Home Class 11 Applied Maths Chapter 9: Probability Exercise 9.4: Conditional Probability
Chapter 9 • Unit IV

Exercise 9.4: Conditional Probability

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

9.4 Conditional Probability: Reduced Sample Spaces, Multiplication Rule, and Independent Events

In Applied Mathematics, corporate credit scoring, medical diagnostic screening, and algorithmic Bayesian models, probability estimates are constantly updated as new empirical information becomes available. Conditional probability formalizes how the probability of an event \(A\) is revised once we know that another event \(B\) has already occurred.

Fundamental Definitions & Core Theorems

  • Definition of Conditional Probability: If \(A\) and \(B\) are two events associated with the same sample space \(S\), the conditional probability of \(A\) given that \(B\) has already occurred (with \(P(B) > 0\)) is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{n(A \cap B)}{n(B)} \] Conceptual Meaning: Event \(B\) replaces \(S\) as the new, reduced sample space.
  • Multiplication Theorem of Probability: \[ P(A \cap B) = P(A) \cdot P(B | A) = P(B) \cdot P(A | B) \] For 3 events: \(P(A \cap B \cap C) = P(A) \cdot P(B | A) \cdot P(C | (A \cap B))\).
  • Independent Events: Two events \(A\) and \(B\) are statistically independent if the occurrence of \(B\) has no influence on the probability of \(A\): \[ P(A | B) = P(A) \iff P(A \cap B) = P(A) \cdot P(B) \]
  • Critical Distinction: Mutually Exclusive vs Independent:
    Property Mutually Exclusive Events Independent Events
    Definition Cannot happen together (\(A \cap B = \emptyset\)) Occurrence of one doesn't affect the other
    Mathematical Condition \[ P(A \cap B) = 0 \] \[ P(A \cap B) = P(A) \cdot P(B) \]
    Core Incompatibility Two events with non-zero probabilities can never be both mutually exclusive and independent!
  • Independence of Complements Theorem: If \(A\) and \(B\) are independent events, then:
    1. \(A\) and \(B'\) are independent: \(P(A \cap B') = P(A) \cdot P(B')\)
    2. \(A'\) and \(B\) are independent: \(P(A' \cap B) = P(A') \cdot P(B)\)
    3. \(A'\) and \(B'\) are independent: \(P(A' \cap B') = P(A') \cdot P(B')\)

Comprehensive Applied Example: Target Reliability Engineering

Scenario: Two independent defense radar units, \(R_1\) and \(R_2\), track an incoming airspace target. The probability that \(R_1\) detects the target is \(P(R_1) = 0.8\), and the probability that \(R_2\) detects the target is \(P(R_2) = 0.7\).

Find the probability that:

  1. Both radar units detect the target.
  2. The target is detected by at least one radar unit.
  3. Exactly one radar unit detects the target.
  4. Neither radar unit detects the target.

Solution:

Because the radar units operate independently: \(P(R_1 \cap R_2) = P(R_1) \cdot P(R_2)\).

  1. Both detect the target: \[ P(R_1 \cap R_2) = 0.8 \times 0.7 = 0.56. \]
  2. At least one detects the target (Addition Theorem): \[ P(R_1 \cup R_2) = P(R_1) + P(R_2) - P(R_1 \cap R_2) = 0.8 + 0.7 - 0.56 = 0.94. \] Alternative via Complements: \[ P(R_1 \cup R_2) = 1 - P(R_1' \cap R_2') = 1 - (1 - 0.8)(1 - 0.7) = 1 - (0.2 \times 0.3) = 1 - 0.06 = 0.94. \]
  3. Exactly one radar detects the target: \[ P(\text{Exactly 1}) = P(R_1 \cap R_2') + P(R_1' \cap R_2) = (0.8 \times 0.3) + (0.2 \times 0.7) = 0.24 + 0.14 = 0.38. \]
  4. Neither radar detects the target: \[ P(R_1' \cap R_2') = P(R_1') \cdot P(R_2') = 0.2 \times 0.3 = 0.06. \]
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