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📡 AAI ATC Exam Prep · Class 12 Maths · Chapter 13

Probability

From Conditional Probability to Bayes' Theorem — master the logic of uncertainty!

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Class 12 Maths
Chapter 13
NCERT + AAI ATC
🎯 Conditional · Bayes · Independence
🎲

Why Probability is Critical for AAI ATC?

In Air Traffic Control, every decision involves uncertainty — Will equipment fail? Is the weather forecast reliable? What is the probability of a runway conflict? Probability theory, especially Conditional Probability and Bayes' Theorem, provides the mathematical backbone for risk assessment, decision-making under uncertainty, and reliability analysis — all key to AAI ATC operations and the Junior Executive exam.

1
Conditional Probability

What is Conditional Probability?

When the occurrence of one event affects the probability of another, we use conditional probability. If event F has occurred, the probability of event E occurring is written as P(E|F) — "probability of E given F".

The sample space effectively reduces from S to F. We only consider outcomes inside F.

P(E|F) = P(E ∩ F) / P(F)    [provided P(F) ≠ 0]
Definition of Conditional Probability — NCERT Class 12, Ch. 13

Properties of Conditional Probability

📌 Property 1

P(S|F) = P(F|F) = 1

📌 Property 2

P((A∪B)|F) = P(A|F) + P(B|F) – P((A∩B)|F)

📌 Property 3

P(E′|F) = 1 – P(E|F)

📌 Range

0 ≤ P(E|F) ≤ 1 always

🗼

ATC Real-World Link

An ATC radar detects an aircraft in a particular sector. Given that weather alert F has been issued, what is P(E|F) — the probability that aircraft E is in distress? This conditional reasoning is exactly how ATC operators update their threat assessments in real-time!

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Image: Add Venn diagram showing sample space S, events E and F, with E∩F highlighted

Recommended: 700×380 px | Label E, F, E∩F clearly with probability values

Practice MCQs

Test Your Understanding — Subtopic 1

1If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, find P(A|B).
✅ Answer: (B) 4/9
P(A|B) = P(A∩B)/P(B) = (4/13) ÷ (9/13) = 4/9.
Always divide P(A∩B) by P(B) — not by P(A)!
2A die is thrown twice. The sum is observed to be 6. What is the conditional probability that number 4 appeared at least once?
✅ Answer: (B) 2/5
Let E = "4 appears at least once", F = "sum is 6".
F = {(1,5),(2,4),(3,3),(4,2),(5,1)} → P(F) = 5/36
E∩F = {(2,4),(4,2)} → P(E∩F) = 2/36
P(E|F) = (2/36)/(5/36) = 2/5
3An ATC system has two radar units A and B. P(A works) = 0.8, P(B works) = 0.6, P(both work) = 0.5. If unit B is working, what is P(A is also working)?
✅ Answer: (C) 5/6
P(A|B) = P(A∩B)/P(B) = 0.5/0.6 = 5/6 ≈ 0.833
This is the probability that radar A is working given that radar B is already confirmed working — a classic conditional probability scenario in ATC reliability!
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2
Multiplication Theorem & Independent Events

Multiplication Rule of Probability

To find the probability that both events E and F occur simultaneously, we use the Multiplication Theorem:

P(E ∩ F) = P(E) · P(F|E) = P(F) · P(E|F)
For three events: P(E∩F∩G) = P(E) · P(F|E) · P(G|EF)
Multiplication Rule — extends to any number of events

Independent Events

Two events E and F are independent if the occurrence of one does NOT affect the other:

E and F are independent ⟺ P(E∩F) = P(E) · P(F)
Equivalently: P(E|F) = P(E) and P(F|E) = P(F)
Independence Condition — do NOT confuse with mutually exclusive!
⚠️ Independent ≠ Mutually Exclusive

Mutually exclusive events can't occur together (P(E∩F)=0). Independent events CAN occur together but don't influence each other.

📌 At Least One Formula

P(at least one of A or B) = 1 − P(A′)·P(B′) when A,B are independent.

✈️

ATC Application

Two independent communication systems in an ATC tower each have reliability 0.95. P(both work) = 0.95 × 0.95 = 0.9025. P(at least one works) = 1 − 0.05 × 0.05 = 0.9975. This is why ATC systems are always redundant — independent backups dramatically raise overall reliability!

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Image: Add tree diagram showing multiplication rule — two-stage experiment with branches and probabilities labeled at each node

Recommended: 700×420 px | Use Venn or tree diagram style

Practice MCQs

Test Your Understanding — Subtopic 2

4An urn contains 10 black and 5 white balls. Two balls are drawn without replacement. What is the probability both are black?
✅ Answer: (B) 3/7
P(1st black) = 10/15 = 2/3
P(2nd black | 1st black) = 9/14
P(both black) = (10/15) × (9/14) = 90/210 = 3/7
5A die is thrown. E = {3,6} (multiple of 3), F = {2,4,6} (even). Are E and F independent?
✅ Answer: (C) Yes, P(E∩F) = P(E)·P(F)
P(E) = 2/6 = 1/3  |  P(F) = 3/6 = 1/2  |  E∩F = {6} → P(E∩F) = 1/6
P(E)·P(F) = (1/3)×(1/2) = 1/6 ✓  →  E and F are independent
6Two ATC radar systems A and B operate independently. P(A fails) = 0.1, P(B fails) = 0.05. What is the probability that at least one system works?
✅ Answer: (C) 0.9955
P(A works) = 0.9, P(B works) = 0.95
P(at least one works) = 1 − P(A fails)·P(B fails) = 1 − 0.1×0.05 = 1 − 0.005 = 0.9955
This is why ATC uses independent backup systems — near-perfect combined reliability!
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3
Theorem of Total Probability & Bayes' Theorem

Partition of Sample Space

Events E₁, E₂, …, Eₙ form a partition of sample space S if they are pairwise disjoint, exhaustive (their union = S), and each has non-zero probability.

Theorem of Total Probability

If {E₁, E₂, …, Eₙ} is a partition of S and A is any event, then:

P(A) = P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂) + … + P(Eₙ)·P(A|Eₙ)
Theorem of Total Probability — sum across all hypotheses

Bayes' Theorem

Given that event A has occurred, what is the reverse probability that it came from hypothesis Eᵢ? Bayes' theorem answers this:

P(Eᵢ|A) = P(Eᵢ)·P(A|Eᵢ) / Σ P(Eⱼ)·P(A|Eⱼ)
Bayes' Theorem — "reverse probability" or "formula for causes"
📌 Prior Probability

P(Eᵢ) — probability of hypothesis Eᵢ before observing event A

📌 Likelihood

P(A|Eᵢ) — probability of observing A if Eᵢ is true

📌 Posterior Probability

P(Eᵢ|A) — updated probability of Eᵢ after observing A

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ATC Application — Bayes in Action

An ATC radar alert fires. Prior probability of actual aircraft conflict = 2%. The radar has 95% accuracy (true positive) but gives 8% false alarms. Bayes' theorem calculates the posterior probability that it's a real conflict given the alert — essential for deciding whether to scramble emergency response!

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Image: Add Bayes' Theorem diagram — Partition E₁, E₂, E₃ of sample space S with event A overlapping all three, showing total probability and reverse arrow for Bayes

Recommended: 750×420 px | Use Fig 13.4 from NCERT as reference

Practice MCQs

Test Your Understanding — Subtopic 3

7Bag I has 3 red, 4 black balls. Bag II has 5 red, 6 black balls. A ball is drawn from a randomly chosen bag and is found red. What is P(it came from Bag II)?
✅ Answer: (A) 35/68
P(E₁) = P(E₂) = 1/2  |  P(Red|E₁) = 3/7  |  P(Red|E₂) = 5/11
P(E₂|Red) = [(1/2)(5/11)] / [(1/2)(3/7) + (1/2)(5/11)]
= (5/22) / (3/14 + 5/22) = (5/22) / (33/154 + 35/154) = (5/22)/(68/154)
= (5/22)×(154/68) = 35/68
8Machines A, B, C produce 25%, 35%, 40% of bolts with defect rates 5%, 4%, 2% respectively. A defective bolt is found. What is P(made by machine B)?
✅ Answer: (B) 28/69
P(defective) = 0.25×0.05 + 0.35×0.04 + 0.40×0.02 = 0.0125+0.0140+0.0080 = 0.0345
P(B|defective) = (0.35×0.04) / 0.0345 = 0.014/0.0345 = 28/69 ≈ 0.406
9A man speaks truth 3/4 of the time. He throws a die and reports it is a 6. What is the probability that it actually IS a 6?
✅ Answer: (B) 3/8
S₁ = six occurs → P(S₁) = 1/6  |  S₂ = six does not → P(S₂) = 5/6
P(reports 6|S₁) = 3/4 (truth)  |  P(reports 6|S₂) = 1/4 (lies)
P(S₁|reports 6) = [(1/6)(3/4)] / [(1/6)(3/4)+(5/6)(1/4)]
= (3/24) / (3/24 + 5/24) = 3/8 = 3/8
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