Example of using binomial distribution Q:  Is my approach correct?
Event:  You toss 5 coins at once. 
A student of mine claimed he got 4T & 1H in 39 out of 40 trials (!!)  
I decided to calc the odds of this...
First, P(4T & 1H) = 5C4 * (1/2)^4 * (1/2)^1 = .16
I did this 2 ways:

1) Binomial Probability
n = 40
r = 39
p = .16
q = .84
P(Exactly 39) = 40 C 39 * (.16)^39 * (.84)^1 = 0%

2) Binomial Distribution:
n = 40
r = 39 (or more)
p = .16
q = .84
E(X) = u = np = (.16)(40) = 6.4
SD = SQRT(npq) = 3.16
Z(39) = (observed - expected) / SD = (39 - 6.4) / 3.16 = 10.3
p = P( Z > 10.3) = 0%

Conclusion:  The odds of getting 4T & 1H in 39 out of 40 trials is negligible.
Student was on drugs at the time.
 A: The probability of observing 4 heads and 1 tail 39 times out of 40 after observing 4 heads and 1 tail 39 times out of 40 is 1.0.
:)
A: How about a simulation based approach?  Here's some R code to generate 100000 students each trying the 40 tosses.
theSum = c()
for (i in 1:100000) {
  coin1 = rbinom(40,1,.5)
  coin2 = rbinom(40,1,.5)
  coin3 = rbinom(40,1,.5)
  coin4 = rbinom(40,1,.5)
  coin5 = rbinom(40,1,.5)
  theSum[i] = sum(coin1+coin2+coin3+coin4+coin5 == 1)
}

summary(theSum)
hist(theSum, xlim = c(0,40), freq = F, main = "", xlab = "")

The range of times the HTTTT combination occurred (in any order): 0-18 (out of 40), with a mean of around 6.  
Below: a histogram of the 100000 attempts and how many times the magical combination occurred.  You'd have to be very lucky indeed to get it 39 times out of 40 with fair coins.  But stranger things have happened by chance (e.g., our evolution).
alt text http://img80.imageshack.us/img80/9268/coinflips.png
A: 
It must be indicative of something
besides the redistribution of wealth.
Heads.
A weaker man might be moved to
re-examine his faith, for nothing else
at least in the law of probability...
Heads.
Consider. One, probability is a
factor which operates within natural
forces. Two, probability is not
operating as a factor. Three, we are
now held within um... sub or
supernatural forces. Discuss!
What?
Look at it this way. If six monkeys...
If six monkeys... The law of averages,
if I have got this right means... that
if six monkeys were thrown up in the
air long enough... they would land on
their tails about as often as they
would land on their...
Heads, getting
a bit of a bore, isn't it?
– Tom Stoppard Rosencrantz and Guildenstern are Dead (1966)

As John Christie pointed out, no matter how unlikely the student's result was, you can't infer anything from a single trial. A clever student might well have tried this gambit knowing it could not be refuted, in which case I might be inclined to commend her.
Incidentally, Rosencrantz (or Guildenstern) tossed at least 157 consecutive heads and it was nothing to write home about.
A: Technically, your case 1 and 2 calculations are not correct as they are not independent trials. You are tossing the same 5 coins 40 times. So, those events are dependent.
If you ignore the above issue then the above seems ok.
On some more reflection I think you can ignore the issue of dependency. Here is my reasoning: The probability of observing 4T & 1H is 0.16. In your case 1 and case 2 calculations you are using this probability across all 40 trials which implicitly accounts for the dependence in the trials. 
Another way to think about the issue is: If you observe 4T and 1H in the first trial what can you say about the probability that you would observe 4T and 1H in the second. It clearly equals 0.16 and thus there is no dependency. Knowledge of the outcome of one trial does not give us any additional information about the events that are likely to happen in the subsequent trial. Thus, the trials are independent.
I think the calculation is fine as it stands.
