Consider a sheet of paper 20 cm square. The paper is divided up into 16 squares side 5cm. A coin ( diameter 2cm ) is randomly thrown onto the sheet of paper. Can we allocate a probability to the outcome of the coin landing completely within one of the 16 squares? ( not touching any of the lines ) Is it the ratio of the area of the coin to the the area of a circle radius 2.5 cm?
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Can this be calculated??
- Thread starter scib59
- Start date
Hi, nice to meet you.
This problem remind me of Buffon's needle.
Well, i took the problem as : the center of the coin has to remain at least at 1cm of all border of the small squares.
Area of one "ok-square"=(5-2)^2=9, total area=16*9=90+54=144 cm^2
Prob(ok)=144/400=72/200=36/100=18/50=9/25 (?)
This problem remind me of Buffon's needle.
Well, i took the problem as : the center of the coin has to remain at least at 1cm of all border of the small squares.
Area of one "ok-square"=(5-2)^2=9, total area=16*9=90+54=144 cm^2
Prob(ok)=144/400=72/200=36/100=18/50=9/25 (?)
Good morning:
this looks like a continuous uniform variable problem. We have a square with an area of 400 cm.
Now we have 16 squares five lines horizontal and five lines vertical (the edge of the paper counts as a line) Measure out two inches from each of the lines as to where the coin could safely fall. When you do this you should wind up with a one inch square in each of the larger squares or 16 one inch squares in total. That's the total "safe area" (sixteen square inches out of the whole piece of paper)
So the possibility of the coin not touching any of the lines is 16/400 or somewhere around 4 percent. that's my story anyway and I'm sticking to it...
if the edges of the paper don't count as lines than you have some more "safe area" on the edges so where it looks like you draw only three lines. three vertical and three horizontal and your chances of not landing on a line are 66/40 or 16.5 percent
do you happen to have the answer by any chance or is this to settle a bar bet?
this looks like a continuous uniform variable problem. We have a square with an area of 400 cm.
Now we have 16 squares five lines horizontal and five lines vertical (the edge of the paper counts as a line) Measure out two inches from each of the lines as to where the coin could safely fall. When you do this you should wind up with a one inch square in each of the larger squares or 16 one inch squares in total. That's the total "safe area" (sixteen square inches out of the whole piece of paper)
So the possibility of the coin not touching any of the lines is 16/400 or somewhere around 4 percent. that's my story anyway and I'm sticking to it...
if the edges of the paper don't count as lines than you have some more "safe area" on the edges so where it looks like you draw only three lines. three vertical and three horizontal and your chances of not landing on a line are 66/40 or 16.5 percent
do you happen to have the answer by any chance or is this to settle a bar bet?
Hello, scib59!
Consider any of the 5-by-5 squares.
The coin is interior to the 5-by-5 square
. . if its center is within the 3-by-3 square (shaded region).
The probability is the ratio of the areas of the two squares.
\(\displaystyle P(\text{coin does not touch a line}) \;=\;\frac{3^2}{5^2} \;=\;\frac{9}{25} \;=\;36\%\)
Consider a sheet of paper 20 cm square. The paper is divided up into 16 squares side 5 cm.
A coin (diameter 2 cm) is randomly thrown onto the sheet of paper.
Can we allocate a probability to the outcome of the coin landing
completely within one of the 16 squares (not touching any of the lines)?
Consider any of the 5-by-5 squares.
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| * *:*:::::::::::::::::::::::::*:* * - - - |
| |:::::::::::::::::::::::::::::| |The coin is interior to the 5-by-5 square
. . if its center is within the 3-by-3 square (shaded region).
The probability is the ratio of the areas of the two squares.
\(\displaystyle P(\text{coin does not touch a line}) \;=\;\frac{3^2}{5^2} \;=\;\frac{9}{25} \;=\;36\%\)