The squares of a 2×500 chessboard are coloured black and white in the standard alternating pattern. k of the black squares are removed from the board at random. What is the minimum value of k such that the expected number of pieces the chessboard is divided into by this process is at least 20 ?
13
wrong
To solve this problem, let's start by understanding the structure of the chessboard and how it is divided.
A 2x500 chessboard consists of 1000 squares in total, arranged in 2 rows and 500 columns. These squares alternate between black and white in a standard pattern, which means the first square is black, the second square is white, the third square is black again, and so on.
When we remove k of the black squares at random, we want to find the minimum value of k such that the expected number of pieces the chessboard is divided into is at least 20.
To determine the expected number of pieces, we need to consider how the removal of black squares affects the division of the chessboard.
Let's analyze the possible scenarios:
1. If we remove 0 black squares, the chessboard remains intact, resulting in 1 piece.
2. If we remove 1 black square, we create two pieces, as the board is divided into two non-adjacent segments.
3. If we remove 2 black squares, we still create two pieces, but the segments are now adjacent.
4. If we remove 3 black squares, we create three pieces, with two adjacent segments.
From these observations, we can deduce that the general pattern is as follows: removing n black squares will result in n+1 pieces, with the exception of n=1, where it results in 2 pieces.
Now, let's calculate the expected number of pieces based on the probabilities of removing different numbers of black squares.
To find the probability of removing a certain number of black squares, we can use combinatorics. There are C(500, k) ways to choose k black squares from a total of 500 black squares on the chessboard.
Since each black square has an equal chance of being removed, the probability of removing any particular set of k black squares is 1 / C(500, k).
To calculate the expected number of pieces, we need to sum up the probabilities of each scenario multiplied by the number of pieces in that scenario.
Let E(k) represent the expected number of pieces when k black squares are removed.
E(k) = (1 / C(500, k)) * (k + 1)
To find the minimum value of k such that E(k) is at least 20, we need to solve the inequality:
E(k) ≥ 20
(1 / C(500, k)) * (k + 1) ≥ 20
Now, we can use this inequality to iteratively find the minimum value of k that satisfies it. Starting from k = 0, we increment k until E(k) is greater than or equal to 20.
Please note that due to the large values involved in these calculations, it might be computationally intensive to find the exact minimum value of k using this method. However, we can use numerical or approximation techniques to estimate it.