Suppose you flipped heads an even number of times on the first 99 flips. Then there is a $\frac{1}{2}$ probability that you will get another heads, and thus and odd number of heads total. So in this case it's 50-50.
Suppose you flipped heads an odd number of times on the first 99 flips. Then there is a $\frac{1}{2}$ probability that you will get another heads, and thus and even number of heads total. So in this situation it's also 50-50.
So regardless of what happens for the first 99 flips, there's a $\frac{1}{2}$ change you end up with an odd number of heads and a $\frac{1}{2}$ chance you end up with an even number of heads.
When you flip the first coin there are two equally probable results: $\rm H$ or $\rm T$. The probability for each is $1/2$.
Now if that result was a head (half of the total probability), you must flip the coin a second time, and again there are two equally probable results branching from that point. This give the probabilities of two of the outcomes $\rm (H,H), (H,T)$ as each being half of the half: $1/4$.
Now if the first toss were a tail, you would toss a die. This time their would be six outcomes branching off that initial result, all equally likely from that point. This give the probabilities of these remaining six outcomes $\rm (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)$ as each being $1/12$.
This is nothing more than the definition of conditional probability.
$$\mathsf P((X,Y){=}(x,y)) ~=~ \mathsf P(X{=}x)~\mathsf P(Y{=}y\mid X{=}x)\\\mathsf P((X,Y){=}({\rm T},6)) ~=~ \mathsf P(X{=}{\rm T})~\mathsf P(Y{=}6\mid X{=}{\rm T})
\\=~\tfrac 1 2\times \tfrac 16$$
And such.
PS: the probability that there die shows greater than 4 given that there is at least one tail is obviously:
$$\mathsf P(Y\in\{5,6\}\mid X{=}{\rm T}\cup Y{=}{\rm T})=\dfrac{\mathsf P((X,Y)\in\{~({\rm T},5),({\rm T},6)~\}) }{\mathsf P((X,Y)\in\{~({\rm H},{\rm T}),({\rm T},1),({\rm T},2),({\rm T},3),({\rm T},4)({\rm T},5), ({\rm T},6)~\})}$$
Best Answer
The answer by drhabs is wrong, his answer misses the null set. 2^n comes from the fact that for each element in your set you have 2 choices, to either include or not include it in a subset. For the example Ω={H,T} you will have the following subsets: