I've created this "tree" for the question:
A single game round is defined as follows: player throws a stone 2 times on a board with numbers from 1-8 (assume that a stone always falls inside the board) If the sum of the results of the parts on which the stone fell, in both throws is:
At most 5: The player proceed to the area T $[I1]$.
At least 12: Go to W $[I2]$.
Else: Go to G $[I3]$.
What's going inside T,G,W doesn't matter (I calculated all already, look in the picture).
$A,B,C$ meaning player won a "star sticker", $notA,notB,notC$ meaning player won a "cup".
$P(winning-star sticker) = 0.485$
The question I don't understand how to solve is: Knowing that the player played $10$ rounds and won $4$ "star stickers", calculate the probability that he got at most $2$ "star stickers" from the rounds 3,4,5,6.
*If anyone is interested, I can post full question.
Let $E$ be the event that at most two star stickers are won in the rounds $3, 4, 5, 6$. Let $F$ be the event that exactly four star stickers are won in the ten rounds. We wish to find $$\Pr(E \mid F) = \frac{\Pr(E \cap F)}{\Pr(F)}$$
Since the probability of winning in each round is $0.485$, the probability of winning exactly four times in ten rounds is $$\Pr(F) = \binom{10}{4}(0.485)^{4}(1 - 0.485)^6$$ where we have used the binomial distribution with $n = 10$ trials, $p = 0.485$, and $k = 4$ successes.
For the favorable cases, notice that if exactly $j$ star stickers are won in rounds $3, 4, 5, 6$, then exactly $4 - j$ star stickers must be won in the remaining six rounds. The probability of winning exactly $j$ star stickers in rounds $3, 4, 5, 6$ is $$\binom{4}{j}(0.485)^j(1 - 0.485)^{4 - j}$$ The probability of winning exactly $4 - j$ star stickers in rounds $1, 2, 7, 8, 9, 10$ is $$\binom{6}{4 - j}(0.485)^{4 - j}(1 - 0.485)^{6 - (4 - j)}$$ Hence, \begin{align*} \Pr(E \cap F) & = \sum_{j = 0}^{2} \binom{4}{j}(0.485)^j(1 - 0.485)^{4 - j}\binom{6}{4 - j}(0.485)^{4 - j}(1 - 0.485)^{6 - (4 - j)}\\ & = \sum_{j = 0}^{2} \binom{4}{j}\binom{6}{4 - j}(0.485)^4(1 - 0.485)^{6} \end{align*}