Computing the chance of 2 dice accessing the very least a $1$ or a $5$

Go in reverse : Calculate the chance that neither of them reveals a 1 or 5. That suggests both show a 2, 3, 4, or 6. Thats $(4/6)^2$.

Therefore the chance that at the very least one reveals a 1 or 5 is $1-(2/3)^2=5/9$.

To aesthetically see the solution offered by balpha over, you can draw up the whole set of dice rolls

[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]

Total variety of feasible dice rolls : 36

Dice rolls which contain 1 or a 5 : 20

20/ 36 = 5/ 9

The various other means to imagine this would certainly be to attract a chance tree thus : alt text http://img.skitch.com/20100721-xwruwx7qnntx1pjmkjq8gxpifs.gif

(apologies for my inadequate criterion of illustration :))