Question

DATA ANALYSIS AND STATISTICSProbabilities involving two rolls of a dieAn 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 9Event B: The sum is an even number.Write your answers as fractions.

Answer 1

Solution

The table below is the required sample space of the to fair die

From the above table

The sample space contain 36 outcomes

Event A: The sum is greater than 9

we will look at the table and count all the elements that are greater than 9

There are 6 elements (they are 10, 10, 10, 11, 11, 12 from the table)

The probability for event A will be

$\begin{gathered} p(A)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(A)=\frac{6}{36} \\ p(A)=\frac{1}{6} \end{gathered}$

P(A) = 1/6

Event B: The sum is an even number.

We will look at the table and count the number of elements that are even

There are 18 elements (notice that there are 3 even number on each of the 6 rows of the table)

The probability for event B will be

$\begin{gathered} p(B)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(B)=\frac{18}{36} \\ p(B)=\frac{1}{2} \end{gathered}$

p(B) = 1/2