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AMC 10 Math Olympiad USA Math Olympiad

Dice Problem | AMC-10A, 2011 | Problem 14

Try this beautiful problem from Probability based on dice from AMC-10A, 2011. You may use sequential hints to solve the problem

Try this beautiful problem from Probability based on dice Problem

Dice Problem- AMC-10A, 2011- Problem 14


A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?

  • \(\frac{1}{12}\)
  • \(\frac{7}{12}\)
  • \(\frac{5}{12}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{9}\)

Key Concepts


Probability

dice

circle

Check the Answer


Answer: \(\frac{1}{12}\)

AMC-10A (2011) Problem 14

Pre College Mathematics

Try with Hints


Given that A pair of standard 6-sided fair dice are rolled once. The sum of the numbers rolled determines the diameter of a circle. The numerical value of the area of the circle is less than the numerical value of the circle’s circumference. Let the radius of the circle is \(r\). Then the area of the circle be \(\pi(r)^2\) and ¬†circumference be \(2\pi r\)

can you finish the problem……..

Now according to the given condition we say that \(\pi(r)^2 <2\pi{r}\)\(\Rightarrow r<2\)

As The sum of the numbers rolled determines the diameter of a circle, therefore $r<2$ then the dice must show \((1,1)\),\((1,2)\),\((2,1)\)

can you finish the problem……..

Therefore there $3$ choices out of a total possible of $6 \times 6 =36$, so the probability is \(\frac{3}{36}=\frac{1}{12}\)

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