Concept: Conditional Probability (Bayes' Theorem)

We need $P(\text{Box III} \mid \text{Black})$.

Step 1 – Prior probabilities

Each box is equally likely: $P(\text{I}) = P(\text{II}) = P(\text{III}) = P(\text{IV}) = \frac{1}{4}$.

Step 2 – Likelihoods (probability of drawing a black ball from each box)

• Box I: 3 black out of $3+4+5+6 = 18$ balls → $P(\text{Black} \mid \text{I}) = \frac{3}{18} = \frac{1}{6}$
• Box II: 2 black out of 8 balls → $P(\text{Black} \mid \text{II}) = \frac{2}{8} = \frac{1}{4}$
• Box III: 1 black out of $1+2+3+1 = 7$ balls → $P(\text{Black} \mid \text{III}) = \frac{1}{7}$
• Box IV: 4 black out of $4+3+1+5 = 13$ balls → $P(\text{Black} \mid \text{IV}) = \frac{4}{13}$

Step 3 – Total probability of black

Compute common denominator (LCM of 6,4,7,13 = 1092):

$\frac{1}{6} = \frac{182}{1092},\; \frac{1}{4} = \frac{273}{1092},\; \frac{1}{7} = \frac{156}{1092},\; \frac{4}{13} = \frac{336}{1092}$

Sum = $\frac{182+273+156+336}{1092} = \frac{947}{1092}$

Thus $P(\text{Black}) = \frac{1}{4} \cdot \frac{947}{1092} = \frac{947}{4368}$

Step 4 – Bayes’ Theorem

Let $B_1,B_2,B_3,B_4$ be the events of selecting boxes I–IV, and $K$ the event of drawing a black ball. A box is chosen at random, so $P(B_i)=\tfrac14$.

Black-ball probability in each box:

• Box I: $3+4+5+6=18$ balls, $3$ black $\Rightarrow P(K\mid B_1)=\tfrac{3}{18}=\tfrac16$
• Box II: $2+2+2+2=8$ balls, $2$ black $\Rightarrow P(K\mid B_2)=\tfrac{2}{8}=\tfrac14$
• Box III: $1+2+3+1=7$ balls, $1$ black $\Rightarrow P(K\mid B_3)=\tfrac17$
• Box IV: $4+3+1+5=13$ balls, $4$ black $\Rightarrow P(K\mid B_4)=\tfrac{4}{13}$

Total probability of a black ball:

Bayes' theorem:

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

Use this when an item is drawn from one of several containers and, given its property, you want the probability of a particular container. You know the chance of the colour given each box, but want the box given the colour — reversed conditioning.

Steps

Step 1: Set up the partition of possible causes.

List the mutually exclusive, exhaustive hypotheses (here: the ball came from box I, II, III or IV) 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: the drawn ball is black) 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: compute each likelihood against that box's own total number of balls (the boxes here hold different totals), and weight by the equal prior $P(\text{box})=\tfrac14$ for random box selection — do not simply pool all black balls across boxes.

Common Mistakes

Mistake 1: Pooling all black balls over all balls.

Why it's wrong: computing $\tfrac{3+2+1+4}{\text{total balls}}$ ignores that a box is chosen first (each with probability $\tfrac14$) and that the boxes hold different totals. Correct approach: use Bayes' theorem over the four equally likely boxes.

Mistake 2: Not dividing each black count by that box's own total.

Why it's wrong: box III has $7$ balls, box I has $18$ — the black probability differs even for similar counts. Correct approach: $P(\text{black}\mid\text{box})=\dfrac{\text{black in box}}{\text{total in box}}$.

Mistake 3: Forgetting the equal prior $\tfrac14$ per box.

Why it's wrong: the box is selected at random, so each prior is $\tfrac14$ and must appear in every term. Correct approach: weight each likelihood by $\tfrac14$ in both numerator and denominator.