Catalan numbers application

Question

"We place 2n balls in line, n balls are black and the other n balls are white. The event En is performed when, on positions 1 to k, the number of white balls observed is greater than the number of black balls observed, with k = 1,. . . , 2n.

Prove that:

1) card( En) = (2n C n) / (n+1)

2) P(En) = 1/(n+1) "

Just to make sure, if I'm not wrong, the white ball has to be at position 1, then card(E1) = 1.

So on, if n=2, then we have 3 possibilities: WWBB, WBBW, WBWB? But it doesn't fit with the formula I have to prove hmm.

Can someone give me a hand and please explain to me what the event En really is. Thank you so much!

Answer 1

To restate the question so it makes sense and give the Catalan numbers as a solution:

We place $2n$ balls in line. $n$ balls are black and the other $n$ balls are white. The event $E_n$ occurs when, for all $k$ in $\{1,2,3,\ldots,2n\}$, the number of white balls among the first $k$ balls is greater than or equal to the number of black balls among the first $k$ balls.

So with $n=2$, the successful event may be WWBB or WBWB, so $E_2$ has a cardinality of $2 = {4 \choose 2}/(2+1)$.

Your suggestion of WBBW fails when $k=3$ since among the first three balls there is one white ball, fewer than the two black balls.

Wikipedia currently gives six proofs that the number of such cases is ${2n \choose n}/(n+1)$.

Since the total number of arrangements is ${2n \choose n}$, this makes the probability $\mathbb P(E_n)=\dfrac{{2n \choose n}/(n+1)}{{2n \choose n}} = \dfrac{1}{n+1}$