When we throw 2 dice together find the probability of getting I the sum of 11 of both the numbers which occurs on the both dice?

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 is a measure of the possibility of how likely an event will occur. It is a value between 0 and 1 which shows us how favorable is the occurrence of a condition. If the probability of an event is nearer to 0, let’s say 0.2 or 0.13 then the possibility of its occurrence is less. Whereas if the probability of an event is nearer to 1, lets say 0.92 or 0.88 then it is much favourable to occur.

Probability of an event

The probability of an event can be defined as a number of favorable outcomes upon the total number of outcomes.

P(A) = Number of favorable outcomes / Total number of outcomes

Some terms related to probability

• Experiment: An experiment is any action or set of action performed to determine the probability of an event. The result of action performed is random or uncertain. e.g. Tossing a coin, rolling dice, etc.
• Event: An event can be defined as certain condition which can happen while performing an experiment. e.g. getting head while tossing a coin, getting even number while rolling dice, etc.
• Sample Space: It is set of all the possible outcomes which happens after performing an experiment. e.g. Sample Space of tossing a coin = {H,T} and Sample Space of rolling a dice = {1,2,3,4,5,6}, so on.
• Sample Point: It is a part of sample space which contains one of the outcomes from Sample Space. e.g. Getting 1 while rolling dice, getting an ace of Spades while drawing a card from pack of cards, etc.
• Types of Events: There are majorly four kinds of events that are-
  • Complimentary events- It is used to find probability of not happening of an event. It is denoted by ( ‘ ) symbol. If event is denoted by A, then complimentary of event is A’. e.g. probability of not getting 2 while rolling a dice. It can be calculated by subtracting normal probability from 1 i.e. P(A’) = 1 – P(A)
  • Impossible event- Impossible event is a type of event which can never happen. The probability of Impossible event is 0. e.g. getting a number 8 while rolling a dice.
  • Certain event- Certain event is a type of event which always happen. The probability of a certain event is 1. e.g. getting a head or a tail after tossing a coin.
  • Equally Likely events- Events whose probability of occurrence are equal i.e. they are equally likely to happen. The value of probability of such events are same. e.g. getting a head and getting a tail both have 50% probability.

When two dice are rolled what is the probability of getting same number on both?

Since, the number of outcomes while rolling a dice = 6

Number of outcomes while rolling two dice = 62

= 36

The Sample Space for rolling a die is given as,

{(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)}

Sample points of getting same number on both dice- (1,1) ,(2,2) ,(3,3) ,(4,4) ,(5,5) & (6,6).

Thus, the number of favourable outcomes = 6

Total number of outcomes = 36

P (getting same number on both dice) = 6/36

= 1/6

Hence, the probability of getting same number on both the dice is 1/6.

Sample Questions

Question 1: Find the probability of getting odd number on first dice and even number on other dice when two dice are thrown simultaneously.

Answer: 

Total number of outcomes = 36

Sample Space :

{(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)}

In 9 outcomes we will get odd number on first dice and even number on second dice.

So, required probability is 9/36 = 1/4

Question 2: If two dice are thrown together then find the probability of getting 1 or 2 on either of the dice.

Answer: 

Total number of outcomes = 36

Sample Space :

{(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)}

From above sample space it is clear that there are total 20 possibilities in which 1 or 2 appears on either of the dice.

So, required possibility = 20/36 = 5/9

Question 3: In an event 2 dice are thrown simultaneously. Find the probability of getting prime number on first dice.

Answer:

Total number of possibilities = 36

Sample Space :

{(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)}

Since, 2, 3 and 5 are prime number which appear on first dice in 2nd, 3rd and 5th row of sample space respectively.

So, number favourable sample points = 18

Required probability = 18/36

Question 4: Three coins are tossed together find the probability of getting at least one head and one tail.

Answer: 

Number of possibilities while tossing a coin = 2

Number of possibilities while tossing 3 coins together = 23

                                                                                    = 8

Sample Space : 

{ (H,H,H) , (H,H,T) , (H,T,H) , (H,T,T) , 

(T,H,H) , (T,H,T) , (T,T,H) , (T,T,T) }

6 sample points are having head and tail both.

P(E) = 6/8 

= 3/4

Question 5: Find the probability of getting at least two tails when a coin is tossed three times.

Answer: 

Total number of outcomes = 8

Sample Space :

{(H,H,H) , (H,H,T) , (H,T,H) , (H,T,T) ,

(T,H,H) , (T,H,T) , (T,T,H) , (T,T,T)}

Number of favourable outcomes = 4

Probability = 4/8

= 1/2