$ and $b$ for you to calculate.
UPDATE:
It's been a while.
$$P(T\in\{2,3\}|\texttt{Flip coin 3 times})=P(TTH)+P(HTT)+P(THT)+P(TTT)=\frac{4}{8}=\frac{1}{2}.$$
$$P(T\in\{2,3,4,5\}|\texttt{Flip coin 5 times})=1-P(T\in\{0,1\}|\texttt{Flip coin 5 times})=1-P(HHHHH)-5P(THHHH)=1-\frac{6}{32}=\frac{13}{16}.$$
$$P(T \ge 2)=\frac{1}{6}\left(\frac{1}{2}+\frac{13}{16}\right)=\frac{7}{32}.$$