Probability Advance

Conditional Probability, Law of Total Probability

Probability of event A given that B has occurred.

$$\boxed{P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad (P(B) > 0)}$$

Meaning (Intuition)

• Sample space shrinks to B
• We ask: within B, how often does A occur?

Rearranged Forms (Very Useful ⭐)

• $P(A \cap B) = P(A \mid B).P(B)$
• $P(A \cap B) = P(B \mid A) . P(A)$

If A and B are independent:

$$\boxed{P(A \mid B) = P(A)}$$

Law of Total Probability ⭐⭐

🔹 Statement

If $A_1, A_2, \dots, A_n$ are mutually exclusive and exhaustive events, then for any event B:

$$ P(B) = \sum_{i=1}^{n} P(B \mid A_i)P(A_i) }$$

Why this works (Reason)

• Event B can occur only through one of the ($A_i$)
• Break B into disjoint parts: $B = (B \cap A_1) \cup (B \cap A_2) \cup \dots$

Diagram View (Mental)

A1 ─┐
A2 ─┼──► B
A3 ─┘

Quick Comparison Table

Concept	Formula	Purpose
Conditional Probability	$P(A\mid B)$	Probability after B
Joint Probability	$P(A\cap B)$)	Both occur
Total Probability	$\sum P(B\mid A_i)P(A_i)$	Find $P(B)$
Bayes’ Theorem	$\frac{P(B\mid A_i)P(A_i)}{\sum P(B\mid A_k)P(A_k)}$	Reverse condition

Exam Memory Tricks 🧠

• Conditional → Restrict sample space
• Total probability → Weighted sum
• Bayes → Reverse the condition
• Independent? → conditioning doesn’t change probability

When to Use What?

• Given “given that” → Conditional probability
• Given multiple sources/causes → Total probability
• Asked cause after effect → Bayes’ theorem

Bayes’ Theorem

Bayes’ theorem gives the probability of a cause given an observed event.

General Form (Multiple Causes) ⭐

If $A_1, A_2, \dots, A_n$ are mutually exclusive and exhaustive events, then:

$$\boxed{ P(A_i \mid B) = \frac{P(B \mid A_i)P(A_i)} {\sum_{k=1}^{n} P(B \mid A_k)P(A_k)} } $$

Standard Two-Event Case

For events AAA and its complement $A’$:

$${ P(B) = P(B \mid A)P(A) + P(B \mid A’)P(A’) }$$

$$\boxed{ P(A \mid B) = \frac{P(B \mid A)P(A)} {P(B \mid A)P(A) + P(B \mid A’)P(A’)} } $$

Why Bayes’ Theorem Works (Intuition)

Using Conditional Probability $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad (P(B) > 0)$$

• But by definition of conditional probability, $$P(A \cap B) = P(B \mid A)P(A)$$

• Substitute into the first equation:

$$\boxed{P(A_i \mid B) = \frac{P(B \mid A_i)P(A_i)}{P(B)}}$$

Using Law of Total Probability $$

P(B) = \sum_{i=1}^{n} P(B \mid A_i)P(A_i)

$$

• Substitute this value of $P(B)$ into Bayes’ formula:

$$\boxed{ P(A_i \mid B) = \frac{P(B \mid A_i)P(A_i)} {\sum_{k=1}^{n} P(B \mid A_k)P(A_k)} } $$

Meaning of Each Term

Term	Meaning
$P(A_i)$	==Prior probability== (before seeing $B$)
$P(B \mid A_i)$	==Likelihood==
$P(B)$	Total probability of (B)
$P(A_i \mid B)$	==Posterior probability==

Posterior ∝ Likelihood × Prior

Common GATE Traps

1. Confusing $(P(A \mid B))$ with $(P(B \mid A))$
2. Forgetting denominator (total probability)
3. Not checking mutually exclusive & exhaustive
4. Using Bayes when simple conditional probability is enough $$