Are there two ways to calculate conjunction probability in the cookie problem?

Question

I'm reading Think Bayes By Allen B. Downey.

He introduces a simple conditional probability problem.

You have two bowl's of cookies.

Bowl 1 has 30 vanilla and 10 chocolate cookies Bowl 2 has 20 vanilla and 20 chocolate cookies

If you select a cookie and it is vanilla, what is the probability that it came from bowl 1? Assuming that you have an equal probability of selecting from either bowl (0.5).

You can use the formula for conditional probability to solve this.

P(Bowl 1 | Vanilla) = P(Bowl 1 ∩ Vanilla) / P(Vanilla)

There appear to be two ways to solve for the numerator: P(Bowl 1 ∩ Vanilla)

I understand that intersecting probabilities (∩) can be calculated by multiplying the probability of each event. Thus, to calculate the numerator you would multiply P(Bowl 1) * P(Vanilla)

Which would be (0.5 * (Probability of selecting vanilla from bowl 1)

Or (0.5 * 0.75) = 0.375

Plugging that into the conditional probability formula gives 0.375/0.625 = 0.6

Claude (the AI Chatbot) presented this calculation to determine the numerator.

Vanilla cookies in Bowl 1 / Total cookies, which is 30/80 or 0.375

Is this a more accurate way to calculate the numerator than what I assumed? P(Bowl 1) * P(Vanilla)

Answer 1

You got the same answer for $P(\text{Bowl }1 \cap \text{ Vanilla})$ using different methods here because the probability of choosing a bowl was equal to the proportion of cookies in that bowl. Trying a different setup could help understand the difference in methods:

Suppose instead Bowl 1 had 6 vanilla and 2 chocolate cookies, while Bowl 2 had 20 vanilla and 20 chocolate cookies.

• If the probability of choosing Bowl 1 was $\frac12$ and the conditional probability of then choosing a vanilla cookie was $\frac{6}{6+2}=\frac{3}{4}$ then the joint probability of having a vanilla cookie from Bowl 1 would be $\frac12 \times \frac34=\frac38 = 0.375$

• If you chose one of the $6+2+20+20=48$ cookies with equal probability then the joint probability of having a vanilla cookie from Bowl 1 would be $\frac6{48}=\frac{1}{8} = 0.125$

The question tells you which of those situations you are in and therefore which method is correct.