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 $\frac{45}{216}$=$\frac{5}{24}$
Other useful links
- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=h4MmDUky7KM