Q.Refer to Question 41 above. If a white ball is selected, what is the probability that it came from

Referenced setup (Exemplar Q41): Bag 1 has $3$ red (no white), Bag 2 has $2$ red and $1$ white, Bag 3 has $3$ white, and bag $i$ is chosen with probability $P(\text{Bag }i)=\dfrac{i}{6}$. We want $P(\text{Bag }i\mid\text{white})$ by Bayes' theorem.

White-ball likelihoods: $P(W\mid B_1)=0$, $P(W\mid B_2)=\dfrac{1}{3}$, $P(W\mid B_3)=1$.

Total probability of white:

$P(W)=\frac{1}{6}\cdot 0+\frac{2}{6}\cdot\frac{1}{3}+\frac{3}{6}\cdot 1=\frac{1}{9}+\frac{1}{2}=\frac{11}{18}.$ …

The referenced problem

From the previous question, the three bags and their selection probabilities are:

We are told a white ball was drawn and asked which bag it likely came from — a reverse-conditioning question, so we use Bayes' theorem.

Step 1: total probability of a white ball

Step 2: Bayes' theorem for Bag 2 …

Method: Bayes' theorem — reversing the conditioning

Use this for "given the result, which source produced it?" questions — you know $P(\text{effect}\mid\text{cause})$ but want $P(\text{cause}\mid\text{effect})$.

Steps

Step 1: Write priors and likelihoods

List $P(\text{source}_i)$ and $P(\text{observed}\mid\text{source}_i)$ for every source.

Step 2: Compute the denominator by total probability

$P(\text{observed})=\sum_i P(\text{source}_i)\,P(\text{observed}\mid\text{source}_i).$ …

Common Mistakes

Mistake 1: Reporting the likelihood instead of the posterior

Why it's wrong: the question asks $P(\text{Bag}\mid\text{white})$, but students often stop at $P(\text{white}\mid\text{Bag})$. Correct approach: these differ; you must flip via Bayes.

Mistake 2: Omitting the denominator

Why it's wrong: dividing by the total probability of white is what normalises the posteriors to sum to $1$; skipping it leaves an un-normalised joint probability. Correct approach: always divide by $P(W)$. …