Three fair six-sided dice are rolled. What is the probability that the

Question

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?

(A) 1/6
(B) 13/72
(C) 7/36
(D) 5/24
(E) 2/9

nick1816 · Answer

Total number of cases possible when 3 Dice roll= 6^3=216
Now we have to find the number of cases when values shown on two of the dice sum to the value shown on the remaining die-
1. When a dice shows 6, other 2 dices must show either (5,1) or (4,2) or (3,3)
 Total number of cases possible in this scenario= 3!+3!+3!/2!=15
2. When a dice shows 5, other 2 dices must show either (4,1) or (3,2)
 Total number of cases possible in this scenario= 3!+3!=12
3. When a dice shows 4, other 2 dices must show either (3,1) or (2,2)
 Total number of cases possible in this scenario= 3!+3!/2!=9
4. When a dice shows 3, other 2 dices must show (2,1)
 Total number of cases possible in this scenario= 3!=6
5. When a dice shows 2, other 2 dices must show (1,1)
 Total number of cases possible in this scenario= 3!/2!=3
Total cases possible= 15+12+9+6+3=45

Prbability= 45/216=5/24

ScottTargetTestPrep · Answer

There are 6^3 = 216 outcomes when three dices are rolled. If the sum of two of the dice is equal to the value of the remaining die. It could be:

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

For the combos above, each has 3!/2! = 3 outcomes (for example, 1-1-2 can also be 1-2-1 and 2-1-1). So the bold combos have a total 3 x 3 = 9 outcomes.

For the italic combos above, each has 3! = 6 outcomes (for example, 1-2-3 can also be 1-3-2, 2-1-3, 2-3-1. 3-1-2 and 3-2-1). So the italic combos have a total 6 x 6 = 36 outcomes.

Therefore, the probability that the values shown on two of the dice sum to the value shown on the remaining die is (9 + 36)/216 = 45/216 = 5/24.

Answer: D

OhsostudiousMJ · Answer

Let the die be A, B and C.
Given = A+B= C or A+C = B or B+C = A. .......  (I)

A dice has maximum number of 6. Therefore, the maximum sum of 2 numbers from the other two die would be 6.

Possibilities are:
1+1, 1+2, 1+3, 1+4, 1+5,
2+1, 2+2, 2+3, 2+4,
3+1, 3+2, 3+3,
4+1, 4+2, 
5+1.
Totally, 15 possibilities.
Hence, probability =  \frac{15}{6^3} = \frac{15}{216}

From  (I) , there are 3 such cases.
Hence, required probability =  \frac{15}{216}*3 = \frac{45}{216} = \frac{5}{24}

Answer  D .

Archit3110 · Answer

total such pairs ;
(1,1,2) ( 1,2,3) , ( 1,3,4) , ( 1,4,5) , ( 1,5,6 ) 
(2,1,3) ( 2,2,4), ( 2,3,5) , ( 2,4,6)
(3,1,4) ( 3,2,5) , ( 3,3,6)
(4,1,5) ( 4,2,6) 
(5,1,6) 
15* 3; 45 pairs possible
total fair plays ; 6^ 3 ; 216
so P ; 45/216 ; 5/24
IMO D

BrushMyQuant · Answer

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

As we are rolling three dice => Number of cases =  6^3  = 216

Now,  out of the three rolls lets pick the two rolls whose numbers will add up to the number in the third dice . We can do that in 3C2 ways
=  \frac{3!}{2!*(3-2)!}  =  \frac{3*2!}{2!*1!}  = 3 ways

Now,  the third dice can show numbers from 1 to 6, but the sum of the other two dices should be equal to the third dice so third dice can show numbers only from 2 to 6 (as minimum we should get 1 in each dice making the sum as 2)

=> Possibilities are as follows for sum between 2 and 6:
 1 in First Dice:  (1,1), (1,2), (1,3), (1,4), (1,5)
 2 in First Dice:  (2,1), (2,2), (2,3), (2,4)
 3 in First Dice:  (3,1), (3,2), (3,3)
 4 in First Dice:  (4,1), (4,2)
 5 in First Dice:  (5,1)
 6 in First Dice:  0 cases
=> 15 cases

=>  Total number of ways = 3 * 15 = 45 ways

=>  Probability that the values shown on two of the dice sum to the value shown on the remaining die  =  \frac{45}{216}  =  \frac{5}{24}

So,  Answer will be D 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

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