Here's an answer to the first question. Since the other two questions are variations of the first one, they are left as challenge for you! :-)
Problem: We have $75$ people with $5$ dollar bills and $25$ people with $10$ dollar bills in line at a ticket counter which has no money and which charges $5$ dollars for admission. If a $10$ dollar bill is presented and there is no change, the line stops. What is the probability that the ticket seller always (after the first person in line, that is) has at least one $5$ dollar bill for change?
Analysis: We have queues of length $100$ with
\begin{align*}
\binom{75+25}{25}\tag{1}
\end{align*}
possible configuration. In order to get a valid configuration it has to be assured, that for each person with a $10$ dollar bill at least one other person with a $5$ bill has to be in front of him.
Lattice paths:
A nice way to describe the situation is using lattice paths. We consider paths of length $100$ starting at $A=(0,0)$ and ending at $B=(75,25)$. The paths consist of $(1,0)$-steps in $x$-direction representing persons with $5$ dollar bills and $(0,1)$-steps in $y$-direction representing persons with $10$ dollar bills. Since for each $10$ dollar bill a person with a $5$ dollar bill has to be in front, the valid paths are those having at each point a $y$-height less or equal the $x$-length.
We observe:
The number of all possible configurations is the number of all paths with length $100$ from $A=(0,0)$ to $B=(75,25)$.
The number of valid configurations is the number of paths with length $100$ from $(0,0)$ to $(75,25)$ which do not touch the line $y=x+1$.
The number of pathes from $(0,0)$ to $(75,25)$ with length $100$ is according to (1): $\binom{100}{25}$ since we can select $25$ $(0,1)$-steps in $y$-direction out of $100$ leaving $75$ left for the $(1,0)$-steps in $x$-direction.
Andre's Reflection Principle: The number of invalid configurations can be calculated using Andre's reflection principle. A path is invalid if it touches or crosses the line $y=x+1$. In any case it touches the line the first time at let's say $P$. Now, we reflect the first part of the path from $A=(0,0)$ to $P$ at the line $y=x+1$ getting a path from $A^\prime=(-1,1)$ via $P$ to $B=(75,25)$. The second part of the path from $P$ to $B$ is the same as the original path.
We observe: Each invalid path from $A=(0,0)$ to $B=(75,25)$ has to touch at least once the line $y=x+1$ and corresponds bijectively to the reflected path $A^\prime=(-1,1)$ via $P$ to $B$, whereby the second part of the path from $P$ to $B$ is identical to the original path.
Since this is a bijection we conclude: The number of invalid paths from $A$ to $B$ is the number of all paths from $A^\prime$ to $B$, namely:
\begin{align*}
\binom{76+24}{24}
\end{align*}
Result:
The resulting probability is the number of valid paths divided by the number of all paths:
\begin{align*}
\frac{\binom{100}{25}-\binom{100}{24}}{\binom{100}{25}}=\frac{51}{76}\approx 0.671
\end{align*}
Note: This problem is a variation of Bertrand's ballot theorem. The relationship with the Catalan Numbers $\frac{1}{n+1}\binom{2n}{n}$ is stated there in the section equivalent problems in this page.
First of all the expected value for the first pick is $(1+5+10+20)/4 = 9$
Then it depends on whether the bill is returned to be selected from.
If it's returned it's quite straight forward. The strategy must be that if you don't pick the 10 or 20 bill you try again. So in 25% of the cases you settle for your 20, 25% for the 10 and otherwise try again with an expected value of 9. So the result would be $20/4 + 10/4 + 9/2 = 12$.
If it's not returned the expected result of the second draw is dependent on the first bill:
- 1 means that the expected value of the next is $(5+10+20)/3 = 35/3$, it's correct to draw the second here
- 5 means $(1+10+20)/3 = 31/3$ that is draw the second here too
- 10 means $(1+5+20)/3 = 26/3$ that is we do not draw the second
- 20 means $(1+5+10)/3 = 16/3$ that is we do not draw the second.
So the result is that in 25% of the case we settle for 20, 25% we settle for 10, 25% we draw a second bill with the expected value of 31/3, and 25% we draw a second with the expected value of 35/3. So the result is
$${20 + 10 + 31/3 + 35/3\over 4} = {60+30+31+35\over12} = 13$$
Best Answer
That is so for a binomial distribution. However, this is not that. So it is just the wrong equation for this situation.
Now you are on the right track. In this instance, that is the correct calculation of the mean, in . (Also known as the weighted average; or the expectation.)
Don't doubt yourself. It appears that you were just trying to apply the wrong shortcut, and that was confusing you, because otherwise you do know what you are about. You have got the basics down.
$\Box$