Question

involving two rolls of a dieESEAn ordinary (falr) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a dle is rolled twice in successionand 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 8.Event B: The sum is an odd number.Write your answers as fractions.Ola(a) P(A) = 1х5?(b) P(B) = 0

Answer 1

`\begin{gathered} a)\text{ }(5)/(18) \\ \\ b)\text{ }(1)/(2) \end{gathered}`

Step-by-step explanation:

Here, we want to compute some probabilities

The first thing to do is to get the count of results in our sample space

In the sample space, the total possible results is 36

Now, let us get the probabilities

a) The event that the sum is greater than 8

We have to count possible results greater than 8 here

3, 6 (3 on the first die, 6 on the second)

6,3 (6 on the first die, 3 on the second)

6,4 (6 on the first die, 4 on the second)

4,6

4,5

5,4

5,5

5,6

6,5

6,6

The number of possible results greater than 8 is 10

Thus, we have the probability as the count of this divided by the total number of possible results

Mathematically, we have that as:

`(10)/(36)\text{ = }(5)/(18)`

b) The sum is an odd number

For the sum to be an odd number, we know that if we add a table of 6 rows for all the sums, the even sum on each line is 3

The total even sum is 6 * 3 = 18

The probability is thus:

`(18)/(36)\text{ = }(1)/(2)`