Probability

$6 \times 6 = 36$ ordered pairs. $(1,6)$ and $(6,1)$ count separately.

Six pairs give 7, out of 36 outcomes: $\dfrac{6}{36} = \dfrac16$. It is the most likely total.

Use the complement: $P(\text{no six}) = \left(\tfrac56\right)^2 = \tfrac{25}{36}$, so the answer is $\tfrac{11}{36}$.

$\dfrac{\binom43}{\binom{10}{3}} = \dfrac{4}{120} = \dfrac{1}{30}$.

75 candidates passed, of whom 45 were women: $\dfrac{45}{75} = \dfrac35$. Note this differs from $P(\text{passed}\mid\text{female}) = 0.75$.

$0.5(0.02) + 0.3(0.03) + 0.2(0.05) = 0.010 + 0.009 + 0.010 = 0.029$, or 2.9%.