There is a deck of $30$ cards, each card labeled a number from $1$ to $15$, with exactly $2$ copies of a card for each number. You draw $8$ cards. What is the probability that you draw the number '$1$' card by the $5$th draw (on the $5$th draw or before that), AND also drawing the number '$2$' card on or before the $8$th draw?
I know how to compute the probability of drawing both the cards on or before the $5$th draw:
$$\frac{\binom{2}{1}\cdot \binom{2}{1} \cdot \binom{26}{3}}{\binom{30}{5}}$$
Since there's $2$ ways to choose from each of the '$1$' and '$2$' cards, and then there's $26$ cards left after those $4$ cards so the other $3$ cards can be any of those $26$, and the total number of combinations you can draw $5$ cards from $30$.
But we want to expand this search to $8$ draws, and also at the same time want to have assumed that we have already drawn the '$1$' card on or before the $5$th draw (if we don't get the '$2$' card by the $5$th draw. How can I combine these ideas? Thanks
I think it is more convenient to evaluate the probability of the complementary event: draw NO number '1' card by the $5$th draw, OR draw NO number '2' card. Here we consider the 2 copies of a card with the same number distinguishable (for example assume that one is red and the other is blue). Let $n^{\underline{k}}:=n(n-1)\cdots(n-k+1)$.
1) If we draw NO number '1' card by the 5th draw then we can have zero, one or two '1's in the $6$th, $7$th or $8$th draw $$p_1=\frac{1}{30^{\underline{8}}}\left(28^{\underline{8}}+(3+3)\cdot 28^{\underline{7}}+3\cdot 2\cdot 28^{\underline{6}}\right)$$
2) If we draw NO number '2' card then $$p_2=\frac{28^{\underline{8}}}{30^{\underline{8}}}$$
3) If we draw NO number '1' card by the $5$th draw AND NO number '2' card then, similarly to case 1), $$p_3=\frac{1}{30^{\underline{8}}}\left(26^{\underline{8}}+(3+3)\cdot 26^{\underline{7}}+3\cdot 2\cdot 26^{\underline{6}}\right)$$
Hence, the desired probability is $$p=1-(p_1+p_2-p_3)=211/1566\approx 0.134738.$$