I had some troubles with the following exercise:
Randomly throw 5 balls into one of the 4 boxes. What is the probability that no boxes are empty?
I think the answer is P(A) = 4 * (5!/1!*1!*1!*2!) * (1/4)^5. (I multiply by 4 because I believe it can happen in 4 different ways)
My question is if the question asked about the probability of having 3 empty box. Would it be P(A) = 4 * (5!/5!) * (1/4)^5
I agree with your answer to the 2nd question - re 3 empty boxes, there are only 4 ways that it can happen, so chances are $$\frac{4}{4^5} = \frac{1}{4^4}.$$
For the 1st question, I can't tell if we agree because you didn't use mathJax. So, I'll just explain my reasoning and post my computation.
Again, the denominator is $$\frac{1}{4^5}.$$
There are 4 ways you can select which box gets 2 balls.
Assume Box 1 will be the box with two balls.
Now, there are 5 choices for which ball went into Box 4.
With that decided, there are 4 choices for which ball went into Box 3.
Then there are 3 choices for which ball went into Box 2.
So the numerator is $4 \times 5 \times 4 \times 3 = 240.$
Final answer is $$\frac{240}{4^5} = \frac{15}{4^3}.$$