So we would be looking at the following cases:
1-3, 2-4, 3-5, 4-6 or 3-1, 4-2, 5-3, 6-4,
that is, there are 8 such cases
prob(of your event) = 8/36 = 2/9
two standard fair 6 sided dice are rolled. What is the probability that their values will differ by exactly two
2 answers
6-4
5-3
4-2
3-1
4-6
3-5
2-4
1-3 so 8 work out of how many
6 * 6 I think or 36
so I get
8/36 = 2/9
see:
http://www.freemathhelp.com/rolling-dice.html
5-3
4-2
3-1
4-6
3-5
2-4
1-3 so 8 work out of how many
6 * 6 I think or 36
so I get
8/36 = 2/9
see:
http://www.freemathhelp.com/rolling-dice.html