$\frac{5}{12}\cdot \frac{4}{11}\cdot \frac{7}{10}\cdot \frac{6}{9}\cdot 6\approx 0.4242$

$1-(\frac{8}{12}\cdot \frac{7}{11}\cdot \frac{6}{10}\cdot \frac{5}{9}+\frac{4}{12}\cdot \frac{8}{11}\cdot \frac{7}{10}\cdot \frac{6}{9}\cdot 4)\approx 0.4061$

$\frac{9}{12}\cdot \frac{8}{11}\cdot \frac{7}{10}\cdot \frac{6}{9}\approx 0.2545$

$4\cdot \frac{5}{12}=1\frac{2}{3}$

A can will not appear twice in the same sample. It therefore is sampling without replacement.

$\frac{100}{1000}\cdot \frac{99}{999}\cdot \frac{98}{998}\cdot \frac{900}{997}\cdot \frac{899}{996}\cdot \frac{898}{995}\cdot \frac{897}{994}\cdot \frac{896}{993}\cdot \left(\begin{array}{c}8\\ 3\end{array}\right)\approx 0.0326$

$\text{P}(B=3|n=8\text{and}p=0.10)\approx 0.0331$. The difference is is 0.0005, or 0.05%.

The difference is very small because you are taking a very small sample (8 cans) out of a large population (1000 cans).

$\text{P}(B\le 3|n=8\text{and}p=0.10)\approx 0.9950$

$4\cdot 0.4=1.6$

$\frac{8}{20}\cdot \frac{7}{19}\cdot \frac{6}{18}\cdot \frac{12}{17}\cdot 4\approx 0.1387$

$\text{P}(X=3|n=4\text{and}p=0.40)\approx 0.1536$. This probability differs from the actual chance by $0.0149$, which is quite a large difference.

$\frac{40}{100}\cdot \frac{39}{99}\cdot \frac{38}{98}\cdot \frac{60}{97}\cdot 4\approx 0.1512$

$\text{P}(X=3|n=4\text{and}p=0.40)\approx 0.1536$. The difference is now much smaller. The larger the sampling population, the smaller the difference.

$\frac{5}{30}\cdot \frac{25}{29}\cdot \frac{24}{28}\cdot \frac{23}{27}\cdot 4\approx 0.4196$

$1-(\frac{25}{30}\cdot \frac{24}{29}\cdot \frac{23}{28}\cdot \frac{22}{27}+\frac{5}{30}\cdot \frac{25}{29}\cdot \frac{24}{28}\cdot \frac{23}{27}\cdot 4)\approx 0.3872$

$\frac{5}{30}\cdot \frac{4}{29}\cdot \frac{3}{28}\cdot \frac{25}{27}\cdot 4\approx 0.0091$