There are 7 balls in urn. $Q$ of them are white and the rest are black. We have hypothesis $H_0:Q=3$ and $H_1:Q=5$. To test this we draw 2 balls (balls don't come back to the urn - i.e. they are drawn without replacement).
We reject $H_0$ when both balls drawn are white. Calculate probability of type I and type II errors.
MY PROBLEM - what are probabilities of $H_0$ and $H_1$? Are they actually needed?
SOLUTION SKETCH:
Let $x$ be number of drawn white balls.
$\alpha=\mathbb{P}(x>1|H_0)=\frac{\frac{3}{7}\cdot\frac{2}{6}}{\frac{3}{7}}=\frac{2}{6}$
or it is
$\alpha=\mathbb{P}(x>1|H_0)=\frac{3}{7}\cdot\frac{2}{6}=\frac{1}{7}$
and for II type error: $\beta=\mathbb{P}(x<2|H_1)=\frac{\frac{5}{7}\cdot\frac{2}{6}+\frac{2}{7}\cdot\frac{1}{6}+\frac{2}{7}\cdot\frac{5}{6}}{\frac{5}{7}}=\frac{11}{15}$
Or is it wrong?