Here we have:
- Population size $N = 52$ (total cards)
- Total successes in population $K = 4$ (there are 4 Aces in a deck)
- Sample size $n = 5$ (cards drawn)
- Desired successes $k = 2$ (we want exactly 2 Aces)

Applying the Hypergeometric formula:
$P(X = 2) = \frac{\binom{4}{2} \binom{52-4}{5-2}}{\binom{52}{5}} = \frac{\binom{4}{2} \binom{48}{3}}{\binom{52}{5}}$

Calculating the combinations:
$\binom{4}{2} = 6$
$\binom{48}{3} = \frac{48 \times 47 \times 46}{3 \times 2 \times 1} = 17,296$
$\binom{52}{5} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2,598,960$

Putting it together:
$P(X = 2) = \frac{6 \times 17,296}{2,598,960} = \frac{103,776}{2,598,960} \approx 0.0399$

There is roughly a 3.99% chance of drawing exactly 2 Aces.

Let $X$ be the number of returns with errors found in the sample of 4.
We want $P(X \ge 1)$. It's much easier to calculate the complement: $1 - P(X = 0)$.

Parameters:
- $N = 20$ (total returns)
- $K = 6$ (returns with errors)
- $n = 4$ (sample size)
- $k = 0$ (finding zero errors)

Calculate $P(X = 0)$:
$P(X = 0) = \frac{\binom{6}{0} \binom{20-6}{4-0}}{\binom{20}{4}} = \frac{\binom{6}{0} \binom{14}{4}}{\binom{20}{4}}$

$\binom{6}{0} = 1$
$\binom{14}{4} = 1001$
$\binom{20}{4} = 4845$

$P(X = 0) = \frac{1001}{4845} \approx 0.2066$

Now, subtract from 1:
$P(X \ge 1) = 1 - 0.2066 = 0.7934$

There is a 79.34% chance the auditor will catch at least one flawed return.

Technically, because we are sampling without replacement from a finite population, this is a Hypergeometric problem:
- $N = 10000, K = 200, n = 50, k = 2$

However, let's check the 5% rule:
$5\% \text{ of } 10000 = 0.05 \times 10000 = 500$.
Our sample size $n = 50$ is much less than 500, so we can safely use the Binomial distribution to approximate this!

For the Binomial approximation:
- $n = 50$ trials
- $p = K/N = 200 / 10000 = 0.02$ (probability of defect)

$P(X = 2) \approx \binom{50}{2} (0.02)^2 (0.98)^{48}$

$\binom{50}{2} = 1225$
$P(X = 2) \approx 1225 \times 0.0004 \times 0.3793 \approx 0.1858$

There is roughly an 18.6% chance of finding exactly 2 defective chips. (For context, the true Hypergeometric probability is 0.1860, showing just how accurate this shortcut is!)