Probability of less than 10 % from the sampled are truck owners A simple random sample n = 25 is being drawn from a population from 320 members, exactly 30% of whom own a truck. Provide answers to the following to three decimal places. What is the probability of less than 10 % from the sampled are truck owners?
My thoughts: Should I use the given condition to find the probability ? Or shouldd I use the combinatoric approach?
 A: $Hypergeometric:$ The population is split into $96$ truck owners and  $320 - 96 = 224$ people who don't own trucks. You are sampling $n = 25.$ If $X$ is the number of truck owners in your sample of 25, then
you want $P(X < 2.5) = P(X = 0) + P(X = 1) + P(X = 2).$
This is
$$P(X \le 2) = \frac{{96 \choose 0}{224 \choose 25} +
{96 \choose 1}{224 \choose 24} + {96 \choose 2}{224 \choose 23}}
{{320 \choose 25}}.$$
Using R, this can be evaluated as follows:
 phyper(2, 96, 320-96, 25)
 ## 0.007081469

$Binomial.$ A 'Success' is getting a truck owner. The proportion of truck owners is $p = 0.3.$ Suppose we select the $n = 25$ people
with replacement. Then use $Y \sim Binom(25, 0.3)$ and find
$P(Y < 2.5)$ as above. This can be evaluated in R as follows:
 pbinom(2, 25, .3)
 ## 0.008960528

This 'pretends' the population of 320 is large enough that any
duplications out of 25 chosen can be ignored. Sometimes one
sees the rule of thumb that this is OK if the sample size $n$ is
less than 10% of the population size $N$, which is true here.
But you want three place accuracy, so it isn't OK. 
Using
a $Poisson$ approximation to binomial is risky because $n = 25$
is not very large and $p = 0.3$ is not very small. (This
method with $\lambda = 7.5$ gives a result around 0.02, compared
with the binomial result about 0.01. Not horrible, but certainly
not three place accuracy.)
The $normal$ approximation to hypergeometric would use
mean $\mu = np = 25(.3) = 7.5$ and variance 
$\sigma^2 = np(1-p)\frac{N-n}{N-1} = 4.8550$. Then the normal
approximation with continuity correction gives the following
result:
 pnorm(2.5, 7.5, sqrt(4.855))
 ## 0.01162767

This is better than Poisson, not far above binomial, but not
within three-place accuracy of hypergeometric.
$Moral\; of\; this\; story:$ Particularly, when software is available, there is no particular
reason to use unnecessary approximations just to avoid tedious
computation. (Easy for me to say when I don't have to fuss
with all those factorials in the hypergeometric expression.)
The plot below compares PDFs of the four distributions mentioned
above (binomial bars to the left of hypergeometric, Poisson right).

