The probability of having pizza on a Saturday is 60%, on Sunday, 30%.
a) What is the probability of having pizza at least once during the weekend?
b) How do the odds change if you are only allowed pizza on one of the two days?
My solutions:
a) is $1 - P(\text{Saturday}^C)*P(\text{Sunday}^C) = 1 - 0.28 = 0.72$
b)
Using the union: $P(\text{at least once}) = P(\text{Saturday}) + P(\text{Sunday}) - P(\text{Saturday} \cap \text{Sunday}) = 0.6 + 0.3 - 0 = 0.9$
Using Bayes: we want $P(\text{at least once} | \text{both days}^C) = P(A |B^C)$.
By the law of total probability we have $P(A) = P(A|B)P(B) + P(A|B^C)P(B^C)$.
This implies that: $P(A|B^C) = \frac{P(A) - P(A|B)P(B)}{P(B^C)}$.
$P(A) = 0.72, P(A|B) = 1, P(B) = 0.18, P(B^C) = 0.82$
$\frac{P(A) - P(A|B)P(B)}{P(B^C)} = \frac{0.72 - 1*0.18}{0.82} = .54/.82 = 27/41$.
Which methodology in part b) is correct (if any?)
Let $\text{Sat}$ ($\text{Sun}$) be the event of having a pizza on Saturday (Sunday).
In general, the probability of having pizza on at least one day is
$$P(\text{Sat}\cup\text{Sun})=P(\text{Sat})+P(\text{Sun})-P(\text{Sat}\cap\text{Sun})$$
If we interpret "you are only allowed pizza on one of the two days" as $P(\text{Sat}\cap \text{Sun})=0$, and the information at the beginning of the question as giving the unconditional probabilities $P(\text{Sat})=0.6$ and $P(\text{Sun})=0.3$, then it is indeed the case that
$$P(\text{Sat}\cup\text{Sun})=P(\text{Sat})+P(\text{Sun})=0.6+0.3=0.9$$
An alternative calculation would be:
$$\begin{align*}P(\text{Sat}\cup\text{Sun})&=P(\text{Sat})+P(\text{Sun}|\text{Sat}^C)P(\text{Sat}^C)\\ &=0.6+0.4P(\text{Sun}|\text{Sat}^C) \tag{1} \end{align*}$$
or
$$\begin{align*}P(\text{Sun}\cup\text{Sat})&=P(\text{Sun})+P(\text{Sat}|\text{Sun}^C)P(\text{Sun}^C)\\ &=0.3+0.7P(\text{Sat}|\text{Sun}^C) \tag{2} \end{align*}$$
Since, $P(\text{Sat}\cap\text{Sun}^C)=P(\text{Sat})=0.6$ and $P(\text{Sun}\cap\text{Sat}^C)=P(\text{Sat})=0.6$, we have
$$\begin{align*}P(\text{Sat}|\text{Sun}^C)&=\frac{P(\text{Sat}\cap\text{Sun}^C)}{P(\text{Sun}^C)}=\frac{0.6}{1-0.3}=\frac{0.6}{0.7}\approx 0.8571\\ P(\text{Sun}|\text{Sat}^C)&=\frac{P(\text{Sun}\cap\text{Sat}^C)}{P(\text{Sat}^C)}=\frac{0.3}{1-0.6}=\frac{0.3}{0.4}=0.75\end{align*} $$
Substituting into $(1)$ and $(2)$ gives the same value as before, i.e. $0.9$.
In my notation, the equation from the law of total probability
$$\begin{align*}P(\text{Sat}\cup \text{Sun})&=\phantom{+}P(\text{Sat}\cup \text{Sun}|\text{Sat}\cap \text{Sun})P(\text{Sat}\cap \text{Sun})\\&\phantom{=}+ P(\text{Sat}\cup \text{Sun}|(\text{Sat}\cap \text{Sun})^C)P[(\text{Sat}\cap \text{Sun})^C]\end{align*}$$
But we have $P(\text{Sat}\cap \text{Sun})=0$ and $P((\text{Sat}\cap \text{Sun})^C)=1$, so the equation becomes just
$$P(\text{Sat}\cup \text{Sun})= P(\text{Sat}\cup \text{Sun}|(\text{Sat}\cap \text{Sun})^C)$$
Earlier in your answer, you have written
$$P(\text{at least once}|\text{both days}^C)$$
In my notation this would be
$$P(\text{Sat}\cup \text{Sun}|(\text{Sat}\cap \text{Sun})^C)$$
But, from above, this is the same as $P(\text{Sat}\cup \text{Sun})$.
Later in your answer you write (in my notation):
$$\begin{align*}P(\text{Sat}\cup \text{Sun})&=0.72,\ P(\text{Sat}\cup \text{Sun}|\text{Sat}\cap \text{Sun} )=1,\\ P(\text{Sat}\cap \text{Sun})&=0.18,\ P[(\text{Sat}\cap \text{Sun})^C]=0.82\end{align*} $$
Here $P(\text{Sat}\cup \text{Sun})$ is the thing we need to calculate. It is not clear why you have assigned to it the value you found in part (a). Also, $P(\text{Sat}\cap \text{Sun} )=0$, so $P(\text{Sat}\cup \text{Sun}|\text{Sat}\cap \text{Sun} )$ is not defined and $P[(\text{Sat}\cap \text{Sun})^C]=1$ (not $0.82$).