Two fair die with sides numbered 1 to 6 are tossed.

Question

Two fair die with sides numbered 1 to 6 are tossed. What is the probability that the sum of the exposed faces on the die is a prime number?

A) 5/11
B) 5/12
C) 5/21
D) 2/9
E) 5/36

(Manhattan made this task without variants of answers, but this is really nice task so I generate some answers to bring this task to standard PS format.)

TimeTraveller · Accepted Answer

Primes less than 12 (6+6) - 2, 3, 5, 7, 11

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.

NickHalden · Answer

Two dices are thrown ! So we may get 1+1 or 1+2 ... or 1+6.. or 2+6 or 3+6 etc

Therefore the total different sum values we can get are :   2,3,4,5,6,7,8,9,10,11,12      ----------------(1)

Out of these values, 5 numbers are primes i.e.   2,3,5,7,11     --------------------(2)

From (1) & (2)
The answer is   5/11   !!

Harley1980  /  Bunuel ,
I am not able to figure out from where the 12th case is coming for my sample space ?
What is the fundamental mistake that I am commiting?

Harley1980 · Answer

Hello  NickHalden

When you solve probability tasks you should take into account all possible variants 
For example if you throw a two coins you can't say that there is three possible variant HH, TT and HT
because from the point of probability there is 4 possible variants: HH, TT, HT and TH

So you make a mistake on this step: 
 Therefore the total different sum values we can get are :   2,3,4,5,6,7,8,9,10,11,12

NickHalden · Answer

Thanks  Harley1980  !!
Now I get it.

The sample space will be : 1,1 or 1,2 or 2,1 or 1,3 or 3,1 & so on.
i.e 6*6 = 36  ---------------- (1)

Similarly the favorable events are getting 
2: 1,1 
3: 1,2 or 2,1 
5: 2,3 or 3,2 or 1,4 or 4,1 
7: 3,4 or 4,3 or 2,5 or 5,2 or 1,6 or 6,1
11: 5,6 or 6,11
i.e.  15  in all ---------------- (2)

Therefore answer is 15/36 =  5/12  !!
 B

Harley1980 · Answer

Hello  NickHalden

For two dices from the point of probability possible 36 variants because variant 1-2 differ from variant 2-1
So each side of first dice will have six variants from second dice.
6*6 = 36

If it was task about combinations when there is only 21 variants because from the point of combinatorics 1-2 and 2-1 are the same variants.

You received only 11 variants because you calculated not all variants of two dice and not combinations but sum that can be from two dices. 
In probability questions you can't omit 1 step (calculating all variations) because it will bring you to the wrong answer.

pacifist85 · Answer

Well... I cannot say I really agree. I thought of the possibility to double the number of pairs that result in a prime number after solving it, as my answer was not there (7/36)..

However, it says how many possibilities are there that the sum (kind of implying that it is one thing, not a part of 2 numbers) will be a prime number. 3+2 or 2+3 (etc etc) both result in a 5. I also understand that it is possible to get this sum in 2 different ways (3+2 or 2+3), but still, it is the same sum...

To me, this would make more sense if I knew that we are throwing the die one by one. But, if we are throwing them together, in one go, to me it doesn't make much sense to count both possible ways for the sum, mostly because you see the result as "one" (sum) and not as 2 different numbers that make up the sum.

TimeTraveller · Answer

Hi pacifist85, I had that doubt also initially, in fact I have had that doubt ever since I took probability class in college. For some reason it is not very apparent to me why we need to count it twice when the result is the same.

However, I think they are counted twice because although the result of the event is the same, the events themselves are different. I think that's why they are counted twice. It's my understanding, I am not sure if it's correct reasoning though.

Harley1980 · Answer

Hello  pacifist85 .

I agree with you. It's really confusing topic. And I did ask the same questions not so long time ago. 
About your question:

If task was about some jar with 11 numbers from 2 to 12 and we should pick one of the number - then we can say that probability of picking prime number is is 5 from 11.
But in this case we can't operate with sums because these sums are created by different events and each such event is completely indepent.

About "one by one". What the difference if we throw dice together or one by one? Does they land ideally simulataneously if we throw them together? I think no (it can be millisecond difference).
So when you see in task that we throw two coins or three dices simultaneously it is equal to that these coins/dices was thrown one by one.

Does that make sense?

pacifist85 · Answer

Hey Harley 1980,

Yes it does make sense in a philosophical way! And this is why I also tried this reasoning and reached the answer in the end. By "one by one", again philosophically, the difference for me is that the events are kind of independent. So, it is sort of "one toss" vs "two tosses" instead of one "die" vs "two dice". In other word it is "one toss" of two dice, vs "two tosses" of two dice (independent events). That because by definition, the sum needs these 2 dice as a unity... So, 3+2 is the same (unity, ie 5) as 2+3.

Just explaining my reasoning! I can very well accept the answer however, I just sometimes think probability can be a lot about someone's perception, so a little bit phlosophical as a topic (not always though).

law258 · Answer

Way to think about this:

Combinations to get #2 = 1+1 --> 1/6 x 1/6 = 1/36
Combinations to get #3 = 1+2 (and 2+1) --> 1/6(1/6)+ 1/6(1/6) = 2/36
Combinations to get #5 = 2+3 (and 3+2) or 1+4 (and 4+1) --> 1/6(1/6)+ 1/6(1/6) +1/6(1/6)+ 1/6(1/6) = 4/36
Combinations to get #7 = 3+4 (and 4+3) or 1+6 (and 6+1) or 2+5 (and 5+2) = 6/36
Combinations to get #11 = 6+5 (and 5+6) --> 2/36

Add up all combinations = 15/36 = 5/12

B.

anairamitch1804 · Answer

Let's first think about the prime numbers less than 12, the maximum sum of the numbers on the
dice. These primes are 2, 3, 5, 7, 11.
The probability of rolling 2, 3, 5, 7, or 11 = the number of ways to roll any of these sums, divided by the
total number of possible rolls. The total number of possible die rolls is 6 × 6 = 36.
Sum of 2 can happen 1 way: 1 + 1
Sum of 3 can happen 2 ways: 1 + 2 or 2 + 1
Sum of 5 can happen 4 ways: 1 + 4, 2 + 3, 3 + 2, 4 + 1
Sum of 7 can happen 6 ways: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1
Sum of 11 can happen 2 ways: 5 + 6, 6 + 5
That's a total of 1 + 2 + 4 + 6 + 2 = 15 ways to roll a prime sum.
Thus, the probability is 15/36 = 5/12.

ScottTargetTestPrep · Answer

The prime numbers between 2 and 12, inclusive, are 2, 3, 5, 7, and 11. The possible pairings with a sum being a prime number are:

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

So the probability is 15/36 = 5/12.

Answer: B

BrushMyQuant · Answer

Two fair die with sides numbered 1 to 6 are tossed.

As we are rolling two dice => Number of cases =  6^2  = 36

What is the probability that the sum of the exposed faces on the die is a prime number?
  
Let's write down the cases in which Sum  = Prime
 Now, Prime numbers between 2 (1+1) and 12 (6+6) are 2, 3, 5, 7, 11

Following are the cases 
Sum = 2 -> (1,1)
Sum = 3 -> (1,2), (2,1)
Sum = 5 -> (1,4), (4,1), (2,3), (3,2)
Sum = 7 -> (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)
Sum = 11 -> (5,6), (6,5)

=>  Total cases  = 1 + 2 + 4 + 6 + 2 = 15

=>  P(Sum    = Prime)  =  \frac{15}{36}  =  \frac{5}{12}

So,  Answer will be B 
Hope it helps!

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

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

AkshayMetha · Answer

This approach to divide favourable sums by sample space of sums below does not consider weights. It considers that all sums from 2 (1+1) to 12 (6+6) are equally likely to occur. Whereas this is not the case as sum 2 or 12 have lesser chance/probability where as sum like 7 can be obtained more often when dices roll to (1,6) or (5,2) or (4,3). As such, I believe this approach shouldnt be used.

Two fair die with sides numbered 1 to 6 are tossed.

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Two fair die with sides numbered 1 to 6 are tossed.

Prep Toolkit

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