A pair of six-sided dice are labeled so that one die has only even num

Question

A pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?

(A) 1/6
(B) 1/5
(C) 1/4
(D) 1/3
(E) 1/2

(A) 1/6
(B) 1/5
(C) 1/4
(D) 1/3
(E) 1/2

GMATinsight · Accepted Answer

Sum = 7 has the following sample spaces

(1,6) (2,5) (3,4)

Probability of (1,6)  = (\frac{2}{6})*(\frac{2}{6}) = 1/9 
Probability of (2,5)  = (\frac{2}{6})*(\frac{2}{6}) = 1/9 
Probability of (3,4)  = (\frac{2}{6})*(\frac{2}{6}) = 1/9

Required probability  = (\frac{1}{9})+(\frac{1}{9})+(\frac{1}{9}) = (\frac{1}{3})

Answer: Option D

Archit3110 · Answer

total fair chances (2,5). ( 4,3) ( 6,1) AND each can come twice each side ; 4 chances each 
so 4* 3 = 12 chances out of 36 total 
12/36; 1/3 IMO D

ScottTargetTestPrep · Answer

We can let the numbers on the die with even numbers be 2, 2, 4, 4, 6, 6 (note: we use bold numbers to distinguish, for example, the two 2s are essentially different from each other), and similarly, the numbers on the die with odd numbers be 1, 1, 3, 3, 5, 5.

The number of possible outcomes for a roll of two dice is 6 x 6 = 36.

The number of ways to get a sum of 7 is:
(2, 5), (2, 5), (2, 5), (2, 5)

(3, 4), (3, 4), (3, 4), (3, 4)

(1, 6), (1, 6), (1, 6), (1, 6)

So we have 12 outcomes of getting a sum of 7, and hence, the the probability is 12/36 = 1/3.

Alternate Solution:

A sum of 7 can be obtained in three ways: (2, 5), (4, 3) and (6, 1).

The probability of rolling a 2 with the die that has only even numbers on it is 2/6 = 1/3. Similarly, the probability of rolling a 5 with the die that has only odd numbers on it is 2/6 = 1/3. Thus, the probability of the outcome (2, 5) is 1/3 x 1/3 = 1/9.

The probabilities of the outcomes (4, 3) and (6, 1) are also 1/9 by similar reasoning. Thus, the probability of obtaining a sum of 7 is 1/9 + 1/9 + 1/9 = 3/9 = 1/3.

Answer: D

OhsostudiousMJ · Answer

Since each number is present twice in each die, the probability of getting any number is  \frac{2}{6} = \frac{1}{3} .
Possible ways of getting a sum of 7 = 2+5, 4+3 and 6+1.

Probability of one possibility, lets say 2+5, is  \frac{1}{3}*\frac{1}{3}=\frac{1}{9} .
There are three such possibilities, Therefore, 
 P = \frac{1}{9}*3 = \frac{1}{3} 
Answer  D .

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A pair of six-sided dice are labeled so that one die has only even num

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