1. ## Coin toss

A horizontal flat square board 6 inches width is ruled with a grid of fine lines, 2 inches apart and has a vertical rim like a carrom board, all around the edge. If a coin of diameter 2/3 inches is tossed on to the board. What is the probability that it rests without crossing a line?

2. Hello, alexmahone!

A horizontal flat square board 6 inches width is ruled with a grid of fine lines,
2 inches apart and has a vertical rim like a carrom board, all around the edge.
If a coin of diameter 2/3 inches is tossed on to the board,
what is the probability that it rests without crossing a line?

Each square is 2 inches by 2 inches; there are 9 such squares.

The coin has a radius of $\displaystyle \tfrac{1}{3}$ inches.
To avoid all lines, its center must be more than $\displaystyle \tfrac{1}{3}$ inch from any line.

The center of the coin can be in this shaded region:
Code:
      : - - - - 2 - - - - :
-   *---+-----------+---*  -
:   |   |           |   | 1/3
:   + - + - - - - - + - *  -
:   |   |:::::::::::|   |  :
:   |   |:::::::::::|   |  :
2   |   |:::::::::::|   | 4/3
:   |   |:::::::::::|   |  :
:   |   |:::::::::::|   |  :
:   + - + - - - - - + - *  -
:   |   |           |   | 1/3
-   *---+-----------+---* -
1/3     4/3     1/3

The shaded area is: .$\displaystyle \left(\frac{4}{3}\right)^2 \:=\:\frac{16}{9}$ inē.

In the nine squares, the shaded area is: .$\displaystyle 9 \times\frac{16}{9} \:=\:16$ inē.

The total area of the board is: .$\displaystyle 6^2 \:=\:36$ inē.

Therefore, the probability is: .$\displaystyle \frac{16}{36}\;=\;\frac{4}{9}$

3. Originally Posted by Soroban
Hello, alexmahone!

Each square is 2 inches by 2 inches; there are 9 such squares.

The coin has a radius of $\displaystyle \tfrac{1}{3}$ inches.
To avoid all lines, its center must be more than $\displaystyle \tfrac{1}{3}$ inch from any line.

The center of the coin can be in this shaded region:
Code:
      : - - - - 2 - - - - :
-   *---+-----------+---*  -
:   |   |           |   | 1/3
:   + - + - - - - - + - *  -
:   |   |:::::::::::|   |  :
:   |   |:::::::::::|   |  :
2   |   |:::::::::::|   | 4/3
:   |   |:::::::::::|   |  :
:   |   |:::::::::::|   |  :
:   + - + - - - - - + - *  -
:   |   |           |   | 1/3
-   *---+-----------+---* -
1/3     4/3     1/3
The shaded area is: .$\displaystyle \left(\frac{4}{3}\right)^2 \:=\:\frac{16}{9}$ inē.

In the nine squares, the shaded area is: .$\displaystyle 9 \times\frac{16}{9} \:=\:16$ inē.

The total area of the board is: .$\displaystyle 6^2 \:=\:36$ inē.

Therefore, the probability is: .$\displaystyle \frac{16}{36}\;=\;\frac{4}{9}$
I don't think you're right because my options are:
A) 0.5625
B) 0.6525
C) 0.6255
D) 0.5562
E) None of these

4. Consider where the centre of the coin can land.

For each square (2x2) it's centre must be at least $\displaystyle \frac {1}{3}$ from each side, this gives an area of

$\displaystyle (2 - \frac {2}{3})^2 = \frac {16}9$ per square.

There are 9 squares which gives a 'safe' area of 16

Total area is

$\displaystyle (6 - \frac {2}{3})^2 = 256/9$

$\displaystyle 16/ \frac {256}{9} = 9/16 = 0.5625$

Edit: Soroban missed that the centre cannot lie in a ring around the edge
The probability is therefore