Question

An ordinary (fair) die is a cube with the numbers through 6 on the sides (represented by painted spots). Imagine that such a die 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 6.Event B: The sum is an even numberWrite your answers as exact fractions.(a) P (A) = 1Х?(b) P(B) = 0

Answer 1

When a fair die is rolled, it's probability of landing on any side is the same.

To calculated the probability P(A) and P(B), Let us write out the possible outcomes of the event that the die was rolled twice. In each roll we can have the possible outcome of 1 to 6. The set of possible outcomes for the two rolls are shown below.

$\lbrace(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,5),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

The above are the possible outcomes of the two rolls of die.

The sum of each outcome are;

$\left\lbrace 2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\right\rbrace$

To get the value of P(A), Let's count the outcomes that have their sum greater than 6 and divide by the total;

$\begin{gathered} P(A)=\frac{n(A)}{n(T)}=\frac{21}{36} \\ P(A)=\frac{21}{36} \end{gathered}$

Then for P(B), Let's count the outcomes that have their sum even and divide by the total;

$\begin{gathered} P(B)=\frac{n(B)}{n(T)}=\frac{18}{36} \\ P(B)=\frac{18}{36} \end{gathered}$

Therefore;

$\begin{gathered} P(A)=\frac{21}{36} \\ P(B)=\frac{18}{36} \end{gathered}$