Thread: When rolling three dice, what is the probability of

When rolling three dice, what is the probability of

a) getting three different numbers?



b) having at least one two and at least one four?



c) getting two threes given that no more than one five was rolled?


Any helpful hints will be greatly appreciated.

There are simple events
and you'll have to write down a few of them to get a feel for what you're looking for in each case.
For three different numbers you do get (6)(5)(4) favorable events.
So you are correct.
In case (b) you can list all the favorable ones, then divide by 216.
HOWEVER, I just reread (c).
That's a conditional, so you will need to be more careful.

Originally Posted by confusingmath

When rolling three dice, what is the probability of

a) getting three different numbers?

Mr F says: The total number of permutations is 6^3. The number of these with all numbers different is (6)(5)(4) = ..... Therefore .....

b) having at least one two and at least one four?

Mr F says: Consider

2, 4, x
4, 2, x
2, x, 4
4, x, 2
x, 2, 4
x, 4, 2

(or you could just arrange 2, 4, x in 3! ways)

where x can be any number from 1 to 6. So there are (4)(6) = 24 ways.

c) getting two threes given that no more than one five was rolled?

Mr F says: Consider

x, 3, 3 can be arranged in 3! = 6 ways and there are 5 values that x can have (x cannot equal 3) ....


Any helpful hints will be greatly appreciated.

..