Q.A bag contains $4$ white and $5$ black balls. Another bag contains $9$ white and $7$ black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

The transferred ball can be white or black, so use the law of total probability: split on the two transfer cases and add the weighted probabilities of then drawing white.

1. Transfer from the first bag (4 white, 5 black, 9 total):

2. Draw from the second bag (starts 9 white, 7 black; becomes 17 balls after the transfer):

• White transferred $\Rightarrow$ 10 white, 7 black: $P(\text{white} \mid \text{white transferred}) = \dfrac{10}{17}$. …

The idea

A ball is moved from the first bag into the second, and only then do we draw from the second bag. The trouble is we do not know which colour was transferred, and that colour changes what the second bag looks like. So we consider both possibilities, weight each by how likely that transfer was, and add the results. That is the law of total probability:

Step 1 — the transfer

The first bag has $4$ white and $5$ black balls, $9$ in all, so the transferred ball is just a random draw from it:

Step 2 — drawing white from the second bag

The second bag starts with $9$ white and $7$ black balls ($16$ total). After one ball is added, it always holds $17$ balls — this updated denominator is the step students most often miss.

• If a white ball was transferred, the second bag has $9+1 = 10$ white and $7$ black:

• If a black ball was transferred, the second bag has $9$ white and $7+1 = 8$ black: …

Method: Law of Total Probability with a Changed Composition (transfer problems)

Use this whenever an item is moved from one collection to another and then something is drawn, so the second stage depends on an unknown first-stage outcome.

Steps

Step 1: Identify the hidden first stage and split it into cases.

The transferred item has an unknown identity. List every possibility (here: a white or a black ball moves), and find the probability of each straight from the source collection.

Step 2: For each case, update the second collection, then find the target conditional probability. …

Common Mistakes

Mistake 1: Keeping the second bag's total unchanged after the transfer.

Why it's wrong: one ball is added, so the total rises (here $16 \to 17$); using $16$ in the denominators gives conditionals like $\tfrac{10}{16}$ that no longer describe the real bag. Correct approach: after any transfer, recount the total and use the updated denominator ($17$) in every conditional.

Mistake 2: Forgetting to weight the two cases by the transfer probability. …