conditional probability when throwing balls multiple times from a bag

Question

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?

I am unable to assign probability to nodes.please help!!

Answer 1

Here's a start

We need to see the two cases that need checked:

• the probability that $2$ is in the bag after throwing out a $1$ at the end.
• the probability that $2$ is in the bag after throwing out another number that isn't $1$ or $2$. (Meaning $1$ was one of the first ten chosen)

So we need to first choose $10$ balls that aren't a $1$ or a $2$, which has $\binom{18}{10}$ ways of being done. Then the $1$ will be forced out at the end.

This is all for the first case, see if you can now figure out the second case and continue from here.

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One more hint: the total number of ways to have this happen (via the two cases) is $\binom{19}{10} * \binom11 + \binom{19}{9} * \binom11 * \binom{10}{1}$

Answer 2

A quick way to solve the problem:

After all is said and done, the $1$ can not be in box. All the others have the same probability of being in the box and it is guaranteed that there are exactly nine non-$1's$. In other words, the situation is equivalent to choosing $9$ out of $19$ balls. Thus the answer is $$\frac 9{19}=0.473684211$$