#### Alex919191

##### New member

- Joined
- Mar 16, 2019

- Messages
- 10

- Thread starter Alex919191
- Start date

- Joined
- Mar 16, 2019

- Messages
- 10

- Joined
- Nov 12, 2017

- Messages
- 3,004

Please read our submission guidelines, which you were told to read first!

- Joined
- Mar 16, 2019

- Messages
- 10

i have an answer already but i just want to double check with you guys

- Joined
- Nov 24, 2012

- Messages
- 994

What I recommend you do is post your answer and all of your reasoning, and then people can tell you if you are correct, and if not, where you may be going astray.i have an answer already but i just want to double check with you guys

- Joined
- Mar 16, 2019

- Messages
- 10

1 43/144 thats my answer but i dont know how to put it in words

- Joined
- Oct 6, 2005

- Messages
- 10,250

Then show your math steps.… thats my answer … dont know how to put it in words

Have you decided that you don't need to read or follow the forum's guidelines?

- Joined
- Nov 12, 2017

- Messages
- 3,004

What question is that the answer to?1 43/144 thats my answer but i dont know how to put it in words

I don't think it's asking for a number at all. Rather, if you calculate a probability, and you get something like 1 43/144, which is greater than 1, what must you conclude about your answer?

As a followup, you might then want to look up the rule for probability of "A or B", and think about whether you missed something.

- Joined
- Jan 29, 2005

- Messages
- 7,797

I think there maybe a translation problem here. Taking what is posted verbatim, twelve sided dice numbered \(\displaystyle 1\text{ to 12}\) and product (not sum) of the faces. There are \(\displaystyle 144\) possible pairs of which we want to count any pair which has a product which is even or has value greater than thirty. Any pair with an even number yields an even product. Thus remove any pair that contains two odd numbers whose product is less than thirty. That count is \(\displaystyle \|(\{1,3,5\}\times\{1,3,5\}\cup\{(1,7)\,(7,1)\,(1,9)\,(9,1)\,(3,7)\,(7,3)\,(1,11)\,(11,1)\}\|=17\) OR \(\displaystyle 144-17=127\) pairs the product of which is even or less than thirty.

- Joined
- Nov 12, 2017

- Messages
- 3,004

The way I read this, it doesn't say "on any roll of a set of two fair 12 sided dice, the probability of obtaining a product that is an even number or a product of greater than 30,adds to a number larger than 1. Given we know the probability of an event must lie between 0 and 1 explain in your own wordsif this is possible

It says,

I suppose I could be wrong; I'm taking the last phrase to mean "explain

- Joined
- Mar 16, 2019

- Messages
- 10

im so confused could someone reply with the answer and say how they did it

- Joined
- Nov 12, 2017

- Messages
- 3,004

As I read it, the answer is merely: "

Or, as I hinted earlier, you could explain specifically why the method, not just the answer, is wrong: You can't just add two probabilities, because the formula for "or" is

P(A or B) = P(A) + P(B) - P(A and B)

The two events overlap, so you have to subtract that overlap, which will bring the answer down below 1. That is, in adding, you will have counted even numbers greater than 30 twice.

On the other hand, if you think it is asking for the probability, you can either do what pka showed (which I imagine may be written in a style a bit over your head), or just make a table showing all 144 possible products, mark those which are either even or greater than 30, and count them. You should get the number pka gave (assuming he didn't miss anything). Then put that over 144, and you have the answer.

I dislike questions that are as hard to interpret as this one is; they lead to unnecessary frustration, often over very simple problems.

- Joined
- Mar 16, 2019

- Messages
- 10

i think its asking if it is possible

- Joined
- Mar 16, 2019

- Messages
- 10

- Joined
- Mar 16, 2019

- Messages
- 10

- Joined
- Mar 16, 2019

- Messages
- 10

- Joined
- Nov 12, 2017

- Messages
- 3,004

I think you meanYes, can you explain why the two events are notindependent?

It's not at all clear how much the student is expected to explain, but this more extended answer can't hurt!

- Joined
- Mar 16, 2019

- Messages
- 10

so would that be the answer

- Joined
- Nov 24, 2012

- Messages
- 994

Yes, mutually exclusive is in fact what I meant. Independence is something else entirely...thanks for the correction!I think you meanmutually exclusive, right? This amounts to the second possible answer I suggested.

It's not at all clear how much the student is expected to explain, but this more extended answer can't hurt!