help! Solve this Algebra 2 problem??

Question

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 6 . Event B : The sum is divisible by 4 or 6 (or both). Write your answers as exact fractions.

Answer 1

First of all, each roll is independent, so whatever happens on the first roll has nothing to do with the second.

 

So that means there 6 different possible outcomes for the first roll and 6 for the second roll, for a total of 36 different combinations.  Each combination is equally likely.

 

You can make a chart of the outcomes, something like this:

 

1/1 -- sum = 2

1/2 -- sum = 3

...

6/6 -- sum = 12

 

If you go through the list, you can determine which combinations qualify as part of event A and which qualify for event B. 

 

You'll see that the following combinations have sum greater than 6:

 

1/6, 2/5, 2/6, 3/4, 3/5, 3/6, 4/3, 4/4, 4/5, 4/6, 5/2, 5/3, 5/4, 5/5, 5/6, 6/1, 6/2, 6/3, 6/4, 6/5, 6/6

 

That's 21 combinations, so the probability of Event A is 21/36 = 7/12

 

The following combinations have a sum divisible by 4 and/or 6:

 

1/3, 1/5, 2/2, 2/4, 2/6, 3/1, 3/3, 3/5, 4/2, 4/4, 5/1, 5/3, 6/2, 6/6

 

That's 14 combinations, so the probability of Event B is 14/36 = 7/18