There Are Two Bags. One Bag Contains Six Green and Three Red Balls, the Second Bag Contains Five Green and Four Red Balls. One Ball Is Transferred From the First Bag to the Second, Then One Ball Is Drawn From the Second Bag. Find the Probability That It Is a Red Ball

ISC 2010. Previous Year Question.

Problem

There are two bags. One bag contains six green and three red balls. The second bag contains five green and four red balls. One ball is transferred from the first bag to the second bag. Then one ball is drawn from the second bag. Find the probability that it is a red ball.

Key insight. The ball transferred from the first bag changes the composition of the second bag before anything is drawn from it — so the draw’s probability genuinely depends on which colour was transferred. Splitting into the two possible transfer outcomes, and weighting each by its own probability, is the Law of Total Probability in action.

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Approach

Consider the two possible outcomes of the transfer: a green ball moves, or a red ball moves. Find the probability of each transfer outcome, and for each case, find the resulting composition of the second bag and the probability of drawing red from it. Combine both cases using the Law of Total Probability.

Solution

Step 1 — Find the probability of transferring each colour

Bag 1 has $6$ green and $3$ red, totalling $9$ balls:

$$P(\text{transfer green}) = \frac{6}{9}, \qquad P(\text{transfer red}) = \frac{3}{9}$$

Step 2 — Find the composition of Bag 2 in each case

If a green ball is transferred, Bag 2 becomes $5+1=6$ green and $4$ red (total $10$).

If a red ball is transferred, Bag 2 becomes $5$ green and $4+1=5$ red (total $10$).

Step 3 — Find P(draw red) in each case

$$P(\text{red}\mid\text{green transferred}) = \frac{4}{10}, \qquad P(\text{red}\mid\text{red transferred}) = \frac{5}{10}$$

Step 4 — Apply the Law of Total Probability

$$P(\text{red}) = P(\text{transfer green})\times P(\text{red}\mid\text{green transferred}) + P(\text{transfer red})\times P(\text{red}\mid\text{red transferred})$$

$$= \frac{6}{9}\times\frac{4}{10} + \frac{3}{9}\times\frac{5}{10}$$

Step 5 — Compute each term

$$\frac{6}{9}\times\frac{4}{10} = \frac{24}{90}, \qquad \frac{3}{9}\times\frac{5}{10} = \frac{15}{90}$$

Step 6 — Add and simplify

$$P(\text{red}) = \frac{24}{90}+\frac{15}{90} = \frac{39}{90} = \frac{13}{30}$$

Answer

Common mistakes

• Forgetting that the transferred ball changes Bag 2’s total count too. Bag 2 starts with $9$ balls, but after the transfer it has $10$ — using the original count of $9$ when computing the final draw’s probability gives the wrong denominator.
• Treating the transfer and the draw as independent events. The probability of drawing red depends entirely on which ball was transferred first — they must be combined via conditional probability (the Law of Total Probability), not multiplied as if unrelated.

Practise next

• Bag $X$ has $4$ white and $5$ black balls; Bag $Y$ has $3$ white and $6$ black balls. One ball is transferred from $X$ to $Y$, then one is drawn from $Y$. Find the probability it is white, using the same total-probability approach.