Question

In the Banach matchbox problem, find the probability that at the moment when the first box is emptied (as opposed to being found empty), the other box contains exactly $k$matches.

Short answer

$P\left(E\right)=2·\left(\begin{array}{c}2N-k-1\\ N-1\end{array}\right){\left(\frac{1}{2}\right)}^{2N-k}$

Step-by-step solution

Step 1:Given information

Given in the question that,we need to find find the probability that at the moment when the first box is emptied (as opposed to being found empty), the other box contains exactly $k$matches in the Banach matchbox problem.

Step 2:Explanation

Let$E$indicate the event that defines the moment when the first box is emptied, the other box holds exactly $k$matches. This event will happen if and only if we make our $N$the selecting of the right-hand box at $(2N-k)$the trial. Using a similar argument as in Example, we have that

$P\left(E\right)=2·\left(\begin{array}{c}2N-k-1\\ N-1\end{array}\right){\left(\frac{1}{2}\right)}^{2N-k}$

The factor $2$is because of the fact that the considered box is not fixed, it might be left-handed or right-handed.

Step 3:Final answer

$P\left(E\right)=2·\left(\begin{array}{c}2N-k-1\\ N-1\end{array}\right){\left(\frac{1}{2}\right)}^{2N-k}$