We have 12 red balls and 18 blue balls, probability of certain events. So we have 12 red balls and 18 blue balls so 30 in total all distinguishable. Now the questions are:


*

*What is the probability to get exactly 10 red balls if you pick 12. Since the order does not matter I think it should be $\binom{12}{10}\binom{18}2/\binom{30}{12}$.

*What is the probability of getting 10 or more red balls if you pick 12. Here I would just $\binom{12}{10}\binom{18}2 +\binom{12}{11}\binom{18}1+\dots$ 

*How many ways are there for the first ball to be red and the last ball to be blue. Here I am stuck, not really sure how to do this one.


Any help is appreciated, thank you.
 A: The first two are okay — as long as you remember to also divide the three terms in the second by $\tbinom{30}{12}$
For the last, if you reserve the first and last ball, then the middle is a selection of $10$ balls from $28$ ($11$ red and $17$ blue).   So just count the ways to reserve those end balls, then multiply by ways to select the middle balls.
(If calculating a probability for this event, then remember the rule of "as above so below".)
A: Since for third case the order is matter than you can approach it as they are all different.
To put red at first there are 12 possible, and for last there are 18 possible, on the second there are 28 possible, on third 27...
$P=\frac{12\cdot18\cdot28\cdot27\cdot...\cdot19}{30\cdot29\cdot...\cdot19}=\frac{12\cdot18}{30\cdot29}$ 
EDIT
As you just mentioned we are not looking for probability for the 3rd case than total number of outcomes that satisfy condition that first ball is red and last is blue.
You have to define first what is an outcome, is R1B1B2B3...B11 same as R2B1B2B3...B11.
If you don't differentiate those cases then there are $2^{10}$ different outcomes. (at each position between first and last there can be R or B)
If you differentiate each single ball then number of outcomes is $12\cdot18\cdot28\cdot27\cdot...\cdot19$
