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AMC 10 Combinatorics Math Olympiad USA Math Olympiad

Dice Problem | AMC 10A, 2014| Problem No 17

Try this beautiful Problem on Probability from AMC 10A, 2014. Problem-17, You may use sequential hints to solve the problem.

Try this beautiful Problem on Probability based on Dice from AMC 10 A, 2014. You may use sequential hints to solve the problem.

Dice Problem – AMC-10A, 2014 – Problem 17


Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

,

  • $\frac{1}{6}$
  • $\frac{13}{72}$
  • $\frac{7}{36}$
  • $\frac{5}{24}$
  • $\frac{2}{9}$

Key Concepts


combinatorics

Dice-problem

Probability

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2014 Problem-17

Check the answer here, but try the problem first

$\frac{5}{24}$

Try with Hints


First Hint

Total number of dice is \(3\) and each dice \(6\) possibility. therefore there are total $6^{3}=216$ total possible rolls. we have to find out the probability that the values shown on two of the dice sum to the value shown on the remaining die.

Without cosidering any order of the die , the possible pairs are $(1,1,2),(1,2,3),(1,3,4)$,$(1,4,5),(1,5,6),(2,2,4),(2,3,5)$,$(2,4,6),(3,3,6)$

Now can you finish the problem?

Second Hint

Clearly $(1,1,1).(2,2,4),(3,3,6)$ this will happen in $\frac{3 !}{2}=3$ way

$(1,2,3),(1,3,4)$,$(1,4,5),(1,5,6),(2,3,5)$,$(2,4,6),$this will happen in $3 !=6$ ways

Now Can you finish the Problem?

Third Hint

Therefore, total number of ways $3\times3+6\times6=45$ so that sum of the two dice will be the third dice

Therefore the required answer is $\frac{45}{216}$=$\frac{5}{24}$

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