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

But try the problem first…

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

AMC-10A (2011) Problem 14

Pre College Mathematics

## Try with Hints

First hint

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……..

Second Hint

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……..

Final Step

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

## Other useful links

- https://www.cheenta.com/tetrahedron-problem-amc-10a-2011-problem-24/
- https://www.youtube.com/watch?v=ruRoSSe1U18

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