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

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

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