How to determine if a probability problem has a binomial distribution or not? I'm studying for my Stats midterm and I am confused about the binomial distribution concept.
Among these 2 problems, why does the second question have a binomial distribution and not the first question?
1) In a large 2 lb bag of candies, 15% of the candies are green. The chances of pulling out at least one green candy in three tries is...
2) An owner suspects that only 5% of the customers buy a magazine and thinks that he might be able to use the display space to sell something more profitable. What is the probability that exactly 2 of the first 10 customers buy magazines?  
We already know about SPIN: S(Success/Failure) P(Probability) I(Independent) N(Fixed number of trials) and I'm curious as to why SPIN doesn't apply to the first problem as well?
Any help would be appreciated. Thank you.
 A: From the wording of the first situation, it sounds like you are not replacing the candy after you pull it out.  Therefore you are not repeating the same fixed experiment three times in a row; rather, you are changing the experiment (and the associated probability of success) with each pick.
A: Remember that the binomial distribution tells you what the odds are of getting exactly $x$ successes in $n$ attempts, given those attempts are bernoulli trials (Or they fit the restrictions of SPIN, as you say).
The first problem with question one is the part where it says "at least". That is not "exactly". This is more reminiscent of the CDF of the geometric distribution then a binomial distribution.
The second problem with question one is that the attempts are not independent. Taking out a bean changes all further tests. You can put this effect in by using the hypergeometric distribution, but this effect might be negligible if there are enough beans.
Still, you can get really close by using the binomial distribution for x = 1, 2, and 3.
A: Here are a few possible ways to look at the first problem.
1) Hypergeometric distribution. Suppose there are $N = 1000$
candies in the bag. Because $p = .15$ of them are green,
that means there are $Np = 150$ green and $1000 - 150 = 850$
non-green ones in the bag. If you take $n = 3$ candies without
replacement, then the probability of getting $X \ge 1$ green ones
would be computed using a hypergeometric distribution for $X.$
Computation would be just a little messy, but the answer from R
to four places is $0.3862:$
 N = 1000;  n = 3;  p = .15;  g = N*p;  x = 1:n
 sum(dhyper(x, g, N-g, n))
 ## 0.3862006

[By hand, you could save a little work computing 
$P(X \ge 1) = 1 - P(X = 0).$]
Using the hypergeometric distribution we must know the number $N$ 
of candies in the bag. Because sampling is without replacement,
the draws are not strictly independent. For example,
the probability the first candy is green is $150/1000 = 0.15,$
but given the first is green, the probability the second
is also green would be $149/999 =  0.1491491.$
If $N = 100,$ not $1000,$ then $P(X \ge 1) = 0.3892.$
Not much of a change.
 N = 100;  n = 3;  p = .15;  g = N*p;  x = 1:n
 sum(dhyper(x, g, N-g, n))
 ## 0.3891775

2) Binomial. If sampling is with replacement, then draws
are independent and the number $N$ of candies in the bag is
irrelevant. Then the number $X$ of green candies in $n = 3$
independent draws is distributed as $Binom(n, p)$ and
$P(X \ge 1) = 0.3859$ is computed in R as:
 sum(dbinom(x, n, p))
 ## 0.385875

There is a rule of thumb that says a hypergeometric probability
can be approximated by a binomial one provided that the
proportion $n/N$ sampled is small (many say smaller than 5%).
So you would be left to guess how many candies might be in 
a "large" bag. With the 5%-rule, it would still be OK to approximate hypergeometric with $N = 100$ by binomial.
Below is a bar graph of the hypergeometric probabilitiy distribution,
assuming only $N = 100$ candies in the bag. The probabilities in the relevant binomial
distribution is shown as small circles. Differences are almost undetectable. (I could have used $N = 1000$
for the bar graph, but then there wouldn't be any visible distinction from
binomial to show, at the resolution of the graph.)

Summary. In summary, the probability is $0.39$ to two places in all of
these computations. Also, I don't think any of these
computations is a good candidate for a normal approximation.
