Respuesta :
In order for the sum to be greater than 7, there are a few possible combinations which are shown below:
[tex]\text{(2,6); (3,6); (6,2);(6,3).}[/tex]The probability for each individual dice to roll a certain number is:
[tex]p\text{ = }\frac{1}{6}[/tex]If we want both to roll an exact number we have:
[tex]p(2\text{ and 6) = }\frac{1}{6}\cdot\frac{1}{6}\text{ = }\frac{1}{36}[/tex]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:
[tex]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}[/tex]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:
[tex]\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}[/tex]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:
[tex]p(\text{total) = 18}\cdot\frac{1}{36}\text{ = }\frac{18}{36}\text{ = }\frac{1}{2}[/tex]