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Conditional Probability

Short Notes

Conditional probability is the probability of an event occurring given that another event has already occurred.

\[P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0\]

This is fundamental to probabilistic reasoning and forms the basis for Bayes' theorem and probabilistic inference.

Key Theories & Formulas

1. Multiplication Rule

\[P(A \cap B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A)\]

Chain Rule for multiple events: $\(P(A_1 \cap A_2 \cap ... \cap A_n) = P(A_1) \cdot P(A_2|A_1) \cdot P(A_3|A_1 \cap A_2) \cdots\)$

2. Conditional Probability Properties

  • \(P(A|B) + P(A'|B) = 1\)
  • \(P(A \cup B | C) = P(A|C) + P(B|C) - P(A \cap B | C)\)
  • \(P(\emptyset | B) = 0\), \(P(S|B) = 1\)

3. Law of Total Probability

If \(B_1, B_2, ..., B_n\) form a partition of S: $\(P(A) = \sum_{i=1}^{n} P(A|B_i) \cdot P(B_i)\)$

4. Independence vs Conditional Probability

  • Independent: \(P(A|B) = P(A)\) (B gives no information about A)
  • Dependent: \(P(A|B) \neq P(A)\)

5. Conditional Expectation

\[E[X|Y=y] = \sum_x x \cdot P(X=x|Y=y)\]

Tower Property: \(E[E[X|Y]] = E[X]\)

6. Conditional Variance

\[Var(X|Y) = E[X^2|Y] - (E[X|Y])^2\]

Total Variance Formula: $\(Var(X) = E[Var(X|Y)] + Var(E[X|Y])\)$

Example Problems

Problem 1: A fair die is rolled. If the result is even, find P(result ≥ 4 | result is even).

Solution:

  • Even outcomes: {2, 4, 6}
  • Outcomes ≥ 4 among even: {4, 6}
  • \(P(\geq 4 | even) = \frac{2}{3}\)

Problem 2: In a class, 60% are boys. 70% of boys and 80% of girls passed an exam. A student passed. Find the probability the student is a boy.

Solution:

  1. P(Boy) = 0.6, P(Girl) = 0.4
  2. P(Pass|Boy) = 0.7, P(Pass|Girl) = 0.8
  3. P(Pass) = 0.7(0.6) + 0.8(0.4) = 0.42 + 0.32 = 0.74
  4. \(P(Boy|Pass) = \frac{0.7 \times 0.6}{0.74} = \frac{0.42}{0.74} = \mathbf{\frac{21}{37}} \approx 0.568\)

Problem 3: Three machines M1, M2, M3 produce 30%, 45%, 25% of items. Defect rates: 2%, 3%, 2%. An item is defective. Find P(from M2).

Solution:

  1. P(D) = 0.02(0.3) + 0.03(0.45) + 0.02(0.25) = 0.006 + 0.0135 + 0.005 = 0.0245

  2. \(P(M2|D) = \frac{0.03 \times 0.45}{0.0245} = \frac{0.0135}{0.0245} = \mathbf{\frac{27}{49}} \approx 0.551\)

Hardest GATE Questions

Topic: Sequential Drawing

Question (GATE 2018 Style): An urn has 5 white and 3 black balls. Balls are drawn one by one without replacement. Find P(third ball is white | first two are of different colors).

Solution:

  1. Given: first two are different colors (one white, one black)
  2. After drawing: 4 white, 2 black OR 5 white, 2 black remain?
  3. Either way: 4 white + 2 black = 6 balls remain
  4. Case 1 (WB first): 4W, 2B remain → P(W) = 4/6
  5. Case 2 (BW first): 4W, 2B remain → P(W) = 4/6
  6. Both cases give same result: \(P(3rd = W | different) = \mathbf{\frac{2}{3}}\)

Topic: Markov Property

Question (GATE 2017 Variant): Rain on consecutive days has probability 0.7. Rain following no-rain has probability 0.4. Today is Monday with rain. Find P(rain on Wednesday).

Solution:

  1. Let R = rain, N = no rain
  2. P(Tue=R | Mon=R) = 0.7, P(Tue=N | Mon=R) = 0.3
  3. P(Wed=R | Tue=R) = 0.7, P(Wed=R | Tue=N) = 0.4
  4. P(Wed=R) = P(Wed=R|Tue=R)P(Tue=R) + P(Wed=R|Tue=N)P(Tue=N)
  5. = 0.7(0.7) + 0.4(0.3) = 0.49 + 0.12 = 0.61

Topic: Conditional Independence

Question: Given \(P(A|C) = 0.6\), \(P(B|C) = 0.4\), and A, B are conditionally independent given C. Find \(P(A \cup B | C)\).

Solution:

  1. Conditional independence: \(P(A \cap B | C) = P(A|C) \cdot P(B|C)\)
  2. \(P(A \cap B | C) = 0.6 \times 0.4 = 0.24\)
  3. \(P(A \cup B | C) = P(A|C) + P(B|C) - P(A \cap B | C)\)
  4. \(= 0.6 + 0.4 - 0.24 = \mathbf{0.76}\)

Topic: Simpson's Paradox

Question: Hospital A treats 100 mild (90% success) and 100 severe (50% success) cases. Hospital B treats 50 mild (80% success) and 150 severe (40% success) cases.

Which hospital has better overall success rate? Better conditional rates?

Solution: Hospital A:

  • Mild: 90, Severe: 50, Total: 140/200 = 70%

Hospital B:

  • Mild: 40, Severe: 60, Total: 100/200 = 50%

Paradox: A has 70% overall vs B's 50%. But:

  • For mild: A = 90% > B = 80% (A better)
  • For severe: A = 50% > B = 40% (A better)

A is better in both categories AND overall! (No paradox here actually, but demonstrates conditional analysis importance)

References