What is the probability that the sum of two dice is prime?

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Answer

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Hint: Calculate the number of possible outcomes for throwing two dice. Calculate the number of favourable outcomes for each of the cases. Use the fact that the probability of any event is the ratio of the number of favourable outcomes and the number of possible outcomes to calculate the probability of each of the events.

Complete step-by-step solution -

We have to calculate the probability of each of the events when two dice are thrown.We know that the total number of possible outcomes when two dice are thrown is $=6\times 6=36$.We know that the probability of any event is the ratio of the number of favourable outcomes and the number of possible outcomes.We will now calculate the probability of events in each case.(a) We have to calculate the probability that the sum of digits is a prime number.We will draw a table showing the sum of digits on rolling both the dice.

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

We observe that the possible values of prime numbers when two digits on the dice are added are 2, 3, 5, 7, and 11.We observe that 2 occurs only once, 3 occurs 2 times, 5 occurs 4 times, 7 occurs 6 times and 11 occurs 2 times.The number of favourable outcomes is the sum of occurrences of all the favourable outcomes. So, the number of favourable outcomes $=1+2+4+6+2=15$.We know that the number of possible outcomes is 36.Thus, the probability of getting the sum of two numbers as prime numbers is $=\dfrac{15}{36}=\dfrac{5}{12}$.(b) We will now calculate the probability of occurrence of a doublet of an even number.We know that the favourable outcomes are (2, 2), (4, 4), and (6, 6).So, the number of favourable outcomes is 3.We know that the number of possible outcomes is 36.Thus, the probability of getting a doublet of an even number is $=\dfrac{3}{36}=\dfrac{1}{12}$.(c) We will calculate the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice.Possible multiples of 2 on dice are 2, 4, and 6.Possible multiples of 3 on dice are 3 and 6.The possible outcomes for multiples of 2 on one dice and multiple of 3 on other dice are (2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (6, 2), (3, 4), (6, 4), and (3, 6).So, the number of favourable outcomes is 11.We know that the number of possible outcomes is 36.Thus, the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice is $=\dfrac{11}{36}$.(d) We will calculate the probability of getting a multiple of 3 as a sum of digits on both the dice.We will draw the table showing possible values of the sum of digits on both the dice.

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

The possible values of multiples of 3 as a sum of digits on dice are 3, 6, 9, and 12.We observe that 3 occurs 2 times, 6 occurs 5 times, 9 occurs 4 times and 12 occurs once.The number of favourable outcomes is the sum of occurrences of all the favourable outcomes. So, the number of favourable outcomes $=2+5+4+1=12$.We know that the number of possible outcomes is 36.Thus, the probability of getting multiples of 3 as a sum of digits on dice is $=\dfrac{12}{36}=\dfrac{1}{3}$.Note: We must calculate the number of favourable and possible outcomes in each case to calculate the probability of each of the given events. We should also be careful that we don’t count the same event repeatedly or we miss some event.

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

(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).

Chris Lele is the Principal Curriculum Manager (and vocabulary wizard) at Magoosh. Chris graduated from UCLA with a BA in Psychology and has 20 years of experience in the test prep industry. He's been quoted as a subject expert in many publications, including US News, GMAC, and Business Because. In his time at Magoosh, Chris has taught countless students how to tackle the GRE, GMAT, SAT, ACT, MCAT (CARS), and LSAT exams with confidence. Some of his students have even gone on to get near-perfect scores. You can find Chris on YouTube, LinkedIn, Twitter and Facebook!

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