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STATEMENT Bowl 1 contains 30 vanilla cookies and 10 chocolate cookies. Bowl 2 contains 20 of each. Now suppose you choose one of the bowls at random and, without looking, choose a cookie at random. The cookie is vanilla. What is the Probability that it came from Bowl 1?

$P(\text {choosing Bowl 1}) = \frac 12$

$P(\text {choosing a vanilla cookie from Bowl 1}) = \frac {30}{40}$

Since events are independent, $ = \frac 12 \times \frac {30}{40} = \frac 38$ .. and this answer is wrong.

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This is a good example of Bayes Rule: $$P(B_1|V)=\frac{P(V|B_1)P(B_1)}{P(V)}$$

Your calculation gives $P(V|B_1)P(B_1)=P(B_1\cap V)$, but the denominator needs to be calculated as well:

$$P(V)=P(V|B_1)P(B_1)+P(V|B_2)P(B_2)$$

A toy example: Bowl 1 contains 1 vanilla, Bowl 2 contains 1 chocolate. Say, we select a random bowl and select a random cookie. If the chosen cookie, is vanilla, what is the probability that it's chosen from Bowl 1?

Your way gives $1/2\times1=1/2$, but actually it's certain, i.e. probability is 1, that the cookie is from Bowl 1. Keep in mind that what you're asked is $P(B_1|V)$, not $P(B_1\cap V)$.

Note: $V|B_1$ is not an event. This notation makes sense inside $P(.)$ expression. So, you should define your events carefully and express the question in terms of these events.

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  • $\begingroup$ I could have calculated it using Baye's Theorem. Even the denominator can be expanded to: $P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$ But the point is which part of the statement shows they are dependent events? $\endgroup$
    – ArnuldOnData
    Commented May 15, 2020 at 8:30
  • $\begingroup$ The probability you're asked is $P(B_1|V)$. But, you're calculating $P(B_1\cap V)$. So, multiplying them ($P(V|B_1)$ and $P(B_1)$) will not give you what is asked on the question. It's not a dependence/independence issue. $\endgroup$
    – gunes
    Commented May 15, 2020 at 10:26
  • $\begingroup$ So, conditional probability can be applied to independent events? $\endgroup$
    – ArnuldOnData
    Commented May 15, 2020 at 14:59
  • $\begingroup$ I can't follow you. What does 'applying conditional probability' mean? $\endgroup$
    – gunes
    Commented May 15, 2020 at 15:01
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    $\begingroup$ $V$ and $B_1$ are not independent. The odds of choosing a vanilla cookie depends on what bowl chosen. $\endgroup$
    – gunes
    Commented May 15, 2020 at 15:45

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