chapter

Independence and Expected Waiting Time Notes for GATE DA

Independence and Expected Waiting Time notes for GATE DA: 10 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions

independence and expected waiting time notes

Chapter Roadmap: Independence and Expected Waiting Time

Your journey through this chapter 1 Independent Trials — the foundation What "no memory" really means, and why it matters. 2 Waiting for a single event The geometric distribution and the clean formula E [T] = 1/ p . 3 Why patterns break the simple formula Two consecutive successes is not the same as two independent successes. 4 The state method — your main weapon Define states by progress, write equations, solve. 5 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.

7 more cards in this chapter

Try a question

Answer it here to see how it works. Nothing is recorded until you sign in.

Question 1

Level 1: Warm-up

A student incorrectly calculates the probability of the first success in the first two trials as p + p = 2 p = 0.4 by overcounting. Using the correct geometric distribution with p = 0.2, what is the minimum number of trials k required for the cumulative probability P (T \leq k) to be at least 0.36 ?

Question 2

Level 1: Warm-up

A student incorrectly calculates the probability of getting at least one success in k trials as k \cdot p . For p = 0.2, what is the minimum number of trials k required for the TRUE probability of getting at least one success to be strictly greater than 0.5 ?

Question 3

Level 1: Warm-up

A process involves independent trials with success probability p . The expected time to the first success is derived by conditioning on the first trial. If the first trial is a success (prob p), the time is 1. If it is a failure (prob 1 - p), the time is 1 + E . Solving E = p (1) + (1 - p) (1 + E) for p = 1/3 yields what value?

Question 4

Level 1: Warm-up

Assertion (A): For a coin with P (H) = p, the expected number of tosses to get two consecutive Heads is (1 + p) / p^{2} . Reason (R): The expected waiting time for any sequence of length k is 1/ p^{k}, which ensures the units of time match the inverse of probability.

Question 5

Level 1: Warm-up

Let E be the expected number of trials to get the first success. By conditioning on the first TWO trials, we get the equation: E = p (1) + (1 - p) p (2) + (1 - p)^{2} (2 + E) For p = 0.5, what is the value of E ?

Question 6

Level 1: Warm-up

Assertion (A): For a biased coin with P (H) = 1/3, the expected number of tosses to get two consecutive Heads is 12. Reason (R): The expected waiting time is 1/ p^{2} = 9 . The actual value is 12 because the formula 1/ p^{2} suffers from a unit mismatch when applied to overlapping patterns.

Question 7

Level 1: Warm-up

A coin with P (H) = 0.6 is tossed repeatedly. Assuming independent trials, what is the maximum possible value of the conditional probability P (Head on 10th toss ∣ first 9 tosses are Tails) ?

Question 8

Level 1: Warm-up

Consider the infinite series S = \sum_{k = 1}^{\infty} k (1 - p)^{k - 1} p which represents E [T] for a geometric distribution. Which statement correctly describes the convergence of this series for 0 < p < 1 ?

Question 9

Level 1: Warm-up

A sequence of independent trials is conducted. The probability of success on any trial is p . What is the maximum possible value of the probability that the first success occurs on the second trial, given that p can be any value in [0, 1] ?

Question 10

Level 1: Warm-up

Consider the recursive derivation E = p (1) + (1 - p) (1 + E) for the expected waiting time. Which statement is true regarding the constraints on p ?

Free preview ends here

Login to view the complete notes

Creating an account is free. You get the rest of this chapter, step-by-step solutions, and a study plan built around the topics you are actually weak at.

Why MastersUp

Personalised first. High quality throughout.

Most platforms hand everyone the same content. Here the content moves with your performance, topic by topic.

Built around you, not around a syllabus PDF

Every answer you give moves your topic-level intelligence rate. The next question, the next revision card and tomorrow's plan all change with it.

Revision that hits your weak spots

We only revise topics you have actually attempted and are still below the safe bar on — never the same chapter on repeat.

Questions calibrated to the real exam

Each question carries a measured toughness. You are served a rung above your current level, so practice keeps stretching you.

Notes written for recall, not for volume

Full lesson cards for first study, curated short-note cards for the last mile — with derivations, traps and exam patterns marked.

One place for everything

Notes, chapter practice, previous-year questions, test series and full-length papers — all feeding one picture of your preparation.

Honest progress

No vanity streaks. Progress here means chapters mastered and accuracy that held up on harder questions.

Unlock the whole course

Full notes and short notes, the complete question bank with worked solutions, mock tests, full-length papers, and an adaptive plan that rebuilds itself as you improve.

View all plans Start free

Independence and Expected Waiting Time Notes for GATE DA

Independence and Expected Waiting Time notes for GATE DA: 10 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.

Independence and Expected Waiting Time: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Probability and Statistics MCQ

A student incorrectly calculates the probability of the first success in the first two trials as $p + p = 2 p = 0.4$ by overcounting. Using the correct geometric distribution with $p = 0.2$ , what is the minimum number of trials $k$ required for the cumulative probability $P (T \leq k)$ to be at least $0.36$ ?

A.
2
B.
1
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: The cumulative distribution function (CDF) for a geometric distribution is $P (T \leq k) = 1 - (1 - p)^{k}$ .

Step 1: Identify the target cumulative probability: $P (T \leq k) \geq 0.36$ .

Step 2: Substitute the CDF formula: $1 - (1 - p)^{k} \geq 0.36$ .

Step 3: Plug in $p = 0.2$ : $1 - (0.8)^{k} \geq 0.36$ .

Step 4: Rearrange: $(0.8)^{k} \leq 1 - 0.36 = 0.64$ .

Step 5: Test integer values for $k$ :

For $k = 1$ : $0. 8^{1} = 0.8$ (Not $\leq 0.64$ )
For $k = 2$ : $0. 8^{2} = 0.64$ (Satisfies $\leq 0.64$ )

Answer: The minimum number of trials $k$ is 2.

Question 2 · Probability and Statistics MCQ

A student incorrectly calculates the probability of getting at least one success in $k$ trials as $k \cdot p$ . For $p = 0.2$ , what is the minimum number of trials $k$ required for the TRUE probability of getting at least one success to be strictly greater than $0.5$ ?

A.
3
B.
4
C.
5
D.
2

Correct Answer:

Step-by-Step Solution

Key idea: The true probability of at least one success in $k$ independent trials is given by the complement rule. We must solve an inequality for $k$ .

Step 1: Write the true probability using the complement rule:

$P (at least one) = 1 - P (none) = 1 - (1 - p)^{k}$ .

Step 2: Set up the inequality with $p = 0.2$ :

$1 - (1 - 0.2)^{k} > 0.5$

$1 - 0. 8^{k} > 0.5$

Step 3: Isolate the exponential term:

$0. 8^{k} < 0.5$

Step 4: Test integer values for $k$ :

For $k = 2$ : $0. 8^{2} = 0.64$ (Not $< 0.5$ )
For $k = 3$ : $0. 8^{3} = 0.512$ (Not $< 0.5$ )
For $k = 4$ : $0. 8^{4} = 0.4096$ (Strictly $< 0.5$ )

Answer: The minimum number of trials $k$ is 4.

Question 3 · Probability and Statistics MCQ

A process involves independent trials with success probability $p$ . The expected time to the first success is derived by conditioning on the first trial. If the first trial is a success (prob $p$ ), the time is 1. If it is a failure (prob $1 - p$ ), the time is $1 + E$ . Solving $E = p (1) + (1 - p) (1 + E)$ for $p = 1/3$ yields what value?

A.
1
B.
1.5
C.
9
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: The recursive definition of expected waiting time conditions on the outcome of the very first trial.

Step 1: Write down the given recursive equation: $E = p (1) + (1 - p) (1 + E)$ .

Step 2: Expand the right side: $E = p + 1 - p + (1 - p) E$ .

Step 3: Simplify: $E = 1 + (1 - p) E$ .

Step 4: Isolate $E$ : $E - (1 - p) E = 1 ⟹ pE = 1 ⟹ E = 1/ p$ .

Step 5: Substitute $p = 1/3$ : $E = 1/ (1/3) = 3$ .

Answer: The expected time is 3.

Question 4 · Probability and Statistics MCQ

Assertion (A): For a coin with $P (H) = p$ , the expected number of tosses to get two consecutive Heads is $(1 + p) / p^{2}$ .

Reason (R): The expected waiting time for any sequence of length $k$ is $1/ p^{k}$ , which ensures the units of time match the inverse of probability.

A.
Both A and R are true and R is the correct explanation of A
B.
A is false but R is true
C.
Both A and R are true but R is NOT the correct explanation of A
D.
A is true but R is false

Correct Answer:

Step-by-Step Solution

Key idea: Waiting for a pattern like two consecutive heads requires the state method, not a simple inverse probability formula.

Step 1: Evaluate Assertion (A). The correct formula for two consecutive successes is indeed $(1 + p) / p^{2}$ . For a fair coin ( $p = 0.5$ ), this gives $(1.5) /0.25 = 6$ . So, A is true.

Step 2: Evaluate Reason (R). The formula $1/ p^{k}$ is incorrect for overlapping patterns because it ignores the fact that a failure doesn't always reset your progress to zero. It falsely assumes the pattern probability $p^{k}$ acts as a simple independent trial rate. So, R is false.

Step 3: Conclude that A is true but R is false.

Answer: A is true but R is false.

Question 5 · Probability and Statistics MCQ

Let $E$ be the expected number of trials to get the first success. By conditioning on the first TWO trials, we get the equation:

$E = p (1) + (1 - p) p (2) + (1 - p)^{2} (2 + E)$

For $p = 0.5$ , what is the value of $E$ ?

A.
2
B.
1.2
C.
1.5
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: The recursive definition of expected waiting time can be extended by conditioning on the first $k$ trials. The algebra must be solved carefully to isolate $E$ .

Step 1: Substitute $p = 0.5$ into the given equation:

$E = 0.5 (1) + (0.5) (0.5) (2) + (0.5)^{2} (2 + E)$

Step 2: Simplify the terms:

$E = 0.5 + 0.5 + 0.25 (2 + E)$

$E = 1.0 + 0.5 + 0.25 E$

$E = 1.5 + 0.25 E$

Step 3: Isolate $E$ by subtracting $0.25 E$ from both sides:

$0.75 E = 1.5$

Step 4: Solve for $E$ :

$E = 1.5/0.75 = 2$ .

Answer: The value of $E$ is 2.

Question 6 · Probability and Statistics MCQ

Assertion (A): For a biased coin with $P (H) = 1/3$ , the expected number of tosses to get two consecutive Heads is 12.

Reason (R): The expected waiting time is $1/ p^{2} = 9$ . The actual value is 12 because the formula $1/ p^{2}$ suffers from a unit mismatch when applied to overlapping patterns.

A.
Both A and R are true and R is the correct explanation of A
B.
A is true but R is false
C.
Both A and R are true but R is NOT the correct explanation of A
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Key idea: Waiting for a pattern like two consecutive heads requires the state method, not a simple inverse probability formula. The reason given relies on a flawed conceptual justification.

Step 1: Evaluate Assertion (A). The correct formula for two consecutive successes is $E [T] = (1 + p) / p^{2}$ .

For $p = 1/3$ : $E [T] = (1 + 1/3) / (1/3)^{2} = (4/3) / (1/9) = 12$ . So, A is true.

Step 2: Evaluate Reason (R). The formula $1/ p^{2} = 9$ is incorrect for overlapping patterns. The difference between 12 and 9 is not due to a "unit mismatch" (probability is dimensionless), but because $1/ p^{2}$ ignores the fact that a failure doesn't always reset your progress to zero. The overlap of states changes the expected time. So, R is false.

Step 3: Conclude that A is true but R is false.

Answer: A is true but R is false.

Question 7 · Probability and Statistics MCQ

A coin with $P (H) = 0.6$ is tossed repeatedly. Assuming independent trials, what is the maximum possible value of the conditional probability $P (Head on 10th toss ∣ first 9 tosses are Tails)$ ?

A.
0.0
B.
0.6
C.
0.4
D.
1.0

Correct Answer:

Step-by-Step Solution

Key idea: Independent trials have no memory. Past outcomes do not influence future probabilities.

Step 1: Identify that the coin tosses are explicitly stated to be independent.

Step 2: Recall the definition of independence: $P (A ∣ B) = P (A)$ .

Step 3: The probability of a Head on any single toss is $P (H) = 0.6$ .

Step 4: Therefore, the conditional probability $P (10th is H ∣ first 9 are T) = P (10th is H) = 0.6$ .

Answer: The maximum possible value is 0.6.

Question 8 · Probability and Statistics MCQ

Consider the infinite series $S = \sum_{k = 1}^{\infty} k (1 - p)^{k - 1} p$ which represents $E [T]$ for a geometric distribution. Which statement correctly describes the convergence of this series for $0 < p < 1$ ?

A.
It converges to $1/ p$ only if $p > 0.5$
B.
It diverges to infinity due to the linear growth of $k$
C.
It converges to $1/ p$ for all $0 < p < 1$
D.
It converges to $p / (1 - p)$

Correct Answer:

Step-by-Step Solution

Key idea: The series is an arithmetico-geometric series that defines the expected value of a geometric distribution.

Step 1: Recognize the series $S = \sum_{k = 1}^{\infty} k (1 - p)^{k - 1} p$ as the formal definition of $E [T]$ for $T \sim Geometric (p)$ .

Step 2: Recall that for any valid probability $0 < p < 1$ , the term $(1 - p)$ is strictly between 0 and 1.

Step 3: The exponential decay of $(1 - p)^{k - 1}$ dominates the linear growth of $k$ , ensuring the series converges.

Step 4: The sum of this specific series is exactly $1/ p$ for all $0 < p < 1$ .

Answer: It converges to $1/ p$ for all $0 < p < 1$ .

Question 9 · Probability and Statistics MCQ

A sequence of independent trials is conducted. The probability of success on any trial is $p$ . What is the maximum possible value of the probability that the first success occurs on the second trial, given that $p$ can be any value in $[0, 1]$ ?

A.
0.25
B.
0.50
C.
0.00
D.
1.00

Correct Answer:

Step-by-Step Solution

Key idea: The probability that the first success occurs on the second trial is given by the geometric PMF for $k = 2$ . We must maximize this quadratic function over the valid range of $p$ .

Step 1: Write the probability mass function for $T = 2$ :

$P (T = 2) = (1 - p)^{2 - 1} p = p (1 - p) = p - p^{2}$ .

Step 2: Recognize this as a downward-opening parabola in terms of $p$ .

Step 3: Find the maximum by taking the derivative with respect to $p$ and setting it to zero, or by using the vertex formula $p = - b / (2 a)$ :

$d / d p (p - p^{2}) = 1 - 2 p = 0 ⟹ p = 0.5$ .

Step 4: Substitute $p = 0.5$ back into the probability expression:

$P (T = 2) = 0.5 (1 - 0.5) = 0.25$ .

Answer: The maximum possible value is 0.25.

Question 10 · Probability and Statistics MCQ

Consider the recursive derivation $E = p (1) + (1 - p) (1 + E)$ for the expected waiting time. Which statement is true regarding the constraints on $p$ ?

A.
The derivation holds only for $p \geq 0.5$
B.
The term $(1 - p) E$ is strictly positive for all valid $p \in (0, 1)$
C.
The equation yields $E < 1$ when $p < 1$
D.
The derivation assumes $p$ can be greater than 1

Correct Answer:

Step-by-Step Solution

Key idea: The recursive proof for the geometric expectation relies on the fundamental properties of probability, specifically that $p$ is strictly between 0 and 1.

Step 1: Analyze the term $(1 - p) E$ . Since $p \in (0, 1)$ , the term $(1 - p)$ is strictly positive. Since $E = 1/ p$ , $E$ is also strictly positive. Thus, their product is strictly positive.

Step 2: Evaluate the other options:

The derivation holds for all $p \in (0, 1)$ , not just $p \geq 0.5$ .
The equation yields $E = 1/ p$ . For $p < 1$ , $1/ p > 1$ , so $E$ is never less than 1.
The derivation assumes $p$ is a valid probability, so $p \leq 1$ .

Answer: The term $(1 - p) E$ is strictly positive for all valid $p \in (0, 1)$ .

independence and expected waiting time notes

Chapter Roadmap: Independence and Expected Waiting Time

The Big Question: How Long Do I Wait?

The setup

Two worlds

Independent Trials: Every Throw Is Fresh

Definition

What this buys you

Try a question

Login to view the complete notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Independence and Expected Waiting Time Notes for GATE DA

Chapter Roadmap: Independence and Expected Waiting Time

The Big Question: How Long Do I Wait?

The setup

Two worlds

Independent Trials: Every Throw Is Fresh

Definition

What this buys you

Waiting for the First Success: Geometric Distribution

The random variable

Probability mass function

Sanity checks

Independence and Expected Waiting Time: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More notes in this unit