A four sided, balanced die has the values 1, 2, 3, and 4. There are three successive throws. Find the probability of throwing 1, 2, and 4 in any order

Question

mathmate · Answer

Hints:
First throw: 3 choices (among 1,2 and 4)
Second throw: 2 choices remaining
Third throw: 1 choice remaining
Success: 3*2*1 possibilities
$\Omega$: 4^3 possible outcomes.

anonymous · Answer

So would you do 3/4 * 2/3 * 1/2? Or would you divide 6/64?

anonymous · Answer

So would it be 3/4 * 2/4 * 1/4 since they are independent?

perl · Answer

$$ 
111\
112\
113\
114\
121\
122\
123\
124\
131\
132\
133\
134\
...
$$

perl · Answer

so we know there are 4*4*4 possible ways to list three integers from 1 to 4, since repetitions are allowed. Thus the denominator will be 4^3. 
For the numerator we have 3 ways to choose the first number, 2 ways for the second number, and 1 way for the last number. 

$$ \large \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{1}{4}$$

anonymous · Answer

Ahhh okay I understand.

perl · Answer

so you can separate the problem into two parts. The numerator is counting how many ways you can rearrange 1,2,4 in any order

1,2,4
1,4,2
2,1,4
2,4,1
4,1,2
4,2,1

perl · Answer

the denominator is how many ways you can count
1,1,1
1,1,2
1,1,3 , etc

anonymous · Answer

So would the answer be 9.375%?