Try this beautiful problem from Probability based on dice

## Probability Dice Problem – AMC-10A, 2009- Problem 22

Two cubical dice each have removable numbers \(1\) through \(6\). The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is \(7\)?

- \(\frac{3}{8}\)
- \(\frac{2}{11}\)
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)
- \(\frac{2}{9}\)

**Key Concepts**

Probability

Combinatorics

Cube

## Check the Answer

But try the problem first…

Answer: \(\frac{2}{11}\)

AMC-10A (2009) Problem 22

Pre College Mathematics

## Try with Hints

First hint

We assume that the colours of the numbers are different.there are two dices and each of them 1 to 6.after throw,the probability of getting some pair of colors is the same for any two colors.

Therefore there are \(\ 12 \choose 2\)=\(66\) ways to pick to of the colours…

can you finish the problem……..

Second Hint

Now given condition is that the sum will be \(7\).So \(7\) can be obtained by \(1 +6\),\(2+5\),\(3+4\) and Each number in the bag has two different colors, Therefore each of these three options corresponds to four pairs of colors.SO \(7\) comes from \(3.4\)=\(12\) pairs…..

can you finish the problem……..

Final Step

So our required probability will be \(\frac{12}{66}=\frac{2}{11}\).

## Other useful links

- https://www.cheenta.com/octahedron-problem-amc-10a-2006-problem-24/
- https://www.youtube.com/watch?v=w0Y2oXoyEEQ

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