Q.Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Idea: This is a Bayes' theorem problem. The transferred ball is the unknown "cause"; drawing a red from Bag II is the observed "effect."

Let $E_1$ = red transferred, $E_2$ = black transferred, $A$ = ball drawn from Bag II is red.

Priors (from Bag I: 3 red, 4 black): $P(E_1)=\dfrac{3}{7}$, $P(E_2)=\dfrac{4}{7}$.

Likelihoods (Bag II starts with 4 red, 5 black; total becomes 10 after transfer):

• Red transferred $\Rightarrow$ 5 red of 10: $P(A\mid E_1)=\dfrac{5}{10}$.
• Black transferred $\Rightarrow$ 4 red of 10: $P(A\mid E_2)=\dfrac{4}{10}$. …

The setup

One ball is moved from Bag I (3 red, 4 black) into Bag II (4 red, 5 black), and then a ball is drawn from Bag II and turns out to be red. We want the chance that the ball we transferred was black. Two hidden possibilities ("red transferred" or "black transferred") could have led to the same visible outcome ("red drawn"), so this is a Bayes' theorem question.

Define:

• $E_1$: the transferred ball is red
• $E_2$: the transferred ball is black
• $A$: the ball drawn from Bag II is red

We want $P(E_2\mid A)$.

Step 1 — prior probabilities

Bag I has $3+4=7$ balls, so

Step 2 — likelihoods after the transfer

Bag II starts with 4 red and 5 black (9 balls); adding one ball makes 10.

If a red was transferred, Bag II has $5$ red and $5$ black:

If a black was transferred, Bag II has $4$ red and $6$ black:

Step 3 — apply Bayes' theorem

$P(E_2\mid A)=\frac{P(E_2)\,P(A\mid E_2)}{P(E_1)\,P(A\mid E_1)+P(E_2)\,P(A\mid E_2)}.$ …

Method: Bayes' Theorem for a transfer-then-draw problem

Use this when an unseen item is moved from one bag to another, a draw is then made from the second bag, and you must judge what the transferred item was.

Steps

Step 1: Make the hidden transfer the set of hypotheses.

The unknown is the colour transferred; list those cases $H_i$ with priors equal to the composition of the source bag.

Step 2: Update the receiving bag for each case, then read the draw likelihood.

Add the transferred item to the second bag — its total count goes up by one — and recompute $P(E \mid H_i)$, the chance of the observed draw for that case. Forgetting to update the count (and the total) is the classic slip. …

Common Mistakes

Mistake 1: Not updating Bag II's count after the transfer.

Why it's wrong: after one ball is added, Bag II holds $10$, not $9$; the draw likelihoods are read from the $10$-ball bag. Correct approach: red transferred $\Rightarrow$ $5$ red of $10$; black transferred $\Rightarrow$ $4$ red of $10$.

Mistake 2: Taking the priors from Bag II instead of Bag I.

Why it's wrong: the transferred ball's colour is governed by Bag I's composition ($3$ red, $4$ black). Correct approach: $P(\text{red transferred})=\tfrac37$, $P(\text{black transferred})=\tfrac47$. …