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

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23 Apr 2019, 13:15
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85% (hard)

Question Stats:

38% (01:51) correct 62% (03:02) wrong based on 29 sessions

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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

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Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

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24 Apr 2019, 00:29
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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
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Re: Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

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25 Apr 2019, 19:18
Bunuel wrote:
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

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.

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Re: Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

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25 Apr 2019, 23:10
Bunuel wrote:
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

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 = $$\frac{15}{216}*3 = \frac{45}{216} = \frac{5}{24}$$

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Re: Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

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27 Apr 2019, 01:23
Bunuel wrote:
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

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
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Re: Three fair six-sided dice are rolled. What is the probability that the   [#permalink] 27 Apr 2019, 01:23
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