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### Probability - Throwing Two Dice

As explained earlier, there are a total of 36 possible outcomes when you throw two dice once, and the outcome of each unique pair of numbers, for instance, 1 on the first and 2 on the second dice, is 1/36.

Different types of questions on probability can be based on throwing two dice. Mainly, these are:
• Probability of getting a particular sum
• Probability of getting a sum greater/lesser than a particular value
• Probability of getting particular numbers on both dice
• Problems involving conditional probability
Each of these question types is discussed below:

### Getting a particular sum

Suppose, on throwing the two dice, the first dice shows the number 4 while the second one shows the number 6. The sum of these two numbers is 10. How many other such possible pairs of numbers are possible on the two dice to get a sum of 10? These are given in the table below:

 dice 1 dice 2 Sum 4 6 10 6 4 10

 dice 1 dice 2 Sum 2 6 8 3 5 8 4 4 8 5 3 8 6 2 8

 Sum Number of favorable outcomes 2 1 3 2 4 3 5 4 6 5 7 6 8 5 9 4 10 3 11 2 12 1

Thus, only two outcomes out of the possible thirty six outcomes show a sum of 10. This means that total number of outcomes favorable is two.  Hence we can say that the probability of getting a sum of 10 on the throw of two dice is:

P("sum of 10") =  "number of favorable outcomes"/"total number of possible outcomes"

P("sum of 10")  = 2/36 = 1/18

Let us consider another sum, eight. How many pairs of numbers can you come up with, both between 1 and 6, that add up to eight? The following table shows the numbers:
There are a total of five outcomes which give you a number of eight. Hence total number of favorable outcomes in this case is five. Probability of getting a sum of eight is given by:

P("sum of 8") =  "number of favorable outcomes"/"total number of possible outcomes"

P("sum of 8")  = 5/36

Similarly you can come up with the probabilities of getting other sums ranging from 2 to 12 (since 2 is the least possible sum and 12 is the greatest). A patter is obvious from the table below which specifies the total number of pairs of numbers on two dice that gives each sum. The number of outcomes increases from 1 to 6 and then decreases to 1 again from the sum of 7.

### Getting sum greater than or lesser than a given number

#### What is the probability of getting a sum greater than or equal to ten on the throw of two dice?

When throwing two dice, total number of pairs of numbers that can come up is 36. Out of these 36 outcomes, total number of outcomes which give a sum of 10 is 3 (refer to the table above), total number of outcomes which give a sum of 11 is 2 and total number of outcomes which give a sum of 12 is 1. Hence total number of outcomes which give a sum of 10 or greater is 3 + 2 + 1 = 6. Thus the number of outcomes favorable is 6. By using the theoretical probability formula, we get
P("outcome") = "number of favorable outcomes"/"total number of possible outcomes"
P("sum greater than or equal to 10")= 6/36 = 1/6

### Getting particular numbers on both the dice

#### What is the probability of getting a 2 on the first dice and a 5 on the second one?

As is clear from the section on throwing a single dice above, the probability of getting any given number on a single throw of a dice is equal to 1/6. Hence the probability of getting a 2 on the first dice is 1/6 and that of getting a 5 on the second dice is 1/6 as well. But what is the probability of getting this combination of numbers, 2 and 5, on the two dice respectively, out of all the other possible pairs of numbers possible?

Notice that whatever number you get on the first dice, the probability of getting a particular number on the second dice does not change. It always remains 1/6. Thus the two outcomes - one, throwing the first dice, and two, throwing the second dice - are independent of each other.

Whenever you have to find the probability of two independent outcomes occurring together, you multiply their respective probabilities.

Thus the probability of getting 2 on the first dice and 5 on the second dice is equal to
P("2 on 1st and 5 on 2nd dice") = 1/6 * 1/6 = 1/36
There is another way of solving the above problem. Out of the 36 possible outcomes when throwing two dice, only one outcome is such that a 2 shows up on the first dice and a 5 shows up on the second dice. Hence by the formula of theoretical probability:
P("outcome") =  "number of favorable outcomes"/"total number of possible outcomes"
P("2 on 1st and 5 on 2nd dice") = 1/36
Modifying this question a bit, what is the probability of getting the pair of numbers 2 and 5 when throwing two dice?

Now this question asks us to find the probability of getting a 2 and a 5 when throwing two dice; it does not tell us whether we expect to get a 2 on the first dice and 5 on the second dice or 2 on the second dice and 5 on the first dice. Thus, we have to take into account both the cases.

We know that there are 36 possible outcomes when two dice are thrown. Out of these 36 possible outcomes there is only one outcome in which a 2 shows up on the first dice and a 5 shows up on the second dice. Furthermore, there is another outcome in which you get 5 on the first dice and 2 on the second dice. Thus, there are two favorable outcomes which give us the pair of numbers 2 and 5.

By the formula for theoretical probability:
P("outcome") =  "number of favorable outcomes"/"total number of possible outcomes"
P("2 and 5") = 2/36 = 1/18
This question can be extended to getting particular number sets on both dices, which is given in the worksheet at the end.

### Problems Involving Conditional Probability

#### What is the probability of getting an even number on the second dice if it is given that the first dice shows up the number 4?

In conditional probability the total number of possible outcomes is limited by some constraint. Here the constraint is that the first dice shows up a 4 necessarily. Out of the 36 possible outcomes possible in the throw of two dice, the first dice shows up 4 in only six outcomes. Hence the total number of possible outcomes here is lessened from 36 to just 6. These are:

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

Now out of these six outcomes, how many does the second dice show an even number on it?

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

Easily, the highlighted ones above show an even number on the second dice. There are three such possible outcomes in which the second dice shows a six on it while the first one shows the number 4. Hence the total number of favorable outcomes are 3.

Thus, by the formula of theoretical probability, we get:
P("outcome") = "Number of favorable outcomes"/"Total number of outcomes"
P("even number on second dice"/"first dice shows 4") = 3/6 = 1/2
The major thing to keep in mind for these type of questions is that normally you would have solved it like this (this is the incorrect way but is given below to clarify the use of conditional probability here)

Probability of getting a four on the first dice and an even number on the second dice is equal to
P("outcome") = "Number of favorable outcomes"/"Total number of outcomes"
Number of favorable outcomes is three since only three outcomes are there in which the first number is 4 and the second number is even: (4, 2), (4, 4) and (4, 6)
Total number of outcomes is 36 since you get 36 different number pairs on throwing two dice.
So
P("4 on first dice and even on second dice") = 3/36 = 1/12
The above method of solving it seems correct but it is incorrect since we did not use conditional probability here. The problem specifically states that it is "given that the first dice shows a 4". Whenever there is a statement like this one, in which the problem specifically tells you of one condition, the question is generally based on conditional probability.

Now, what is the meaning of conditional probability? Conditional probability is the probability of one outcome given that the another outcome yielded some specific outcome given in the problem. In the problem discussed above, the specific outcome given is that you get a four on the first dice. This implies that you have to narrow the sample space to include only those outcomes in which the given condition is satisfied. That is what we did in the first approach: We narrowed the total number of possible outcomes, which is 36 for two dice, to only six outcomes in which the number four shows up on the first dice. Hence, in the approach involving conditional probability, the total number of possible outcomes is narrowed down to include the given specific condition.

These where the main types of problems of probability that can be made on throwing two dice. If you have any other question on it, please post it in the comments. If it seems alright then we may try to include it in this or another post.

Thank you.