Probability question!!!?

With one die, the probability is 1/6 obviously.

With two dice, there are 36 possible outcomes; as one of them is 'TWO threes' and two times five ways to get ONE three, the answer is

( 2 x 5 + 1 ) / 36 = 11 / 36

Or alternatively ( 2 x 6 - 1 ) / 36 of course.

( RAWR is correct, but .... actually LISTING the possible outcomes? MAN! ---- still, one thumby uppy )

It's actually pretty easy logic, you see, there are six sides on each side, but there's only two threes. So you write a fraction of 2(for the number of chances you could get a three from both dice) over 12(comes from twelve sides of both dice) Then you simplify and get 1/6, I think your answer is incorrect, you better double check it.

Possible outcomes: 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 (36, 6*6)

If you eliminate every possibility without a three in it anywhere, you get:

1,3 2,3 3,1 3,2 3,3 3,4 3,5 3,6 4,3 5,3 6,3 (11 1*5 + 6)

Your chances of rolling at least one three using two dice is roughly 11 / 36.

how many possible outcome?

any one of 21 outcomes (order is not important 2,1 is 1,2)

1,1

1,2

1,3*

1,4

1,5

1,6

2,2

2,3*

2,4

2,5

2,6

3,3*

3.4*

3,5*

3,6*

4,4

4,5

4,6

5,5

5,6

6,6

sucesses are

3.1

3,2

3.3

3,4

3,5

3,6

p(at least 1 three) = 6/21 = 2/7