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, probability is 0.28
Now we need to count the instances where the values on two dice add up to the value on the other die, which is one. If a die shows 6, the remaining two dice must either show (5,1) or (4,2), or (3,3)
Total number of scenarios that could occur in this situation: 3!+3!+3!/2!=15
2. If a die shows 5, the other two dice must either display (4,1) or (3,2)
The total cases that could occur in this scenario are 3!+3! =12.
3. If a die shows 4, the other two dice must either display (3,1) or (2,2)
Total number of scenarios that could occur in this situation: 3!+3!/2!=9
4. If a die shows 3, the other two dice must also show 3. (2,1)
The total number of cases possible in this scenario= is 3!/2! =3
Total cases possible= 15+12+9+6+3=45
Probability= 45/216=5/24= 0.208
probability is 0.208
correct option is (D).
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The question was incomplete. Check below the complete 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
