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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.