Coin toss

• Oct 20th 2008, 10:37 AM
alexmahone
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?
• Oct 20th 2008, 12:13 PM
Soroban
Hello, alexmahone!

Quote:

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 $\tfrac{1}{3}$ inches.
To avoid all lines, its center must be more than $\tfrac{1}{3}$ 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: . $\left(\frac{4}{3}\right)^2 \:=\:\frac{16}{9}$ inē.

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

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

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

• Oct 20th 2008, 12:23 PM
alexmahone
Quote:

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 $\tfrac{1}{3}$ inches.
To avoid all lines, its center must be more than $\tfrac{1}{3}$ 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: . $\left(\frac{4}{3}\right)^2 \:=\:\frac{16}{9}$ inē.

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

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

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

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
• Oct 21st 2008, 04:57 AM
SimonM
Consider where the centre of the coin can land.

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

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

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

Total area is

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

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

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