Simple use of a permutation rule in calculating probability I have the following problem from DeGroot:
A box contains 100 balls, of which 40 are red. Suppose that the balls are drawn from the box one at a time at random, without replacement. Determine (1) the probability that the first ball drawn will be red. (2) the probability that the fifteenth ball drawn will be red, and (3) the probability that the last ball drawn will be red.
For the first question:

*

*the probability that the first ball drawn will be red should be:
$$\frac{40!}{(100-40)!}$$


*the probability that the fifteenth ball will be red:
$$\frac{60! \times 40!}{(60-14)! \times (40-1)}$$


*the probability that the last ball drawn will be red:
$$\frac{60! \times 40!}{100}$$
Am I on the right track with any of these?

I find it instructive to lead students through the horrible, brute
force calculation before teaching  them that the end result is best 
understood  using  symmetry.
Let's calculate the probability that the 15th ball is red, taking
into account all the balls drawn previously. For example, one of 
the outcomes that make up our event is 
$$RNNRNNNRRRNRNRR$$
where $R$ means a red ball and $N$ a non-red ball. 
The probability of getting this particular outcome is 
$${40\over 100}\cdot{60\over 99}\cdot{59\over 98}\cdots {33\over 86}={(40)_8 \, (60)_7\over (100)_{15}}.$$
We simplify the product of fractions  using the Pochhammer symbol. The "7" and "8" are the number of $N$s and $R$s in the outcome.
Adding the probabilities of all such outcomes gives 
$$\mathbb{P}(\mbox{15th ball is red})={1\over (100)_{15}}\sum_k {14\choose k} (40)_{15-k}\ (60)_k $$ 
$$={1\over (100)_{15}}\ 40\ \sum_k{14\choose k} (39)_{14-k}\ (60)_k ={1\over (100)_{15}}\ 40\ (99)_{14} ={40\over 100}.$$ Amazing!
