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Suppose a box contains 20 balls: each ball has a distinct number in {1,…,20} written on it. We pick 10 balls (without replacement) uniformly at random and throw them out of the box. Then we check if the ball with number "1" on it is present in the box. If it is present, then we throw it out of the box; else we pick a ball from the box uniformly at random and throw it out of the box.

What is the probability that the ball with number "2" on it is present in the box?

  1. 9/20
  2. 9/19
  3. 1/2
  4. 10/19
  5. None of the above

My attempt: We have two cases in the event here.

Case1) If in the first 10 draws if we have thrown out 1 then we go for randomly picking again (this will be the 11th time) from the remaining 10 balls and throw out the randomly selected ball.

For this the number of ways = $\binom{19}{9}$ $\times$ $\binom{10}{1}$

Case2) If in the first 10 draws if we did not get 1 then the remaining 10 balls in the box will definitely have a '1', so we need not go for random picking again. We just have to pick the 1 left in the box and throw it out.

For this the number of ways = $\binom{19}{10}$ $\times$ $\binom{1}{1}$

Now, we will calculate the number of ways of randomly choosing so that 2 remains in the box.

Even for this we have 2 cases.

Case1) If both '1' and '2' are not thrown out after the first 10 random pickings. In this case we need not go for 11th random picking.

For this the number of ways = $\binom{18}{10}$ $\times$ $\binom{1}{1}$

Case2) If '1' is thrown out and '2' is not thrown out in the first 10 random pickings. This case requires 11th random picking, even here we will not be choosing '2' to be thrown out.

For this the number of ways = $\binom{18}{9}$ $\times$ $\binom{9}{1}$

So, finally the required probability = [$\binom{18}{10}$ $\times$ $\binom{1}{1}$ + $\binom{18}{9}$ $\times$ $\binom{9}{1}$] $\div$ [$\binom{19}{10}$ $\times$ $\binom{1}{1}$ + $\binom{19}{9}$ $\times$ $\binom{10}{1}$]

= (9/19) .....(I have skipped numerical calculations here :P)

Please let me know if anything is incorrect. :)

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  • $\begingroup$ @saulspatz, I guess I did some numerical mistake. The calculated value is also 9/19 and not 1/11. I have added this now. :) $\endgroup$
    – rohith
    Nov 24, 2019 at 16:15

1 Answer 1

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Your approach looks okay and the outcome is somehow an indication that you did not make mistakes. This because the outcome is correct as is shown in this answer.


The probability that ball $1$ is present in the box is $0$ and the other $19$ balls have equal probability to be present in the box.

There are $9$ balls present in the box so you could say that from these $19$ balls there are $9$ selected.

Then the probability for each ball (also the one with number $2$) to be selected is $\frac9{19}$.

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  • $\begingroup$ So, we have 19 balls that have equal probability (call it p = 1/19). So, probability that we pick 9 of them and '2' occurs in it = C(18,8) / C(19,9) = 9/19. Is this understanding correct? But the equal probability number p = 1/19 is not used while calculating probability here, how to use it? $\endgroup$
    – rohith
    Nov 25, 2019 at 1:00
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    $\begingroup$ Yes, your understanding is correct. Another way is picking $9$ balls one by one. Then the probability that ball 2 is picked at the $i$-th pick equals $\frac1{19}$. There are $9$ picks and the events are mutually exclusive so we get probability $\frac1{19}+\cdots+\frac1{19}=\frac9{19}$. Often students find it hard to believe that all picks give the same probability for ball 2 to be picked ("...but after picking one ball the situation has changed....", is what they bring in as objection) but I assure you that it is true and hope that you have the insight already. $\endgroup$
    – drhab
    Nov 25, 2019 at 8:49
  • $\begingroup$ Actually I have the same doubt which you mentioned "...but after picking one ball the situation has changed...." :P In case of dice problems, the probability of a value to occur is 1/6 in each throw. But how here probability is 1/19 in each pick? $\endgroup$
    – rohith
    Nov 25, 2019 at 12:59
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    $\begingroup$ For instance take the second pick. If the in the first already 2 was picked (probability on that is $\frac1{19}$) then the probability that it will be picked at second pick is $0$. If the first is not picked (probability on that is $\frac{18}{19}$) then the probability that it will be picked as second pick is $\frac1{18}$). So the probability that ball 2 is picked at second pick equals: $$\frac1{19}\times0+\frac{18}{19}\times\frac1{18}=\frac1{19}$$ $\endgroup$
    – drhab
    Nov 25, 2019 at 14:11
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    $\begingroup$ Let me add that it is not smart to look at it the way I sketched in my former comment. Unnecessarily complicated with all these conditional probabilities. Just think of the selection as taking $9$ balls out at the same time and then placing these balls randomly on spots with numbers $1,2,\dots,9$ where you are allowed to identify the ball placed on e.g. spot $3$ with the ball picked at the third picking. Do you agree that all spots have the same probability to become the receiver of ball 2?... $\endgroup$
    – drhab
    Nov 25, 2019 at 14:12

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