Dick throws a die once. If the upper face shows $j$, he then throws it a further $j − 1$ times and adds all $j$ scores shown. If this sum is $3$, what is the probability that he only threw the die
(a) Once altogether?
(b) Twice altogether?
MY ATTEMPT
Let us denote by $F_{i}$ the event "the result from the first throw is given by $i$". Moreover, let us also set that $S_{k}$ represents event "the resulting sum equals $k$". Hence we are interested in the event $\textbf{P}(S_{3}|F_{1})$. But I am unable to proceed from here. Any help is appreciated. Thanks in advance.
In the first case, it's not possible as for the die to be only thrown altogether once, $j$ would necessarily have to be $1$. So the sum $3$ with one throw is not possible.
In the second case, we start with $j=2$. This means that we throw the dice once more $(j-1=2-1=1)$. Now for the sum to be $3$, this number has to be $3-2=1$. Probability of this event is
$$P_1=P(2)\cdot P(1) =\frac1{36}$$
Moving further, we take $j=3$. Here we see that we have to throw the die $3-1=2$ more times and since the die cannot get a $0$, the sum $3$ is not possible. Same reasoning can be applied to all $j\ge3$.
EDIT- Since we're supposed to find out the probability the number of times the die was thrown given that the sum is $3$ Thus the answer for $(b)$ is (where $E_2$ represents the event that the die was thrown twice)
$$P=P(E_2|S_3)=\frac{P(E_2\cap S_3)}{P(S_3)}=1$$
as there is only one case in which sum is $3$ ,i.e. $(2,1)$