Probability of breakfast at work The problem
25% of the time there is free breakfast at work. You just woke up and are really hungry, so you call your coworker who is already at work, and he tells you there is free breakfast. However, he lies to you 1/3 of the time. How likely is it that there is free breakfast at work today?
My solution(s)
The probability is 0.25 -> I arrived at this answer in three ways, and I have a nagging feeling I am wrong; could someone validate my answer or point out my mistakes?
Logical reasoning: 
The probability remains the same regardless of what my friend tells me -> hence P(Breakfast) = 0.25
Applying Bayes Rule
P(Breakfast & Lie) = 0.25 * 0.33 = 8.25%

P(Breakfast & Truth) = 0.25 * 0.67 = 16.75%

Therefore, P(Breakfast)= 25%

P(No Breakfast & Lie) = 0.75 * 0.33 = 24.75%

P(No Breakfast & Truth) = 0.75 * 0.67 = 50.25%

Therefore, P(No Breakfast)= 75%

Data Table

 A: The question in plain english:

What is the probability that there is free breakfast today given that your coworker said there is?


We can write this using the following notation:
$$ P(free\ breakfast\ |\ coworker\ said\ yes) $$

Baye's Theorem states the following:


$$
 P(free\ breakfast\ |\ coworker\ said\ yes)
$$
$$
= \frac{P(coworker\ said\ yes\ |\ free\ breakfast)\ *\ P(free\ breakfast)}{P(coworker\ said\ yes)}
$$

Now let's address each of these in turn:

$$
P(free\ breakfast) = \frac{1}{4}
$$

$$
P(coworker\ said\ yes\ |\ free\ breakfast) = P(coworker\ is\ honest) = \frac{2}{3}
$$

$$
P(coworker\ said\ yes)
$$
$$
= \bigl(P(breakfast)\ *\ P(coworker\ is\ honest)\bigr)\ +\ \bigl(P(no\ breakfast)\ *\ P(coworker\ is\ dishonest)\bigr)
$$
$$
= \Bigl(\frac{1}{4}\Bigr)\Bigl(\frac{2}{3}\Bigr)\ +\ \Bigl(\frac{3}{4}\Bigr) \Bigl(\frac{1}{3}\Bigr) = \frac{5}{12}
$$

Now put it all together!

$$
 P(free\ breakfast\ |\ coworker\ said\ yes)
$$
$$
= \frac{P(coworker\ said\ yes\ |\ free\ breakfast)\ *\ P(free\ breakfast)}{P(coworker\ said\ yes)}
$$

$$
= \frac{\bigl(\frac{2}{3}\bigr)\bigl(\frac{1}{4}\bigr)}{\frac{5}{12}} = \frac{2}{5}
$$
$$
= 40\%
$$

So there is a 40% chance that there is free breakfast today.
A: Tip: lying is a composite event, dependent on what is actual and what is said. Focus on those events.
Let event $B$ be that there is free breakfast.
Let event $S$ be that the co-worker says there is free breakfast. 
You are told the probability for there being breakfast, and for the coworker lying. $$\begin{align}\mathsf P(B)&=1/4\\ \mathsf P(S\mid B)&=2/3\\ \mathsf P(S\mid B^\complement)&=1/3\\\text{et cetera}&~~_\ddots\end{align}$$
So we use Bayes' Rule to find the conditional probability for there being breakfast given that the co-worker says so:
$$\begin{align}\mathsf P(B\mid S)&=\dfrac{\mathsf P(S\mid B)\mathsf P(B)}{\mathsf P(S\mid B)\mathsf P(B)+\mathsf P(S\mid B^\complement)\mathsf P(B^\complement)}\\&=\dfrac{2}{2+3}\\&=\dfrac 25\end{align}$$
