Q.Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Concept: Conditional Probability — we need $P(\text{other coin gold} \mid \text{first coin gold})$.

Step 1: Define events.

Let $B_1$, $B_2$, $B_3$ be the events of choosing box I, II, III respectively. Each box is equally likely: $P(B_1)=P(B_2)=P(B_3)=\frac13$.

Step 2: Probability of drawing a gold coin from each box:

• Box I: both gold → $P(\text{gold} \mid B_1) = 1$
• Box II: no gold → $P(\text{gold} \mid B_2) = 0$
• Box III: one gold → $P(\text{gold} \mid B_3) = \frac12$

Step 3: By Bayes' theorem,

$P(B_1 \mid \text{gold}) = \frac{P(\text{gold} \mid B_1) P(B_1)}{P(\text{gold})}$

where $P(\text{gold}) = 1\cdot\frac13 + 0\cdot\frac13 + \frac12\cdot\frac13 = \frac13 + \frac16 = \frac12$.

Thus $P(B_1 \mid \text{gold}) = \frac{1 \cdot \frac13}{\frac12} = \frac{2}{3}$.

If the first coin is gold, the other coin is also gold only if the box is Box I. So the required probability is $\frac{2}{3}$.

Why conditional probability is the right tool

The question asks: Given that the drawn coin is gold, what is the probability that the other coin in the same box is also gold? This is a textbook conditional probability problem — we are restricting our universe to only those outcomes where the first coin is gold, and then asking what fraction of those outcomes also satisfy the condition “the other coin is gold.”

A common mistake is to think that since you picked a gold coin, you must be in either box I or box III, and since those are two boxes, the answer is $\frac12$. That reasoning is wrong because the two boxes are not equally likely after you see the gold coin. Box I has two gold coins, so it is twice as likely to produce a gold coin as box III, which has only one. Conditional probability corrects for this imbalance.

Step-by-step solution

1. Define the events clearly

Let:

• $B_1$ = event that box I (two gold coins) is chosen
• $B_2$ = event that box II (two silver coins) is chosen
• $B_3$ = event that box III (one gold, one silver) is chosen
• $G$ = event that the drawn coin is gold

We want $P(\text{other coin is gold} \mid G)$. But “other coin is gold” is exactly the same event as “the chosen box is $B_1$” — because only in box I are both coins gold. So we want $P(B_1 \mid G)$.

2. Write down the prior probabilities

Since the box is chosen at random:

3. Write down the likelihoods — the probability of drawing a gold coin from each box

• From box I: both coins are gold, so $P(G \mid B_1) = 1$
• From box II: both coins are silver, so $P(G \mid B_2) = 0$
• From box III: exactly one gold coin out of two, so $P(G \mid B_3) = \frac12$

4. Apply Bayes’ theorem

Bayes’ theorem says:

We already have the numerator: $1 \cdot \frac13 = \frac13$.

Now find $P(G)$, the total probability of drawing a gold coin. By the law of total probability:

So:

5. Interpret the result

Given that you drew a gold coin, there is a $\frac23$ chance that you are in box I, meaning the other coin is also gold. Only $\frac13$ of the time are you in box III, where the other coin is silver.

Method: Bayes' Theorem (finding the cause from the observed outcome)

Use this whenever you observe an outcome and are asked for the probability of the underlying cause — the conditioning is reversed (you know the chance of a gold coin given each box, but want the chance of a particular box given that a gold coin appeared).

Steps

Step 1: Set up the partition of possible causes.

List the mutually exclusive, exhaustive hypotheses (here: the coin came from box I, II or III) and write their prior probabilities $P(E_i)$ (these must sum to $1$).

Step 2: Write each likelihood.

For each cause, state $P(A\mid E_i)$ — the probability of the observed outcome (here: drawing a gold coin) under that cause.

Step 3: Get the total probability of the outcome (the denominator).

By the law of total probability,

Step 4: Apply Bayes' theorem for the cause you want.

The numerator is just the one term of the denominator that belongs to the cause you are asked about — so the answer is that cause's share of the total chance of the outcome.

Key subtlety: the causes are not equally likely after the observation. A box holding more gold is more likely to have produced the gold coin, so weight each cause by both its prior and its likelihood — never assume the surviving boxes are 50–50.

Common Mistakes

Mistake 1: "The gold coin is in box I or box III, so the answer is $\tfrac12$."

Why it's wrong: after seeing a gold coin the two boxes are not equally likely — box I (two gold coins) is twice as likely to have produced a gold coin as box III (one gold coin). Correct approach: weight by the likelihoods, giving $\tfrac23$.

Mistake 2: Setting $P(\text{gold}\mid\text{box III})=1$.

Why it's wrong: box III has one gold and one silver, so the chance of drawing its gold coin is $\tfrac12$. Correct approach: use $P(\text{gold}\mid\text{box III})=\tfrac12$.

Mistake 3: Answering the prior $P(\text{box I})=\tfrac13$ instead of the posterior.

Why it's wrong: the question conditions on having drawn a gold coin. Correct approach: compute $P(\text{box I}\mid\text{gold})$ via Bayes' theorem.