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Probability Dice Problem | AMC-10A, 2009 | Problem 22

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




Check the Answer

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

AMC-10A (2009) Problem 22

Pre College Mathematics

Try with Hints

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........

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........

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

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