Question

Question #6cb27

Answer 1

There are 23 out of 36 outcomes that satisfy our conditions - a probability of `63 8/9%`

Explanation:

The way to think of these probability problems is to find a way to count the number of times each thing can occur and compare that the the total number of things that are possible. In this case, we are dealing with only 36 possible outcomes, so we could use a brute-force approach and write out each possibility:

Then we can test each of the criteria:

Is die 1 even?
Is the sum 6?
Is the sum 8?

Sometimes two of these are true at the same time, but this still only counts once toward our total. This gives us 23 out of 36 outcomes that satisfy our conditions - a probability of `63 8/9%`

If there were many more outcomes, this approach would be too complicated, and we'd want a mathematical way to do the job. For instance, we could start with the first condition, how many times would the first die be even. We know this is a probability of 50% since half of the sides have even numbers on them. This would be 18 of the possible 36 outcomes. Note that it doesn't matter if the sum is 6 or 8 for any of these cases, since it is already a wanted outcome.

Next, we want to know for the odd cases, how many would sum to 6? There are only 3 possibilities to check and there is only one number of the second die that produces this outcome.

`D1 = 1; D2 = 5`
`D1 = 3; D2 = 3`
`D1 = 5; D2 = 1`

So there are 3 more outcomes. Finally, we check the cases that D1 is odd and the sum is 8:

`D1 = 1; "not possible"`
`D1 = 3; D2 = 5`
`D1 = 5; D2 = 3`

Giving us another 2 outcomes. The sum of these 3 cases represents the "or" of the conditions

`"number of positive outcomes" = 18 + 3 + 2 = 23`

So the probability is

`P = 23/36 = 0.63bar(88) = 63 8/9%~= 63.9%`