Bayes' Theorem - Example: A disjoint union

Introduction to Extended Bayes Theorem
📌 The video introduces an upgrade to the standard Bayes theorem formula, addressing scenarios where the sample space is divided into distinct "buckets" or partitions ($B_1, B_2, \dots$).
🎯 The core problem demonstrated involves two buckets ($B_1$ and $B_2$) containing different distributions of blue and yellow balls, aiming to find the probability of having chosen a specific bucket given the color of the drawn ball (e.g., $P(B_1|A)$, where $A$ is drawing a blue ball).

Bucket Setup and Initial Probabilities
🏀 Bucket $B_1$: Contains 3 blue and 3 yellow balls, resulting in $P(A|B_1) = 3/6 = 1/2$.
🎾 Bucket $B_2$: Contains 2 blue and 4 yellow balls, resulting in $P(A|B_2) = 2/6 = 1/3$.
⚖️ The probability of selecting either bucket is equal: $P(B_1) = P(B_2) = 1/2$.

Calculating the Total Probability of Event A ($P(A)$)
🧮 The probability of drawing a blue ball ($A$) is calculated using the law of total probability, leveraging the disjoint union of the sample space created by $B_1$ and $B_2$:
$$P(A) = P(A \cap B_1) + P(A \cap B_2)$$
🧮 Using the definition of conditional probability, this becomes:
$$P(A) = [P(A|B_1) \cdot P(B_1)] + [P(A|B_2) \cdot P(B_2)]$$
📊 Substitution of values yields: $P(A) = (1/2 \cdot 1/2) + (1/3 \cdot 1/2) = 1/4 + 1/6 = 3/12 + 2/12 = 5/12$.

Applying Bayes' Theorem
💡 The goal is to find $P(B_1|A)$, the probability that the draw came from Bucket 1 given a blue ball was selected.
📜 The generalized Bayes' theorem used is:
$$P(B_1|A) = \frac{P(A|B_1) \cdot P(B_1)}{P(A)}$$
✅ Plugging in the calculated values:
$$P(B_1|A) = \frac{(1/2) \cdot (1/2)}{5/12} = \frac{1/4}{5/12} = \frac{1}{4} \cdot \frac{12}{5} = \frac{12}{20} = 3/5$$

Key Points & Insights
➡️ The extended Bayes theorem allows updating beliefs ($P(B_i)$) based on new evidence ($A$) when the underlying process involves several distinct, mutually exclusive scenarios ($B_i$).
➡️ Since Bucket 1 ($B_1$) had a higher prior probability of yielding a blue ball ($1/2$) compared to Bucket 2 ($1/3$), observing a blue ball increased the likelihood of it coming from $B_1$ to $3/5$ (or 60%), which is greater than the initial 50% chance of selecting $B_1$.
➡️ The method for calculating the total probability $P(A)$ scales directly: for $N$ buckets, $P(A)$ becomes a summation of $N$ product terms, $P(A) = \sum_{i=1}^{N} P(A|B_i) P(B_i)$.

📸 Video summarized with SummaryTube.com on Mar 04, 2026, 09:26 UTC