Calculating conditional probability when the definition is hard(?) to apply

Question

I am reading a solution to the famous Induction Paradox. The solution contains the expression $P(W|W') = \frac q2$ where $P(W) = P(\text{a white cow was observed}) = q$ and $P(W') = P(\text{50% of all cows are white}) = \frac 12.$

How did they get $P(W|W') = \frac q2$? By definition and product rule, $P(W|W') = \frac{P(W \cap W')}{P(W')} = \frac{P(W)P(W'|W)}{P(W')} = \frac{q \cdot x}{\frac 12}$ where $x$ must be $\frac 14$ for $P(W|W')$ to have a correct value. Then we have $P(W'|W) = \frac 14$ which I am not sure how that happens. In short, I'd like to know how they calculate expressions like $P(W|W')$ when the definition of conditional probability is hard(?) to apply?

Edit (statement of the problem):

Let $V$ be an event where all cows are white and $W'$ be an event where $50$% of all cows are white s.t. $V, W'$ are complimentary. Let $P(V) = p$. We make an observation of a cow or a crow, with probability $q$ and $1 - q$, respectively, independent of whether $V$ occurs or not. Assume $0< p, q < 1$ and that all crows are black. What are the values of $P(V|\text{a black crow was observed})$ and $P(V|\text{a white cow was observed})$?

Answer 1

Your statement that $P(W)=q$ does not match the quoted text. If $C$ is the event that you observe a cow, the quoted text says $P(C) = q.$

So we have

$$P(W|W') = \frac{P(W \cap W')}{P(W')} = \frac{P(W \cap W' \cap C) + P(W \cap W' \cap C^\complement)}{P(W')}, $$

where

$$ P(W \cap W' \cap C) = P(W \mid W' \cap C) P(W' \cap C). $$

Note that $W \subseteq C,$ so $W \cap W' \cap C^\complement = \emptyset$ and therefore $P(W \cap W' \cap C^\complement) = 0.$

At this point we're going to have to apply some common sense and reasonable assumptions. For example, it seems reasonable to assume that the event of seeing a cow is independent of whether cows are all white or only half are white. Therefore $P(W' \cap C) = P(W')P(C) = \frac q2.$ Moreover, it seems reasonable to assume that if you observe a cow, the probability it is white is the same as the relative proportion of cows that are white. That is, given $C$ and $W'$, you have a $50\%$ chance to see a white cow: $P(W \mid W' \cap C) = \frac12.$

Plugging this all in,

$$P(W|W') = \frac{\frac12 \times \frac q2 + 0}{\frac12} = \frac q2.$$

A simpler way is to observe that

$$P(\text{crow}\mid W') + P(\text{non-white cow}\mid W') + P(W\mid W') = 1,$$

and assuming that observing a crow is independent of what color cows are, $P(\text{crow}\mid W') = P(\text{crow}) = 1 - q.$ Also, since the number of white cows and non-white cows are equal (given $W'$), we reasonably can assume that $$ P(\text{non-white cow}\mid W') = P(W\mid W').$$

From this and from the fact that $P(\text{non-white cow}\mid W') + P(W\mid W') = q,$ we conclude that $P(W\mid W') = \frac q2.$

Answer 2

I think unless you know $P(W'|W^{c})$, i.e., probability that 50% of all cows are white when the observed cow was not white, you won't be able to determine $P(W|W')$. So that is already stated (either explicitly or implicitly) in your problem.