# Probability - Discrete Distribtions - Practice Exam Question #2

Hi Team,

Thanks for your hard work. I'm really struggling to understand how to apply the knowledge from the course thus far to solve the problem of determining how many outcomes there are for the sum of the dice being even. What rule do I use? The multiplication rule, the additive rule, etc? How do I use the tools I've learned thus far to solve this part of the problem? I just don't get it.

Thanks,

Justin

6 answers ( 0 marked as helpful)

Hi Justin, thanks for reaching out with this. I'm happy to reason with you. Which question was challenging to you, please?

Best,

Also wanted to ask - did you try the new practical exercises while taking the course? Their goal is to allow you to practice every 1 or 2 lectures.

Best,

Ned

Hey Ned,

Thanks so much. I'm attaching the screenshot that contains the question I was struggling to determine the answer to: how do you determine the probability of getting an even sum from two fair 6 sided dice?

I don't think there is any rule that we have learned thus far to solve this problem. The best I can figure, I just have to do it by brute force counting. This is what I did. But please tell me if there is a better method because this seems very cumbersome to me.

Here's what I did. I listed each even sum and then thought about the different combinations that could create it. Again, this seems cumbersome.

2: 1+1

4: 3+1, 2+2, 1+3

6: 2+4, 3+3, 4+2, 5+1, 1+5

8: 4+4, 6+2, 2+6, 5+3, 3+5

10: 5+5, 4+6, 6+4

12: 6+6

12: 6+6

When you count them, there are 18 combinations that are even. This means the probability of rolling an even pair of dice is 18/36.

Is there a better way to do this other than brute force counting?

-Justin

Hi Justin,

Here's some reasoning from me.

First, we need to consider the probability of the Event C - sum of the dice will be even. For the sum of the dice to be even, you need to either have both dice even, or both dice odd. Correct? Therefore, the probability of the Event C is 1/2. In all other cases, C is not satisfied.

Then let's take a look at event D. The chance for D1 to land on 5 is 1/6. (because the dice has 6 possible outcomes).

The joint probability P(C∩D) (Sum is even and D1 lands on 5) is given by 1/6*1/2 = 1/12

Then we are asked about the conditional probability of P(C/D).

We'll use the conditional probability formula and we'll have:

P(C∣D)=P(C∩D)/P(D) = (1/12)/(1/6) = 1/12*6 = 1/2

If we know that one dice will fall on 5, then the chance of having an even outcome as a sum of both dice is 1/2 (we'll have an even outcome only if the other dice is odd).

Hope this helps!

This is a tricky one. Keep up the great work!

Best,

Ned

Thanks Ned!

First, we need to consider what counts as an "even outcome" when rolling two dice. An outcome is even if the sum of the two dice is an even number.

The possible even sums are 2, 4, 6, 8, 10, and 12.

Now, let's count how many ways we can get each of these sums:2: only one way (1+1)

4: three ways (1+3, 2+2, 3+1)

6: five ways (1+5, 2+4, 3+3, 4+2, 5+1)

8: five ways (2+6, 3+5, 4+4, 5+3, 6+2)

10: three ways (4+6, 5+5, 6+4)

12: one way (6+6)

If we add these up: 1 + 3 + 5 + 5 + 3 + 1 = 18

So there are 18 ways to roll an even sum with two dice.

This is indeed half of the total outcomes (18 is half of 36), but we arrived at this answer by counting the favorable outcomes, not by simply dividing the total number of outcomes by 2.

It's a coincidence in this case that exactly half of the outcomes are even. This won't always be true for other similar problems, which is why it's important to count the favorable outcomes rather than assume half will always be correct.

Initially, I also thought that I could simply divide the sample space in half to determine the number of even combinations based on the reasoning you shared above. But then I ran that idea through Claude and it said I could not do that. Did this thing give me bad advice? AI is a blessing and a curse; often times it's great, but it struggles with math, so maybe it led me astray here.

**Here's my question:**

to determine number of possible even outcomes, do i divide 36 by 2 in order to cut the number of outcomes in half?

**Here's it's output:**

Not quite. Let's think about this step-by-step:

First, we need to consider what counts as an "even outcome" when rolling two dice. An outcome is even if the sum of the two dice is an even number.

The possible even sums are 2, 4, 6, 8, 10, and 12.

Now, let's count how many ways we can get each of these sums:2: only one way (1+1)

4: three ways (1+3, 2+2, 3+1)

6: five ways (1+5, 2+4, 3+3, 4+2, 5+1)

8: five ways (2+6, 3+5, 4+4, 5+3, 6+2)

10: three ways (4+6, 5+5, 6+4)

12: one way (6+6)

If we add these up: 1 + 3 + 5 + 5 + 3 + 1 = 18

So there are 18 ways to roll an even sum with two dice.

This is indeed half of the total outcomes (18 is half of 36), but we arrived at this answer by counting the favorable outcomes, not by simply dividing the total number of outcomes by 2.

It's a coincidence in this case that exactly half of the outcomes are even. This won't always be true for other similar problems, which is why it's important to count the favorable outcomes rather than assume half will always be correct.

Ned,

Next, I can see that the approach you used to solve the problem is slightly different than the method deployed by the instructor in the solution part of the quiz. It seems there are multiple paths to arrive at the correct conclusion for these problems. I'm still new to learning all this, so it's good for me to know that there are different ways to deploy the techniques we are learning to get the right answer and that there is not one path only to the correct destination.

Hey Justin,

I think you said it very well: "AI is a blessing and a curse". In this situation, I think you trusted the AI more than your own reasoning and got discouraged. I'm convinced you were going to solve the problem on your own if you didn't get distracted with brute force counting.

Anyways, keep up the good work and I'm sure you'll see amazing results from your learning journey here.

Best,

Ned