Oftentimes, math problems will require that you know more than one concept. Sometimes, you’ll have to know the difference between an integer and a number in order to get the question right. Below, I’ve combined prime numbers with probability, two subjects already covered in this blog.

The question below isn’t easy and actually takes a little bit of work. Let’s see if you can crack it.

What is the probability that the sum of two rolled dice will equal a prime number?

(A) 1/3

(B) 5/36

(C) 2/9

(D) 13/36

(E) 5/12

First let’s list the prime numbers that pertain to this problem.

The pertinent prime numbers are 2, 3, 5, 7, and 11. Notice I stopped at 11. Why? Well, the greatest number you can roll on two six-sided dice is 12.

Next, we have to remember this is a probability question. Therefore, we have to divide the number of total outcomes by the number of desired outcomes. First, let’s find the number of desired outcomes. To do this, we have to make sure that each desired outcome conforms to the problem, i.e. how many different ways can we sum two dice to get a prime.

Let’s start with 2. There is only one way to roll a 2, and that is with a 1 and a 1. Therefore, we have one desired outcome.

What about the number 3? Well, we can get 2,1 and 1,2, or two possible ways.

Mind you, to do this problem you will have to make sure you write down each outcome. Do not try to do this in your head, for you’ll most likely get the problem wrong and induce dizziness.

Next we have the number 5, which we can get by rolling the following combinations:

1,4

4,1

2,3

3,2

We add these four possible outcomes to the prior three, giving us a total of 7.

Next, we look to see which numbers sum to 7 and find a total of six possibilities.

1,6

2,5

3,4

4,3

5,2

6,1

Our total is now 13.

Finally, don’t forget 11 as final prime number.

6,5

5,6

We add these two possibilities to the 13 possibilities, giving us a total of 15.

Finally, we want to find the total number of outcomes. Remember, we divided the total outcomes by the total number of desired outcomes, which we just found out was 15.

The total number of outcomes equals the number of different ways we can roll two six-sided dice. We find this number by multiplying 6 x 6. The logic is there are six sides to each die, so for each number on one die you can pair with six different numbers on the other die.

Therefore, the probability of rolling a prime number on two dice is 15/36, which reduces to 5/12 (E).

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why do you make your questions so hard? i dont feel like I’m actually getting any realistic practice in because I’ll never get up to that level and moreover, that’s not even my intention. I just want to not fail the math section entirely… but I do your video lessons and I understand everything but then I try to look into this blog and its totally wtf and incredibly confusing. its like asking a kindergartner to perform brain surgery this is depressing

That’s definitely not our intention here :). Usually when there is a tough question, I mark it as challenge. This question is tough, esp. because it includes probability, which is not an easy subject to wrap one’s head around.

So you’re right–you want to focus on the more basic math to make sure you get a math score that is right for you. I’d avoid combinations/permutations, rates, and probability questions altogether, as they only show up in harder questions. Even you encounter one in the easy/medium sections you can just skip it and focus on what you do know.

With this blog, if there is a tough question, just ignore it. Go through the concepts in our lessons videos and work through problems at your level (including questions in the official guide). Get good at the basics and you’ll do well test day :).

With this question make sure you make it as simple as possible:

Prob (not a multiple of 3) x Prob (not a multiple of 3) = 2/3 x 2/3 = 4/9.

With your solution, you forgot to consider the cases in which you roll a multiple of 3 in one toss and not a multiple of 3 in the other toss. You did the math for the probability of rolling a dice twice and getting a multiple of 3 on both rolls.

Hey Chris ,
The problem says rolling 2 die gives.you prime no …. Guess it should be like the sum of numbers ,as one can interpret it as getting prime no in 1 st die and prime no in 2 nd die….

My apologies: I see why 5/11 is wrong. I simply used a matrix and circled all prime sums, and, moreover, you can see that the denominator must be 36 or some simplified form of 36, and you can’t get 11 from 36. I was looking for a faster way to solve the problem (I don’t like listing and counting very much, and I’m always thinking the GRE has a shortcut in mind that the test-makers want you to see). Though, I do wonder why my answer was so close to the actual solution, given that I was employing some faulty math/logic…

It is interesting that your initial answer was close to the actual one. It looks like it is based on coincidence more than anything else.

I don’t think there is a shorter way to do it. However, counting the sums doesn’t take too long. Even ‘7’, which has the most, you can quickly surmise has six different ways to roll 1+6, 2+5….6+1.

In general, if you have an approach that you think will take you to the solution, don’t hold out for some magic bullet approach that you may or may not see, one that may not even be there.

Thanks for your response. You’re right: the problem doesn’t take too long to solve that way.

I appreciate your tip on the “magic bullet” approach. I think what I’ve seen so far in my GRE Math preparation, particularly on Magoosh, has just conditioned me to almost always look for a shortcut if the question looks like it is going to require a good amount of scrap paper (even if not too much time) to solve by the most apparent means. I’ll try to be more cognizant of my approach in the future…

You bring up another good point – I think that most problems on Magoosh and just about every prep site, are biased towards problems that lend themselves to shortcuts. For one, they allow us – the content writers – to show our magic. Pedagogically, teaching shortcuts helps tune students students’ brains to a different way of thinking, one that we typically don’t learn in school.

On the whole I would say most GRE questions offer a short cut, but the so-called longer way still allows one to solve the problem in about a minute.

Thanks for the question/solution. I know this question is from quite a while ago, but hopefully you will see this post.

I could be totally off base here for some reason unbeknownst to me (Counting/Probability questions are not my forte), but this seems like a relatively straightforward solution using the Basic Probability Formula:

P(the sum of the two dice rolls is a prime #) = # of two dice rolls where sum is prime
———————————————–
total # of sums from two dice rolls

The numerator must be 5 because there are only 5 prime sums, as you mention: 2, 3, 5, 7, 11. It seems to me that the denominator must be 11, since there are only 11 possible sums of two dice rolls: 2-12. The answer would therefore be 5/11, which is close to the solution above (5/12). One might perhaps get 5/12 if they thought that there were 12 possible sums–1 through 12–but 2 is the lowest sum since the lowest you can roll on either dice is 1, and 1+1=2 (math revelation of the day).

Please let me know if there is something I am not seeing here.

I think the explanation above is a little misleading. The way I’ve written it out makes it appear exactly as you mentioned – long and tedious (and definitely not pragmatic from a testing standpoint). However, doing the problem above doesn’t take too long. For instance for the number 5 you can figure out that there are 4 ways (1,4, 2,3, 3,2, 4,1), or one less than 5. For the number 7, 6 ways (1,6….6,1). The other prime numbers have less possibilities and are easier to figure out. Add them together and divide by 36. There’s your answer.

Unfortunately, one can’t use combinations because you are not picking all the possibilities of 2 rolls from 6. Also, order is important (1,6 is different from 6,1). Even permutations wouldn’t work because you are not choosing total possible combinations, i.e. 6,4 or 3,5 are not valid dice rolls because they do not yield prime numbers.

Hi Chris.
I was looking at this and was wondering if there is an easier way because this would take quite a bit of time on the test. I understand that the prime number 2, 3, 5, 7, 11 are only possible and i believe it is a combination question because order does not matter. So is there a way to use the combination formula ((n!)/(r!)(n-r)) for this question? Thanks!

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What is the probability that the total score is prime when 3 dices are rolled?

Is there any shortcut to solve these kinds of sums?

This was very helpful. Thank you.

why do you make your questions so hard? i dont feel like I’m actually getting any realistic practice in because I’ll never get up to that level and moreover, that’s not even my intention. I just want to not fail the math section entirely… but I do your video lessons and I understand everything but then I try to look into this blog and its totally wtf and incredibly confusing. its like asking a kindergartner to perform brain surgery this is depressing

Hi Liz,

That’s definitely not our intention here :). Usually when there is a tough question, I mark it as challenge. This question is tough, esp. because it includes probability, which is not an easy subject to wrap one’s head around.

So you’re right–you want to focus on the more basic math to make sure you get a math score that is right for you. I’d avoid combinations/permutations, rates, and probability questions altogether, as they only show up in harder questions. Even you encounter one in the easy/medium sections you can just skip it and focus on what you do know.

With this blog, if there is a tough question, just ignore it. Go through the concepts in our lessons videos and work through problems at your level (including questions in the official guide). Get good at the basics and you’ll do well test day :).

. What is the probability to get a prime number that is even at the same time if you roll the dice?

Hi Juan,

The only number that is even and a prime is ‘2’. The probability of rolling a ‘2’, if you roll one die, is 1/6.

Hope that helps!

And if the question is that a die is rolled . If the outcome is an even no . , what is the probability that it is a prime number ?

Hi,

I have a confusion in the following question:

A dice is rolled twice, the probability that neither role is a multiple of 3.

I’m not getting the answer through this method,

P(not a multiple of 3)= 1-p(multiple of 3)

1- (2/6)(2/6)

= 8/9

why I’m not getting 4/9 which is the correct probability ?

Muhammed,

With this question make sure you make it as simple as possible:

Prob (not a multiple of 3) x Prob (not a multiple of 3) = 2/3 x 2/3 = 4/9.

With your solution, you forgot to consider the cases in which you roll a multiple of 3 in one toss and not a multiple of 3 in the other toss. You did the math for the probability of rolling a dice twice and getting a multiple of 3 on both rolls.

Hope that helps

You can calculate separately.

First find

P(multiple of 3)=2/6=1/3

P(not a multiple of 3)=1-13/1=2/3

Again do the same

P(multiple of 3)=2/6=1/3

P(not a multiple of 3)=1-13/1=2/3

and now final result is 2/3*2/3. =4/9

Yes, that also works! Thanks for sharing your solution :).

Hey Chris ,

The problem says rolling 2 die gives.you prime no …. Guess it should be like the sum of numbers ,as one can interpret it as getting prime no in 1 st die and prime no in 2 nd die….

Yes, you are right – that is a little ambiguous. I’ve corrected it with the word ‘sum’.

Chris,

My apologies: I see why 5/11 is wrong. I simply used a matrix and circled all prime sums, and, moreover, you can see that the denominator must be 36 or some simplified form of 36, and you can’t get 11 from 36. I was looking for a faster way to solve the problem (I don’t like listing and counting very much, and I’m always thinking the GRE has a shortcut in mind that the test-makers want you to see). Though, I do wonder why my answer was so close to the actual solution, given that I was employing some faulty math/logic…

Hi Sam!

It is interesting that your initial answer was close to the actual one. It looks like it is based on coincidence more than anything else.

I don’t think there is a shorter way to do it. However, counting the sums doesn’t take too long. Even ‘7’, which has the most, you can quickly surmise has six different ways to roll 1+6, 2+5….6+1.

In general, if you have an approach that you think will take you to the solution, don’t hold out for some magic bullet approach that you may or may not see, one that may not even be there.

Hope that was helpful!

Chris,

Thanks for your response. You’re right: the problem doesn’t take too long to solve that way.

I appreciate your tip on the “magic bullet” approach. I think what I’ve seen so far in my GRE Math preparation, particularly on Magoosh, has just conditioned me to almost always look for a shortcut if the question looks like it is going to require a good amount of scrap paper (even if not too much time) to solve by the most apparent means. I’ll try to be more cognizant of my approach in the future…

Sam

Sam,

I’m glad my advice was helpful.

You bring up another good point – I think that most problems on Magoosh and just about every prep site, are biased towards problems that lend themselves to shortcuts. For one, they allow us – the content writers – to show our magic. Pedagogically, teaching shortcuts helps tune students students’ brains to a different way of thinking, one that we typically don’t learn in school.

On the whole I would say most GRE questions offer a short cut, but the so-called longer way still allows one to solve the problem in about a minute.

Chris,

Thanks for the question/solution. I know this question is from quite a while ago, but hopefully you will see this post.

I could be totally off base here for some reason unbeknownst to me (Counting/Probability questions are not my forte), but this seems like a relatively straightforward solution using the Basic Probability Formula:

P(the sum of the two dice rolls is a prime #) = # of two dice rolls where sum is prime

———————————————–

total # of sums from two dice rolls

The numerator must be 5 because there are only 5 prime sums, as you mention: 2, 3, 5, 7, 11. It seems to me that the denominator must be 11, since there are only 11 possible sums of two dice rolls: 2-12. The answer would therefore be 5/11, which is close to the solution above (5/12). One might perhaps get 5/12 if they thought that there were 12 possible sums–1 through 12–but 2 is the lowest sum since the lowest you can roll on either dice is 1, and 1+1=2 (math revelation of the day).

Please let me know if there is something I am not seeing here.

Thanks!

Very well explained , Chris ..thanks !

You’re welcome!

A tricky problem, esp. from the angle of thinking why permutations/combinations wouldn’t work.

I think the explanation above is a little misleading. The way I’ve written it out makes it appear exactly as you mentioned – long and tedious (and definitely not pragmatic from a testing standpoint). However, doing the problem above doesn’t take too long. For instance for the number 5 you can figure out that there are 4 ways (1,4, 2,3, 3,2, 4,1), or one less than 5. For the number 7, 6 ways (1,6….6,1). The other prime numbers have less possibilities and are easier to figure out. Add them together and divide by 36. There’s your answer.

Unfortunately, one can’t use combinations because you are not picking all the possibilities of 2 rolls from 6. Also, order is important (1,6 is different from 6,1). Even permutations wouldn’t work because you are not choosing total possible combinations, i.e. 6,4 or 3,5 are not valid dice rolls because they do not yield prime numbers.

Hope that helps!

Yeah this helps a bunch, thanks!

Hi Chris.

I was looking at this and was wondering if there is an easier way because this would take quite a bit of time on the test. I understand that the prime number 2, 3, 5, 7, 11 are only possible and i believe it is a combination question because order does not matter. So is there a way to use the combination formula ((n!)/(r!)(n-r)) for this question? Thanks!