Hi everyone!

I'm a University Student in Germany, and we are getting ready to take our final exams. I come from the US, but am taking courses in German - so excuse me if the translation is a little bit funny (the questions are in German)!

Right now I'm doing a practice final exam, and I'm at a point where I have no clue where to start.

Here is the question:

3a) In the trainstation, there is a bench with 10 seats with 10 people sitting on it. From these 10 people, 2 don't have Tickets (illegal riders). In how many situations will the 2 without Tickets be sitting next to eachother?

3b) How many illegal riders do there have to be, so that there is a 99% probability that of 100 Riders, at least 1 is an illegal rider?

3c) 97% of all Riders have tickets. One Controller checks 5% of the Riders. The probability that one Rider is an illegal rider and who will be checked is 0.20%. Does the Controller have an eye for who is an illegal rider, or has he picked them out by chance?

3d) The probability that a rider is riding illegally is 5%. 200 controlls will be taking place. What is the probability that at least 7 and at most 12 illegal riders will be caught?

If anybody can give me any help on how to even start, I will be forever grateful!

Thank you

Melanie

2. Originally Posted by lamiastella
3a) In the trainstation, there is a bench with 10 seats with 10 people sitting on it. From these 10 people, 2 don't have Tickets (illegal riders). In how many situations will the 2 without Tickets be sitting next to eachother?
Tie the two illegals together and pretend they're one person(a huge obese person or a typical American). Then there are 8! ways to arrange the other 8. They can be arranged in 2! ways.
There are 9!*2! ways

3d) The probability that a rider is riding illegally is 5%. 200 controlls will be taking place. What is the probability that at least 7 and at most 12 illegal riders will be caught?
$\sum_{k=7}^{12}C(200,k)(\frac{1}{20})^{k}(\frac{19 }{20})^{200-k}\approx{0.673}$

3. Hello, Melanie!

Here's the first one . . .

3a) There is a bench with 10 seats with 10 people sitting on it.
From these 10 people, 2 don't have Tickets (illegal riders).
In how many situations will the 2 without Tickets be sitting next to each other?

Duct-tape the two illegal riders together.
. . Then there are 9 "people" to arrange: . $\left\{\boxed{AB},\,C,\,D,\,E,\,F,\,G,\,H,\,I,\,J\ right\}$
There are: . $9!$ ways to seat them.

But the two illegal riders could sit in another order: . $\left\{\boxed{BA},\,C,\,D,\,E,\,F,\,G,\,H,\,I,\,J\ right\}$
. . and they can be seated in another $9!$ ways.

Therefore, there are: . $2 \times 9! \:=\:\boxed{725,760\text{ arrangements.}}$

4. You guys are fantastic!!!!

Thank you so much. It really helped me to understand and to visualize the question - now I understand much better. I'll have to keep fat & ducktaped people in mind when I'm taking my final exam