Question

2. Two fair dice are thrown,
i. Event A is the score differs by 3 or more. Find the probability of event A.

Answer 1

There are 36 pairs possible, write them and choose...

Answer 2

(1,4),(1,5),(1,6), ...

Answer 3

Yes, I know. I've written them out... But the answer is\[ ({1\over36})^{12}\]

Answer 4

I mean, that's what I got

Answer 5

lol really ?

Answer 6

Yes... I just need some kind of confirmation.?

Answer 7

I'm not sure of what the answer is..

Answer 8

(2)

1 1
1 2
1 3
1 4
1 5
1 6
2 1
2 2
2 3
2 4
2 5 1/36
2 6 1/36
3 1
3 2
3 3 1/36
3 4 1/36
3 5 1/36
3 6 1/36
4 1
4 2
4 3 1/36
4 4 1/36
4 5 1/36
4 6 1/36
5 1
5 2 1/36
5 3 1/36
5 4 1/36
5 5 1/36
5 6 1/36
6 1
6 2 1/36
6 3 1/36
6 4 1/36
6 5 1/36
6 6 1/36


would it be something like this for if the product of the score was greater than 8?

Answer 9

So the answer would be for the first, 12/36, and the second, 20/36?

Answer 10

>>Event A is the score differs by 3 or more. Find the probability of event A.

P(A) = 12/36

which is 1/3. I tend not to simplify probability ratios because the original data count is masked. By that, I mean, for example, that 36 represents the possible outcomes of tossing two die but once simplified, the 3 seemingly does not directly connect to the problem

If you click on the link above, you can view the sample space for tossing two die.

There are 36 possible outcomes. 12 of them are favorable for the spots to differ by 3 or more.

Answer 11

@ Order

>> So the answer would be for the first, 12/36, and the second, 20/36?

Yes. That is correct.

Answer 12

Thank you!