How Cheenta works to ensure student success?

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

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?

,

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

combinatorics

Dice-problem

Probability

Pre College Mathematics

AMC-10A, 2014 Problem-17

$\frac{5}{24}$

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?

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?

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 $\frac{45}{216}$=$\frac{5}{24}$

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=h4MmDUky7KM

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

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?

,

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

combinatorics

Dice-problem

Probability

Pre College Mathematics

AMC-10A, 2014 Problem-17

$\frac{5}{24}$

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?

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?

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 $\frac{45}{216}$=$\frac{5}{24}$

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=h4MmDUky7KM

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?

Google