What is the probability that when two dice are rolled the product of the numbers in the two dice is 6?

One popular way to study probability is to roll dice. A standard die has six sides printed with little dots numbering 1, 2, 3, 4, 5, and 6. If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on. But what happens if we add another die? What are the probabilities for rolling two dice?

To correctly determine the probability of a dice roll, we need to know two things:

  • The size of the sample space or the set of total possible outcomes
  • How often an event occurs

In probability, an event is a certain subset of the sample space. For example, when only one die is rolled, as in the example above, the sample space is equal to all of the values on the die, or the set (1, 2, 3, 4, 5, 6). Since the die is fair, each number in the set occurs only once. In other words, the frequency of each number is 1. To determine the probability of rolling any one of the numbers on the die, we divide the event frequency (1) by the size of the sample space (6), resulting in a probability of 1/6.

Rolling two fair dice more than doubles the difficulty of calculating probabilities. This is because rolling one die is independent of rolling a second one. One roll has no effect on the other. When dealing with independent events we use the multiplication rule. The use of a tree diagram demonstrates that there are 6 x 6 = 36 possible outcomes from rolling two dice.

Suppose that the first die we roll comes up as a 1. The other die roll could be a 1, 2, 3, 4, 5, or 6. Now suppose that the first die is a 2. The other die roll again could be a 1, 2, 3, 4, 5, or 6. We have already found 12 potential outcomes, and have yet to exhaust all of the possibilities of the first die.

The possible outcomes of rolling two dice are represented in the table below. Note that the number of total possible outcomes is equal to the sample space of the first die (6) multiplied by the sample space of the second die (6), which is 36.

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

The same principle applies if we are working on problems involving three dice. We multiply and see that there are 6 x 6 x 6 = 216 possible outcomes. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. For two dice, there are 62 possible outcomes. For three dice, there are 63 possible outcomes. In general, if we roll n dice, then there are a total of 6n possible outcomes.

With this knowledge, we can solve all sorts of probability problems:

1. Two six-sided dice are rolled. What is the probability that the sum of the two dice is seven?

The easiest way to solve this problem is to consult the table above. You will notice that in each row there is one dice roll where the sum of the two dice is equal to seven. Since there are six rows, there are six possible outcomes where the sum of the two dice is equal to seven. The number of total possible outcomes remains 36. Again, we find the probability by dividing the event frequency (6) by the size of the sample space (36), resulting in a probability of 1/6.

2. Two six-sided dice are rolled. What is the probability that the sum of the two dice is three?

In the previous problem, you may have noticed that the cells where the sum of the two dice is equal to seven form a diagonal. The same is true here, except in this case there are only two cells where the sum of the dice is three. That is because there are only two ways to get this outcome. You must roll a 1 and a 2 or you must roll a 2 and a 1. The combinations for rolling a sum of seven are much greater (1 and 6, 2 and 5, 3 and 4, and so on). To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18.

3. Two six-sided dice are rolled. What is the probability that the numbers on the dice are different?

Again, we can easily solve this problem by consulting the table above. You will notice that the cells where the numbers on the dice are the same form a diagonal. There are only six of them, and once we cross them out we have the remaining cells in which the numbers on the dice are different. We can take the number of combinations (30) and divide it by the size of the sample space (36), resulting in a probability of 5/6.

Probability means Possibility. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty.

The higher or lesser the probability of an event, the more likely it is that the event will occur or not respectively. For example – An unbiased coin is tossed once. So the total number of outcomes can be 2 only i.e. either “heads” or “tails”. The probability of both outcomes is equal i.e. 50% or 1/2.

So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event).

P(Event) = N(Favorable Outcomes) / N (Total Outcomes)

Note: If the probability of occurring of an event A is 1/3 then the probability of not occurring of event A is 1-P(A) i.e. 1- (1/3) = 2/3

What is Sample Space?

All the possible outcomes of an event are called Sample spaces.

Examples-

  • A six-faced dice is rolled once. So, total outcomes can be 6 and 
    Sample space will be [1, 2, 3, 4, 5, 6]
  • An unbiased coin is tossed, So, total outcomes can be 2 and 
    Sample space will be [Head, Tail]
  • If two dice are rolled together then total outcomes will be 36 and 
    Sample space will be  (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)    (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)   (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)   (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)   (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 

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

Types of Events

Independent Events: If two events (A and B) are independent then their probability will be

P(A and B) = P (A ∩ B) = P(A).P(B) i.e. P(A) * P(B)

Example: If two coins are flipped, then the chance of both being tails is 1/2 * 1/2 = 1/4

Mutually exclusive events:

  • If event A and event B can’t occur simultaneously, then they are called mutually exclusive events.
  • If two events are mutually exclusive, then the probability of both occurring is denoted as P (A ∩ B) and 
    P (A and B) = P (A ∩ B) = 0
  • If two events are mutually exclusive, then the probability of either occurring is denoted as P (A ∪ B) 
    P (A or B) = P (A ∪ B)    
                     = P (A) + P (B) − P (A ∩ B)         
                     = P (A) + P (B) − 0    
                     = P (A) + P (B)

Example: The chance of rolling a 2 or 3 on a six-faced die is P (2 or 3) = P (2) + P (3) = 1/6 + 1/6 = 1/3

Not Mutually exclusive events: If the events are not mutually exclusive then

P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B)

What is Conditional Probability?

For the probability of some event A, the occurrence of some other event B is given. It is written as P (A ∣ B)

P (A ∣ B) = P (A ∩ B) / P (B)

Example- In a bag of 3 black balls and 2 yellow balls (5 balls in total), the probability of taking a black ball is 3/5, and to take a second ball, the probability of it being either a black ball or a yellow ball depends on the previously taken out ball. Since, if a black ball was taken, then the probability of picking a black ball again would be 1/4, since only 2 black and 2 yellow balls would have been remaining, if a yellow ball was taken previously, the probability of taking a black ball will be 3/4.

What’s the probability of rolling a sum of 6 on two dice?

Solution:

When two dice are rolled together then total outcomes are 36 and 
Sample space is [ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)     (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)    (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)    (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)     (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)  

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

So, pairs with sum 6 are (1,5) (2, 4) (3, 3) (4, 2) (5, 1)  i.e. total 5 pairs

Total outcomes = 36
Favorable outcomes = 5

Probability of getting pair with sum 6 = Favorable outcomes / Total outcomes                                                                                                                          = 5 / 36 

So, P(6) = 5/36.

Similar Questions

Question 1: What is the probability of getting 6 on both dice?

Solution:

When two dice are rolled together then total outcomes are 36 and 
Sample space is [ (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)   (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)    (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)    (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)    (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 

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

So, pairs with both 6 are (6,6) i.e. only 1 pair

Total outcomes = 36
Favorable outcomes = 1

Probability of getting pair with both 6 = Favorable outcomes / Total outcomes                                                                                                                         = 1 / 36 

So, P(6,6) = 1/36

Question 2: What is the probability of getting pair with 6 on only one dice?

Solution:

When two dice are rolled together then total outcomes are 36 and 
Sample space is[ (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)    (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)    (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)    (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)   (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 

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

So, pairs with only one 6 are (1,6) (2,6) (3,6) (4,6) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5)  i.e. total 10 pairs

Total outcomes = 36
Favorable outcomes = 10

Probability of getting the pair with only one 6 = Favorable outcomes / Total outcomes                                                                                                                         = 10/36 = 5/18

So, P(pair with one 6) = 5/18

Question 3: What is the probability of getting a pair with at-least one 6 on two dices?

Solution:

When two dice are rolled together then total outcomes are 36 and 
Sample space is [ (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)    (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)   (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)   (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)     (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)  

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

So, pairs with at-least one 6 are (1,6) (2,6) (3,6) (4,6) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) i.e. 11 pairs

Total outcomes = 36
Favorable outcomes = 11

Probability of getting a pair with at-least one 6 = Favorable outcomes / Total outcomes                                                                                                                         = 11 / 36 

So, P(at-least one 6) = 11/36