Question

An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.Compute the probability of each of the following events.Event A : The sum is greater than 6.Event B : The sum is divisible by 2 or 3 (or both).Write your answers as fractions.

Answer 1

Explanation

So first of all we must find the probability that the sum of the values given by two dices is greater than 6. Here it's important to note that when we roll two dices we have two possible outcomes:

- The sum is greater than 6 which we'll call s > 6.

- The sum is equal or smaller than 6 which we'll call s ≤ 6.

Since those are the only possible outcomes the sum of their probabilities is equal to 1:

$P(s>6)+P(s\leq6)=1$

Which means that:

$P(s\gt6)=1-P(s\leqslant6)$

So we can find the probability of s > 6 (which is the probability of Event A) using that of s ≤ 6. This probability is given by the quotient between these two quantities:

- Number of possible combinations where the sum of the dices is equal or smaller than 6.

- Total number of possible combinations.

Now we are going to find the first number. Let's use a notation (a,b) where the first number indicates the result of the first die and the second number indicates that of the second die. For example, (1,2) means that the result of the first roll is 1 and that of the second is 2.

Now that we have defined the notation let's list all the possible combinations with a sum equal or smaller than 6:

(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2) and (5,1).

So we have 15 possible combinations where s ≤ 6.

The total number of possible combinations is 36 because we are rolling a die with 6different values twice and 6²=36.

Then the probability that s ≤ 6 is:

$P(s\leq6)=\frac{15}{36}$

So the probability that s > 6 is given by:

$\begin{gathered} P(s\gt6)=1-P(s\leqslant6)=1-\frac{15}{36} \\ P(s\gt6)=\frac{21}{36}=\frac{7}{12} \end{gathered}$

We also need to find the probability that the sum is divisible by 2 or 3 or both. We should list all the possible values of the sum of both dices first. These are:

2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12

Now let's divdide this list in two: numbers that are divisible by 2 or 3 or both and those that are not.

For the first condition we get:

2, 3, 4, 6, 8, 9, 10 and 12

And for the second:

5, 7 and 11

We must find the probability P(1) that the sum of the two rolls is a number of the first group. However, it will be easier to find the probability that the sum is a number from the second group P(2) since it has less elements. Just like before we have only two possible outcomes: the sum is a number from the first group or it's a number from the second group. Then we get:

$\begin{gathered} P(1)+P(2)=1 \\ P(1)=1-P(2) \end{gathered}$

So let's find P(2) i.e. the probability that the sum of the two rolls is either 5, 7 or 11. We first need to find all the possible combinations that give a sum equal to any of these three values. These combinations are:

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

These are 12 combinations and we find P(2) just like we found P(s ≤ 6) before:

$P(2)=\frac{12}{36}=\frac{1}{3}$

So P(1) is:

$P(1)=1-\frac{1}{3}=\frac{2}{3}$Answer

Then the probabilities of events A and B are:

$\begin{gathered} P(A)=\frac{7}{12} \\ P(B)=\frac{2}{3} \end{gathered}$