But for (b) i am having problems with using Baye's rule...

I have defined the event A=1 pair dancing in the second

event B=no pair dancing in the first.

So

$\displaystyle P(B)=\frac{9}{24}$ and $\displaystyle P(A)=P(\mbox{1 pair in first from 4} )*$$\displaystyle P( \mbox{0 pair in second dance from 3})$$\displaystyle +P(\mbox{0 pair in first from 4})*P(\mbox{1 pair in second dance from 4})$$\displaystyle =\frac{8}{24}\cdot \frac{1}{3}+\frac{9}{24}\cdot \frac{8}{24}=\frac{17}{72}$

So for bayes rule, $\displaystyle P(B|A)=\frac{P(B) \cdot P(A|B)}{P(A)}$

with $\displaystyle P(A|B)=\frac{P(A \cap B)}{P(B)}$

So I am unsure what $\displaystyle P(A \cap B)$ should equal. I think from the definition of finding the probability of A (Using the law of total probability?) $\displaystyle P(A)=P(A \cap B)+P(A \cap \bar{B})$ in which case $\displaystyle P(A \cap B)=\frac{9}{24}\cdot \frac{8}{24}=\frac{1}{8}$

Which means that $\displaystyle P(A|B)=\frac{1/8}{9/24}=\frac{1}{3}$

So $\displaystyle P(B|A)=\frac{9/24 \cdot 1/3}{17/72}=\frac{9}{17}$

I really feel like i have done something wrong with the intercept of A and B...