Brilli the ant randomly placed a token into a square on a 2×100 chessboard according to a probability distribution P. The token is then moved uniformly at random to one of the horizontally, vertically, or diagonally adjacent squares. The probability that the token is in a particular position after it has been moved also satisfies the distribution P. Let q be the probability that the token is placed into one of the columns of C={5,6,…44} and after being moved is still in one of those columns. The value of q can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?
Please provide a solution along with the answer
To find the probability q, we need to consider the possible movements of the token on the chessboard.
Let's start by thinking about the initial placement of the token. Since it is randomly placed according to the probability distribution P, we can assume that the probability of placing the token in any particular square is the same for all squares.
Now, let's consider the movement of the token. It can move to any of the horizontally, vertically, or diagonally adjacent squares with equal probability. Since the token is initially in one of the columns in C={5,6,…44}, we need to determine the probability that it remains in one of these columns after being moved.
To calculate this probability, we can analyze the possible movements of the token.
If the token is in column 5 or 44, it can only move to the adjacent columns (4 or 6, respectively). This means that the probability of staying in C after being moved is equal to the probability of being initially placed in column 5 or 44, multiplied by the probability of moving to the adjacent column.
If the token is in any other column within C, it can move to three adjacent columns (left and right) or one of the diagonally adjacent columns. The probability of staying in C after being moved is equal to the probability of being initially placed in one of these columns, multiplied by the probability of moving to one of the adjacent columns or diagonally adjacent columns.
To calculate the probability q, we need to sum up the probabilities of staying in C for all possible initial placements within C. Since the probability of each initial placement is the same, we can calculate the sum by multiplying the probability of being initially placed in any column of C by the probability of staying in C after being moved.
Let's denote the probability of being initially placed in column i as p(i) (for i = 5 to 44). The probability of staying in C after being moved from column i is 2/5 if i=5 or i=44 (since there are only two adjacent columns) and 3/5 for all other columns i (since there are three adjacent columns and one diagonally adjacent column).
The probability q is given by:
q = p(5) * 2/5 + p(44) * 2/5 + ∑(p(i) * 3/5) for i = 6 to 43
Since the probability distribution P is not specified, we don't have enough information to calculate the precise values of p(i). However, we know that the sum of probabilities over all possible initial placements should be equal to 1. Therefore, we can assume that p(i) = 1/40 for all i from 5 to 44.
Substituting these values into the expression for q, we get:
q = (1/40) * 2/5 + (1/40) * 2/5 + ∑((1/40) * 3/5) for i = 6 to 43
Simplifying the expression, we have:
q = (1/20) + (1/20) + (1/40) * 3/5 * (43-6+1)
q = 1/10 + 1/10 + (1/40) * 3/5 * 38
q = 1/5 + 3/200 * 38
q = 1/5 + 3/200 * 38/1
q = 1/5 + (3 * 38) / (200 * 1)
q = 1/5 + 114/200
q = 1/5 + 57/100
q = 20/100 + 57/100
q = 77/100
Therefore, the value of a+b is 77+100 = 177.