Q.Suppose we have four boxes A,B,C and D containing coloured marbles as given below: $\begin{array}{|c|c|c|c|} \hline \text{Box} & \text{Marble colour} \\ \hline & \text{Red} & \text{White} & \text{Black} \\ \hline \text{A} & 1 & 6 & 3 \\ \hline \text{B} & 6 & 2 & 2 \\ \hline \text{C} & 8 & 1 & 1 \\ \hline \text{D} & 0 & 6 & 4 \\ \hline \end{array}$ One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Concept: Conditional Probability — we need $P(\text{Box} \mid \text{Red})$, found using Bayes' theorem.

Step 1: Prior probabilities. Each box is equally likely:

$P(A)=P(B)=P(C)=P(D)=\frac14$.

Step 2: Likelihoods (probability of drawing a red marble from each box).

Total marbles per box: A has $1+6+3=10$, B has $6+2+2=10$, C has $8+1+1=10$, D has $0+6+4=10$.

$P(\text{Red}\mid A)=\frac{1}{10},\; P(\text{Red}\mid B)=\frac{6}{10},\; P(\text{Red}\mid C)=\frac{8}{10},\; P(\text{Red}\mid D)=0$.

Step 3: Total probability of drawing a red marble.

$P(\text{Red}) = \frac14\left(\frac{1}{10}+\frac{6}{10}+\frac{8}{10}+0\right) = \frac14 \cdot \frac{15}{10} = \frac{15}{40} = \frac{3}{8}$.

Step 4: Apply Bayes' theorem.

$P(A\mid\text{Red}) = \frac{P(\text{Red}\mid A)P(A)}{P(\text{Red})} = \frac{\frac{1}{10}\cdot\frac14}{\frac{3}{8}} = \frac{1/40}{3/8} = \frac{1}{40}\cdot\frac{8}{3} = \frac{8}{120} = \frac{1}{15}$. …

The problem gives us four boxes with different compositions of red, white, and black marbles. One box is chosen at random, then one marble is drawn from it. We are told the marble is red, and we need the probability that it came from each specific box.

This is a classic case of inverse probability — we know the probability of drawing a red marble given a particular box, but we want the probability of that box given that the marble is red. The tool for this is Bayes’ theorem, which is built on conditional probability.

Why Bayes’ theorem works here:

We start with the prior probability of each box being chosen (all equal, since selection is random). Then we update that probability using the likelihood of observing a red marble from that box. The denominator normalises by the total probability of getting a red marble from any box.

Let’s go step by step.

1. Define the events and priors

Let $R$ be the event that the drawn marble is red.

Let $A$, $B$, $C$, $D$ be the events that the chosen box is A, B, C, D respectively.

Since one box is selected at random from four, each has equal prior probability:

2. Find the probability of drawing a red marble from each box

From the table:

• Box A: 1 red out of $1+6+3 = 10$ marbles → $P(R \mid A) = \frac{1}{10}$
• Box B: 6 red out of $6+2+2 = 10$ marbles → $P(R \mid B) = \frac{6}{10} = \frac{3}{5}$
• Box C: 8 red out of $8+1+1 = 10$ marbles → $P(R \mid C) = \frac{8}{10} = \frac{4}{5}$
• Box D: 0 red out of $0+6+4 = 10$ marbles → $P(R \mid D) = 0$

3. Compute the total probability of drawing a red marble

Using the law of total probability:

Substitute:

Convert to a common denominator (20 works nicely):

• $\frac{1}{4} \cdot \frac{1}{10} = \frac{1}{40}$
• $\frac{1}{4} \cdot \frac{3}{5} = \frac{3}{20} = \frac{6}{40}$
• $\frac{1}{4} \cdot \frac{4}{5} = \frac{4}{20} = \frac{8}{40}$

So:

4. Apply Bayes’ theorem for each box

Bayes’ theorem says:

$P(\text{Box} \mid R) = \frac{P(\text{Box}) \cdot P(R \mid \text{Box})}{P(R)}$ …

Method: Bayes' Theorem across several boxes (watch the empty branch)

Use this when one of several containers is chosen at random, an item is drawn, and you must find the posterior probability for each container.

Steps

Step 1: Assign equal priors to the containers.

$P(H_i) = \tfrac{1}{n}$ when the box is chosen at random.

Step 2: Compute each container's likelihood of the observed colour.

$P(E \mid H_i) = \dfrac{\text{favourable items in box } i}{\text{total items in box } i}$. A box with none of the drawn colour has likelihood $0$ — dismiss it immediately, its posterior is $0$.

Step 3: Total probability of the evidence. …

Common Mistakes

Mistake 1: Spending effort on Box D instead of seeing $P(R\mid D)=0$.

Why it's wrong: Box D has no red marbles, so its posterior is immediately $0$. Correct approach: drop D from the numerator work once you note zero red.

Mistake 2: Forgetting to normalise by $P(R)$.

Why it's wrong: the joint probabilities $P(\text{box}\cap R)$ are not the posteriors until divided by the total $P(R)=\tfrac38$. Correct approach: divide each box's joint by $\tfrac38$.

Mistake 3: Misreading a box's red count or total. …