An ordinary (fair) die is a cube with the numbers

1

through

6

on the sides (represented by painted spots). imagine that such a die is rolled twice in succession and 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

9

.
event

b

: the sum is divisible by

5

or

6

(or both).

There are a few ways to look at this. If you wanted to look at all the events, you could draw a 6 x 6 grid and label the left side from 1 to 6 and the bottom from 1 to 6. The intersection of a row and a column would be a possible sum. You can check to see if the event you want to occur occurs on a cell by cell basis. Like if you wanted to see how many possibilities are there to get a sum of 12. We know there is only one way, that is to roll a 6 and a 6. Then you would shade the top right corner cell of the grid and check the other ones (for more complicated events). Then the probability is just the number of shaded cells divided by the total number of cells.

But, since you can probably just count the events, how many ways are there to roll any two numbers, if the order matters, there are 36 possibilities (6 x 6 grid). You could roll 4 then 6, 6 then 4, 5 then 5, 5 then 6, 6 then 5, and 6 then 6. There are 6 possibilities with a sum more than 9 and 36 total, so the probability of that one event is 6/36 or 1/6. Try out the last one ;)

Ithink the answer is c just trying to others