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Suppose that we drop a round coin with a diameter of 1 cm onto a gigantic sheet of paper with red lines drawn vertically every 10 cm and blue lines drawn horizontally every 5 cm.

(i) What is the probability the coin will land on a red line?


The coin crosses a line when at least 1/2 of it is through it. So since the red lines are 10 cm apart you would have to add the 0.5 space from the left line and the 0.5 from the right line $\frac{0.5}{10} + \frac{0.5}{10} = \frac{1}{10}$. (This is straight from the solutions, Below is a picture if the enter image description here


(ii) Find the probability the coin lands on both a red and blue line.


I was trying to do this but I get $\frac{1}{4}$ which is too big considering in the next question I have to divide this answer with the answer from (i). My reasoning was if horizontal lines were drawn every 5 cm, there would be squares and since this is asking for when the coin crosses BOTH lines I took the area of a small square that would be at each corner which would be (0.5)(0.5) which is 1/4.


(iii) Given the coin has landed on a red line what is the conditional probability it will land on a blue line? Are the two events of landing on a blue line and landing on a red line independent?

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  • $\begingroup$ You have a good start in finding the area of the corner square. To find the probability you want, however, you will need to divide by the area of the big rectangle determined by the blue and red lines. (You're not done though--each rectangle has 4 such corners.) $\endgroup$
    – Joey Zou
    Sep 1, 2016 at 5:40
  • $\begingroup$ BTW, I think the explanation "The coin crosses a line when at least 1/2 of it is through it" makes no sense. At least 1/2 of what is through what? A better way to phrase it is to say that a coin crosses the line if the distance of its center from the line is less than the radius of the coin, i.e. less than half its diameter. For a coin with diameter 1 cm, this means that the coin crosses the line if its center is less than 0.5 cm away from the line. The explanation is fine afterwards. $\endgroup$
    – Joey Zou
    Sep 1, 2016 at 5:42
  • $\begingroup$ @JoeyZou thank you for clearing up my explanation $\endgroup$
    – Deegeeek
    Sep 1, 2016 at 22:02

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