Probability of finding item in a box I'm trying to find if my approach to this kind of problems is correct. 
For example: You have 3 boxes, and you have a 33% chance of finding an item in a box. What is the probability of finding items in: 0, 1, 2, 3 (all) boxes?
My answer:
$P=0.33$ $(!P =1 - 0.33=0.67)$
for 0 boxes: $(!P * !P * !P) * 3 = (0.67 ^ 3) * 3= 0.9$
for 1 box: $P * !P * !P + !P * P * !P + !P * !P * P  = (0.33 * 0.67 ^ 2) * 3 = 0.44$
for 2 boxes: $P * P * !P + P * !P * P + !P * P * P = (0.33 ^ 2 * 0. 67) * 3 = 0.21$
for 3 boxes: $(P * P * P) * 3 = (0.33 ^ 3) * 3 = 0.10$
The reasoning (for example for the 1 box case) is that we need to take the probability that the first chest contains an item AND $(*)$ the other don't, OR $(+)$ the second chest contains an item and the other don't, OR the third contains an item and the other don't.
Is this the correct way to calculate the probability for this type of problem?
 A: Almost correct.
The probabilities corresponding with $0$ and $3$ boxes are wrongly multiplied by $3$. Do you understand why this is wrong?
Also note that the probabilities should add up to $1$.
Checking that often helps you to find your own mistakes.
A: Your reasoning for the one-box case is good,
in particular, the order of the boxes does matter.
You correctly account for this in the one-box case by counting
$(\text{something}, \text{nothing}, \text{nothing})$,
$(\text{nothing}, \text{something}, \text{nothing})$, and
$(\text{nothing}, \text{nothing}, \text{something})$
as three separate events.
(I'm using the notation $(x_1,x_2,x_3)$ to indicate that the
contents of the first, second, and third boxes are $x_1$, $x_2$, and $x_3$,
respectively.)
But even when order is considered, there is still only one way to have three empty boxes: $(\text{nothing}, \text{nothing}, \text{nothing})$,
so you should not multiply this probability by $3$ (or by anything except $1$).
Multiplication by $3$ says there are three distinct,
equally likely events that need to be counted, which is not true
for "items in $0$ boxes".
Likewise there is only one way to have "items in $3$ boxes":
$(\text{something}, \text{something}, \text{something})$.
Again, do not multiply this by $3$ as if there were three ways it could occur.
A: The chance of finding an item in the first box is $\dfrac{1}{3}$
The chance of also finding an item in the second box is $\dfrac{1}{3}*\dfrac{1}{3}=\dfrac{1}{9}$
The chance of finding an item in all $n$ boxes is:
$$\dfrac{1}{3^n}$$
