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?
- 9/20
- 9/19
- 1/2
- 10/19
- 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. 🙂
Best Answer
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 $\frac9{19}$.