A coin with radius r = \(\displaystyle\sqrt{16\over 4π}\) rolled on a normal chessboard. The length of the side of the smallest square of the chessboard is 5. After rolling the coin comes to rest after some time such that its center is inside the chessboard. Find the probability that the coin covers one of the corners of any of the 64 squares on the chessboard.

Option 2 : 16/25

**Calculation:**

⇒ Let the length of the side of the smallest square be s.

⇒ Let us imagine the situation first. The figure below shows the smallest square of the chessboard and one of the positions of the coin such that the coin covers the corner of the square.

⇒ Let the distance between the center of the circle and the overlapping corner is d.

⇒ Now, it is clear that a coin can land on any of the corners only if the center of the coin is at a distance less than or radius from the corner. i.e., d

⇒ The probability of the coin landing on any of the corners is equal.

⇒ So, whenever the center is in the region shown below, the coin will cover the corner.

Thus, required probability = 4 × (Area of the quarter circle)/(Area of a square)

⇒\(\displaystyle(πr^2)\over S^2\)

⇒ \(\displaystyle(π × 16/π)\over 5^2\)

⇒ 16/25

__Additional Information__

⇒ The number of outcomes favorable to A is denoted by n(A) The total number of outcomes in sample space is denoted by n(S). Hence, the formula becomes

P(A) = n(A)/n(S).

⇒ Similarly, The favorable Area to A is denoted by n(A) and the total area is denoted by n(S). Hence, the formula becomes P(A) = n(A)/n(S).