first step analysis-tossing two coins example

Question

1. I am having trouble to understand this easy solution, why does only $A,B,C,D$ could partition the sample space?

2. How come the $\mathbb{P}(B)=\mathbb{P}(C)=\frac{33}{36}$? Isn't that suppose to be $\frac{4}{36}$ and $\frac{2}{36}$,repectively.

Thank you for any help and comments.

Answer 1

There are typographical errors

If $B$ was "$10$ or less on first roll" then clearly $B,C,D$ partition the sample space, since they cover all possibilities without intersecting. You would have $P(B)=\frac{33}{36}$ and $P(C)=\frac{2}{36}$ and $P(D)=\frac{1}{36}$, which add up to $1$.

You would then have $$P(A)=\frac{33}{36}P(A)+\frac{2}{36}\times 0 +\frac{1}{36}\times 1$$ and get $P(A)=\frac13$.