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

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    $\begingroup$ number 2 before 8-th draw... Then the 8th draw is irrelevant? Don't you mean (again) "on or before"? $\endgroup$ – drhab Sep 3 '17 at 9:07
  • $\begingroup$ @user152294 Just to check my try, have you the result of this exercise? $\endgroup$ – Robert Z Sep 3 '17 at 9:46
  • $\begingroup$ @drhab Yes, on or before $\endgroup$ – user152294 Sep 3 '17 at 17:29
  • $\begingroup$ @RobertZ No, I don't have the solution unfortunately $\endgroup$ – user152294 Sep 3 '17 at 17:30
  • $\begingroup$ @user152294 "before 8-th draw" means "on or before 8-th draw"? In case I have to modify my solution. P.S. Where does his exercise come from? $\endgroup$ – Robert Z Sep 3 '17 at 17:33

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 $$p=1-(p_1+p_2-p_3)=211/1566\approx 0.134738.$$

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  • $\begingroup$ Is it also possible to just do these two cases? 1) Draw a '1' and '2' before the 5th draw, or 2) Draw a '1' before the 5th draw AND draw a '2' before the 8th draw? I'm not sure how to express 2) but would it be more cumbersome than the complementary events? $\endgroup$ – user152294 Sep 3 '17 at 17:34
  • $\begingroup$ @user152294 Yes you can consider two cases, but I think that with 3 cases is simpler. $\endgroup$ – Robert Z Sep 3 '17 at 17:36
  • $\begingroup$ How come in this solution, we don't have to use binomial coefficients? $\endgroup$ – user152294 Sep 3 '17 at 17:43
  • $\begingroup$ @user152294 Let me know if the official solution is the one that I have found. $\endgroup$ – Robert Z Sep 3 '17 at 17:44
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    $\begingroup$ @user152294 $3$ is the number of ways to place the red $1$ (position 6,7, 8), $3$ is the number of ways to place the blue $1$ (position 6,7, 8) and $3\cdot 2$ is the number of ways to place the red $1$ and blue $1$ (positions (6,7), (6,8), (7,8), (7,6), (8,6), (8,7)). $28^{\underline{8}}$, $28^{\underline{7}}$ and $28^{\underline{6}}$ are the number of ways to fill the remaining positions (cards different from 1). $\endgroup$ – Robert Z Sep 3 '17 at 18:21

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