The squares of a 2×60 chessboard are coloured black and white in the standard alternating pattern. At random, exactly half of the black squares are removed from the board. The expected number of white squares that have no neighbours after this process can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?
Details and assumptions:
>Note: The black squares are not removed with probability 12. Rather, it is given that exactly 30 black squares are removed.
>A square is a neighbour if it is located directly to the left, right, top or bottom of the initial square. Squares that are connected by exactly 1 vertex are not neighbors. I've started doing this, let's see who will come first. A Challenge !!!
To solve this problem, we need to calculate the expected number of white squares that have no neighboring black squares after exactly half of the black squares are removed from the chessboard.
First, let's determine the number of white squares on the original chessboard. Given a 2×60 chessboard with alternating black and white squares, there are a total of (2×60)/2 = 60 white squares.
Next, let's consider the number of white squares that have no neighboring black squares. On a standard alternating chessboard, each white square has four potential neighbors: two on the left and right, and two on the top and bottom. However, some of these potential neighbors may be black squares.
To count the number of white squares that have no black neighbors, we need to consider the pattern of the chessboard. Since the chessboard is 2×60, there are 30 rows of white squares. In each row, the first and last squares have three potential neighbors, while the remaining squares have four potential neighbors. This is because the first and last squares have only one neighbor on the side facing the black squares. In the remaining rows, all white squares have four potential neighbors.
To calculate the number of white squares that have no neighboring black squares, we have:
- 2 white squares in the first row with 3 potential neighbors each,
- 58 white squares in the remaining rows with 4 potential neighbors each.
Thus, the total number of white squares with no neighboring black squares is given by:
2 × 3 + 58 × 4 = 2 + 232 = 234.
Now, since exactly half of the black squares are removed from the board, we have removed 30 black squares. This means that there are now 30 black squares remaining on the chessboard.
To calculate the expected number of white squares that have no neighboring black squares, we need to consider the probability that each black square is removed. Since there are 30 black squares remaining and 60 white squares in total, the probability of removing any individual black square is given by:
Probability of removing a black square = 30 black squares / 60 white squares = 1/2.
Now, we can calculate the expected number of white squares that have no neighboring black squares as follows:
Expected number = Probability of removing a black square × Number of white squares with no black neighbors
= (1/2) × 234
= 117/2.
The expected number of white squares with no neighboring black squares is 117/2. Therefore, the value of a + b is 117 + 2 = 119.