Two number cubes with six faces numbered with the integers from 1 thro

Question

Two number cubes with six faces numbered with the integers from 1 through 6 are tossed. What is the probability that the sum of the exposed faces on the cubes is a prime number?

A. 1/36
B. 1/8
C. 1/12
D. 5/12
E. 7/12

Archit3110 · Answer

Favourable events - (1,1), (1,2), (1,4), (2,3), (2,5), (5,6), (3,4), (1,6). Except the event (1,1) every other event must be counted twice due to the fact that the number might appear on either of the dice, so total number of favourable events is 15.

Total events = 6^2 = 36

Probability = 15/36 = 5/12.'
IMO D

anchal19 · Answer

Isn’t this wrong? Sets should be; 
1,1
1,2
1,4
1,6
2,3
2,5
3,4
5,6

Posted from my mobile device

kylebecker9 · Answer

Prime numbers between 2 and 12 and the number of ways they can be rolled:
2 (1)
3 (2)
5 (4)
7 (6)
11 (2)

So, 15 total ways to roll a prime. 15/36 = 5/12

ANSWER D

jcerdae · Answer

Thank you for the explanation!! its helpful 
Is there a way to know when the same numbers in different order should be considered as one or two numbers?
My initial thought was that since the problem asks for the SUM, then (1,2) - (2,1) should be considered as 1 case since the SUM is the same (3).
knowing that my answer is wrong, could you please tell me a way when (1,2) - (2,1) are the same or different?

Thank you so much!! GmatClub rules!!

rishab0507 · Answer

1,2 and 2,1 should be considered different, as different numbers on 2 dices. 
All numbers except 1,1 should be considered twice as can occur in reverse order, But since 1,1 pair is similar to be considered once. Even if you consider 1.1 twice total will be 16, and answer shall be 16/36= 4/9, which is not one of options. 
Hope i am clear.

Kudos if it helps

BrushMyQuant · Answer

Given that Two fair dice with sides numbered 1 to 6 are tossed and We need to find What is the probability that the sum of the exposed faces on the dice is a prime number? 
 
As we are rolling two dice => Number of cases =  6^2  = 36

We know that out of the numbers from 2(1+1) to 12 (6+6) we have only five prime numbers and they are 2, 3, 5, 7 and 11
( Watch this video  if you are not aware of  Prime Numbers )

Sum of the exposed faces has to be prime. So, following are the possible cases
(1,1), (1,2), (1,4), (1,6)
(2,1), (2,3), (2,5)
(3,2), (3,4)
(4,1), (4,3)
(5,2), (5,6)
(6,1), (6,5)

=> 15 possibilities

=> Probability that the sum of the exposed faces on the dice is a prime number =  \frac{15}{36}  =  \frac{5}{12}

So,  Answer will be D 
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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

Two number cubes with six faces numbered with the integers from 1 thro

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