Understanding how to solve Balls and Bins Questions

Question

In my university statistics class, we have been going over the balls and bins problems which usually involves throwing some n balls into k bins and calculating probabilities or expectation from that. However, recently, I've been stumped on a lot of these problems because I believe I am not getting the concept down correctly.

During a lecture, there were two questions I could not answer at the end.

1.) Suppose you throw 4 balls into 5 bins. Suppose the bins are labelled 1, 2, 3, 4, and 5. What's the expected number of balls that land in the bin labelled 1?

I'm sure that the bin being labelled 1 here isn't really something to pay attention too. However, although I know how to find the expected value of bins that only contain one ball, I don't know how to find the expected value of just one bin.

2.) Suppose you throw 4 balls into 5 bins. What's the probability that you end up with every bin having 0 or 1 balls?

$${5 \choose 4}(\frac{1}{5})^4$$ is what I believe the answer should be since I am selecting four bins, each with a probability of 1/5 of having a ball tossed into it. I would appreciate it, however, if some light could be shed onto the wrong-doings in my intuition.

Answer 1

1) Let $X$ be the count of the $4$ balls which fall into the specified bin, when they do so independently and with identical rate of $1/5$.   I'd expect a fifth of the balls to fall into the bin; wouldn't you?

More formally, the count follows a recognisable kind of distribution with a known expectation.

(Or, as Lulu commented: the expected count of balls in each of the $5$ bins is identical, by symmetry, and the total of their counts must equal $4$.)

$\mathsf E(X)=4/5$

2) This is another way to ask: "What is the probability that each of the $4$ balls falls into a different one of the $5$ bins, when each ball falls into a bin independently and without bias?"

You were on the right track: you just need to measure the ways to arrange four balls among five places without repeatition when the balls may be arranged among five places with repeatition. (When order is counted in the denominator it must also be counted in the numerator.)

$\mathsf P(A)= {^5\mathrm P_4} ~/~ 5^4 = 4!/5^3$