Question

An ordinary (fair) dle 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 7.Event B: The sum is not divisible by 2.Write your answers as exact fractions.(a) P(A) = 0Da5?(b) P(B) = 1

Answer 1

In order for the sum to be greater than 7, there are a few possible combinations which are shown below:

$\text{(2,6); (3,6); (6,2);(6,3).}$

The probability for each individual dice to roll a certain number is:

$p\text{ = }\frac{1}{6}$

If we want both to roll an exact number we have:

$p(2\text{ and 6) = }\frac{1}{6}\cdot\frac{1}{6}\text{ = }\frac{1}{36}$

Any of the four probabilities can happen for our requirements to be met, therefore the total probability of at least one of them happening is the sum of all the probabilities:

$p(\text{total) = }\frac{1}{36}\text{ + }\frac{1}{36}\text{ + }\frac{1}{36}\text{ + }\frac{1}{36}\text{ = }\frac{4}{36}\text{ = }\frac{1}{9}$

For the sum of the dices to be not be divisible by 2, the result must be an odd number. The combinatios that could result in that are:

$\begin{gathered} (1,\text{ 2), (1,4), (1, 6)} \\ (2,1),\text{ (2,3), (2,5)} \\ (3,2),\text{ (3,4), (3,6)} \\ (4,1),\text{ (4,3), (4,5)} \\ (5,2),\text{ (5,4), (5,6)} \\ (6,1),\text{ (6,3), (6,3)} \end{gathered}$

Which gives us a total of 18 possibilities. Since the probability of each of this happening is equal to 1/36, then the total probability is given by 18 times that, which results in the below:

$p(\text{total) = 18}\cdot\frac{1}{36}\text{ = }\frac{18}{36}\text{ = }\frac{1}{2}$