HI! I need help to find some answers!
What is the probability that if you roll a single standard die four times you will roll four different numbers?
5/18, 240/1296, 1/3 or none of these?

2. Originally Posted by K.S. Turner
HI! I need help to find some answers!
What is the probability that if you roll a single standard die four times you will roll four different numbers?
5/18, 240/1296, 1/3 or none of these?
The first die can be any of six values, the second any of the remaining five
and the third any of the remaining four and so on. So there are $6\times 5 \times 4 \times 3$
ways of rolling four dice so they are all different, and $6^4$ ways of rolling the
die without constraint.

So the required probability is $6 \times 5 \times 4 \times 3/6^4 =5/18$

RonL

3. the ways u may get different numbers are: 6 x 5 x 4 x 3 which is actually the permutation " 6P4 " and it equals 360.

The total number of ways is 6^4 =1296

P = 360/1296 = 5/12.

4. Originally Posted by PHANTOM
the ways u may get different numbers are: 6 x 5 x 4 x 3 which is actually the permutation " 6P4 " and it equals 360.

The total number of ways is 6^4 =1296

P = 360/1296 = 5/12.
I'm sure you meant 5/18.
Cheers.