Please can You Help?
A die is rolled three times. i-) Find the chance that three different faces appear.
My approach using complement rule:
Probability p(same face pops up is)= 1- P. You roll first one and record what you get. The probability of next (2) two rolls give you same face is...? My chances of rolling any given face is 1/6. It does not matter what you get first role the next two rows have to the same as the first one role. Therefore= 1-(1/6)^2)= 1-(1/6^2)=

Question

Please can You Help?
A die is rolled three times. i-) Find the chance that three different faces appear.
My approach using complement rule:
Probability p(same face pops up is)= 1- P. You roll first one and record what you get. The probability of next (2) two rolls give you same face is...? My chances of rolling any given face is 1/6. It does not matter what you get first role the next two rows have to the same as the first one role. Therefore= 1-(1/6)^2)= 1-(1/6^2)=

anonymous · Answer

i think the approach is right, but the answer may be wrong

perl · Answer

for the first die there are 6 possibilities, then for the second die there will be 5, then the third die there will be 4. since we are doing probability, the denominator will be 6^3

( 6x5x4)/(6x6x6)

anonymous · Answer

oh maybe not, maybe the answer is right too
roll one die.  you get something
roll another, the probability it is different it $\frac{5}{6}$ then roll a third, the probability it is different from the first two is $\frac{4}{6}$

anonymous · Answer

multiply and get what @perl  got, namely 
$$\frac{5\times 4}{6\times 6}$$

perl · Answer

using the complement would be more complicated

anonymous · Answer

Show me the multiple rule?