Independence and Expected Waiting Time Notes for GATE DA: Concepts, Formulas, Worked Examples & Practice

Independence and Expected Waiting Time notes for GATE DA: 13 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Independence and Expected Waiting Time

Your journey through this chapter

Independent Trials — the foundation

What "no memory" really means, and why it matters.

Waiting for a single event

The geometric distribution and the clean formula

E [T] = 1/ p

Why patterns break the simple formula

Two consecutive successes is not the same as two independent successes.

The state method — your main weapon

Define states by progress, write equations, solve.

Exam-ready patterns and traps

Recognising the question type, avoiding the memoryless trap, and the HH vs HT surprise.

End goal: given any "repeat until you see pattern X" question, you will set up the states and solve for the expected time in under two minutes.

The Big Question: How Long Do I Wait?

The setup

An experiment is repeated, independently, forever.
Each trial has a success probability $p$ (constant across trials).
You are waiting for some target: a single success, two successes in a row, a specific sequence, etc.

The question
Let

T

be the number of trials until the target is first achieved. What is

E [T]

Two worlds

Waiting for...	Tool	Typical answer shape
A single success	Geometric distribution	$E [T] = \frac{1}{p}$
A pattern (e.g. two in a row)	State equations	$E [T] = \frac{1 + p}{p ^{2}}$

The first world is one line. The second world is where the real exam questions live.

Independent Trials: Every Throw Is Fresh

Definition

Trials $X_{1}, X_{2}, X_{3}, \dots$ are independent if for every $k$ ,

P (X_{k} = x ∣ X_{1}, \dots, X_{k - 1}) = P (X_{k} = x) .

What this buys you

The success probability is $p$ on every trial, no exceptions.
The process has no memory: the past does not bend the future.
You can multiply probabilities across trials freely: $P (A and B) = P (A) P (B)$ .

A clean mental picture Imagine a die. You throw it. You write down the result. You throw again. Each throw is a brand-new roll of a fair die. The universe has not changed. This is the world we work in.

If a problem says "thrown repeatedly" or "tossed until", assume independence unless told otherwise.

Waiting for the First Success: Geometric Distribution

The random variable

$T =$ number of trials until the first success.

Probability mass function

P (T = k) = (1 - p)^{k - 1} p, k = 1, 2, 3, \dots

Read this as: $(k - 1)$ failures, then one success.

The key expectation

E [T] = \frac{1}{p}

Sanity checks

If $p = 1$ (success is certain), $E [T] = 1$ . You wait one trial. Correct.
If $p = 1/2$ (fair coin, waiting for a head), $E [T] = 2$ . Makes sense.
If $p = 0.01$ (rare event), $E [T] = 100$ . You wait a long time. Correct.

When does this formula apply?
Only when you are waiting for a single success on independent trials with constant

p

. The moment the target becomes a pattern, this formula no longer gives the answer directly.

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Independence and Expected Waiting Time Notes for GATE DA: Concepts, Formulas, Worked Examples & Practice

Independence and Expected Waiting Time notes for GATE DA: 13 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions

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