APPSC Statistics – Advanced Sampling & Probability PYQ Problems, Numerical Shortcuts and Exam Traps

APPSC Statistics – Advanced Sampling & Probability PYQ Problems, Numerical Shortcuts and Exam Traps

APPSC Statistics questionsలో Sampling మరియు Probability నుంచి కేవలం definitions మాత్రమే కాకుండా, statement-based questions, sample identification, probability inequalities, with/without replacement, systematic sampling sequences, conditional probability, sampling errors వంటి tricky patterns కూడా ముఖ్యమైనవి.

ఈ Advanced Practice Postలో:

✓ Circular Systematic Sampling
✓ SRSWR & SRSWOR
✓ Finite Population Correction
✓ Standard Error of Sample Mean
✓ Probability Bounds
✓ Conditional Probability Bounds
✓ Sampling Frame & Selection Error
✓ Sampling vs Non-Sampling Error
✓ Two-Way Table Probability
✓ Independence from a Contingency Table
✓ Bayes' Theorem
✓ Repeated Attempts
✓ Sampling Method Identification
✓ APPSC Statement Traps
✓ Advanced Numerical MCQs

Important: Standard Error and Finite Population Correction are treated here as PYQ-oriented extension material. They should not be confused with the basic syllabus wording “Methods of Sampling (Random, Non-Random)”.

PART–1: Circular Systematic Sampling

Ordinary systematic samplingలో population unitsను ఒక ordered listలో arrange చేసి, మొదట random startను select చేసి, తరువాత ప్రతి kth unitను select చేస్తాము.

Sampling Interval k = N/n

Circular systematic samplingలో end of the population list చేరిన తరువాత selection తిరిగి beginning నుంచి కొనసాగుతుంది.

Main Idea:

If the next selected number exceeds N,

New Position = Obtained Position − N

Example 1 – Circular Selection

A population contains 12 units numbered 1 to 12. A sample of 4 units is required by circular systematic sampling. Take sampling interval k = 3 and random start = 11. Find the sample.

Start with 11.

11 → 14 → 17 → 20

But the population contains only 12 units. Therefore wrap around the list.

14 − 12 = 2

17 − 12 = 5

20 − 12 = 8

Selected Sample = 11, 2, 5, 8

⚡ Circular Shortcut:

Keep adding k.
Whenever value > N, subtract N.

Example 2 – Identify the Correct Circular Sample

Population consists of:

A, B, C, D, E, F, G, H, I, J, K, L

Sample size = 4 and interval = 3.
If the random start is K, identify the sample.

Positions:

K → B → E → H

Answer: K, B, E, H

Example 3 – Which Sequence is Impossible?

Suppose the interval in a circular systematic sample is fixed at 3. Which pattern cannot represent the selected units?

A) A, D, G, J
B) C, F, I, L
C) K, B, E, H
D) A, E, H, K

Check the movement between successive units.

A → D → G → J = +3 each time

C → F → I → L = +3 each time

K → B → E → H = +3 circularly

But:

A → E = +4

Answer: D

APPSC Shortcut:

For an “impossible sample” question, do not reconstruct the whole procedure. Check whether the circular gap between consecutive selected units remains constant.

PART–2: Simple Random Sampling With and Without Replacement

Feature	SRSWR	SRSWOR
Meaning	With Replacement	Without Replacement
Selected unit returned?	Yes	No
Population available for next draw	Same	Reduced
Successive draws	Independent under random selection	Generally dependent
Finite Population Correction	Not required	Applicable

Example 4 – Probability With Replacement

A population contains 8 units, of which 3 possess characteristic A. Two units are selected independently with replacement. Find the probability that both possess A.

P(A on first draw) = 3/8

P(A on second draw) = 3/8

P(Both) = 3/8 × 3/8 = 9/64

Answer: 9/64

Example 5 – Same Population Without Replacement

From the same population of 8 units, of which 3 possess A, two units are selected without replacement. Find the probability that both possess A.

P = 3/8 × 2/7

Answer: 3/28

Exam Trap:

With replacement → 3/8 × 3/8

Without replacement → 3/8 × 2/7

PART–3: Standard Error – SRSWR

For a simple random sample of size n selected with replacement from a population having standard deviation σ:

SE(X̄) = σ / √n

Example 6

Population standard deviation σ = 12 and sample size n = 36. Find the standard error of the sample mean under SRSWR.

SE(X̄) = 12/√36

= 12/6 = 2

Answer: 2

Example 7 – Reverse Problem

Population standard deviation is 20. If SE(X̄)=2 under sampling with replacement, find sample size.

2 = 20/√n

√n = 10

n = 100

Answer: 100

⚡ Important Relationship:

SE ∝ 1/√n

If sample size becomes 4 times,
SE becomes 1/2.

If sample size becomes 9 times,
SE becomes 1/3.

PART–4: SRSWOR and Finite Population Correction

When sampling is without replacement from a finite population, the standard error is reduced by the Finite Population Correction (FPC).

FPC = √[(N − n)/(N − 1)]

SE(X̄) = (σ/√n) × √[(N − n)/(N − 1)]

Example 8 – PYQ-Type Standard Error Problem

A population contains N = 101 units. Population standard deviation σ = 12. A sample of n = 20 units is selected.

Find the standard error of the sample mean:
(i) With replacement
(ii) Without replacement

(i) With Replacement

SE = 12/√20

Since √20 = 2√5:

SE = 12/(2√5) = 6/√5

Rationalising:

6/√5 = 6√5/5 = 1.2√5

SRSWR: SE = 1.2√5

(ii) Without Replacement

SE = 1.2√5 × √[(101−20)/(101−1)]

= 1.2√5 × √(81/100)

= 1.2√5 × 9/10

= 1.08√5

SRSWOR: SE = 1.08√5

Very Important APPSC Trap:

SE without replacement is smaller than SE with replacement because:

0 ≤ FPC ≤ 1

Example 9 – Find FPC Directly

N = 100 and n = 20. Find the finite population correction.

FPC = √[(100−20)/(100−1)]

= √(80/99)

Answer: √(80/99) ≈ 0.899

Example 10 – Census as an Extreme Case

What happens to the sampling standard error under SRSWOR when n=N?

FPC = √[(N−N)/(N−1)] = 0

SE = 0

Answer: Sampling error due to sample-to-sample variation disappears when the entire population is enumerated.

PART–5: Probability Bounds – Very Important Objective Pattern

For any two events A and B:

max[0, P(A)+P(B)−1] ≤ P(A∩B) ≤ min[P(A),P(B)]

Example 11 – Minimum Intersection

P(A)=0.8 and P(B)=0.6. Find the minimum possible value of P(A∩B).

Minimum = max[0, 0.8+0.6−1]

= 0.4

Answer: 0.4

Example 12 – Maximum Intersection

P(A)=0.8 and P(B)=0.6. Find the maximum possible value of P(A∩B).

Maximum = min(0.8,0.6)

Answer: 0.6

Example 13 – Range of Intersection

If P(A)=0.7 and P(B)=0.5, determine the possible range of P(A∩B).

Minimum:

0.7 + 0.5 − 1 = 0.2

Maximum:

min(0.7,0.5)=0.5

0.2 ≤ P(A∩B) ≤ 0.5

⚡ Memorise:

Minimum Intersection
= P(A)+P(B)−1, but never below 0.

Maximum Intersection
= Smaller of P(A), P(B).

PART–6: Conditional Probability Bounds

P(A|B) = P(A∩B)/P(B)

If only P(A) and P(B) are known, first determine possible bounds for P(A∩B).

Example 14

P(A)=0.7 and P(B)=0.5. Find the minimum possible value of P(A|B).

First find minimum intersection:

P(A∩B) ≥ 0.7+0.5−1 = 0.2

Therefore:

P(A|B) ≥ 0.2/0.5

Answer: Minimum P(A|B)=0.4

Example 15

P(A)=0.6 and P(B)=0.8. Find the maximum possible value of P(A|B).

Maximum possible intersection:

P(A∩B) ≤ min(0.6,0.8)=0.6

P(A|B) ≤ 0.6/0.8

Answer: Maximum = 0.75

PART–7: Sampling Frame and Selection Error

A Sampling Frame is the list of population members from which the sample is actually selected.

Target Population
All subjects about whom information is required.

↓

Sampling Frame
Operational list used for selecting the sample.

↓

Sample
Units actually selected.

Example 16 – Identify the Error

A researcher wants information about all households in a city but selects households only from a telephone directory. Some households are not listed in the directory. What is the main problem?

Answer: Selection Error due to an incomplete/unrepresentative Sampling Frame.

Example 17

A college wants to estimate the average GPA of all students but a professor uses only students attending her own Statistics class. Identify the likely sampling method.

Answer: Convenience Sampling – a Non-Random Sampling method.

PART–8: Sampling Error vs Non-Sampling Error

Feature	Sampling Error	Non-Sampling Error
Main cause	Chance variation from observing a sample	Human/procedural errors
Occurs in sample survey?	Yes	Yes
Occurs in census?	No sampling error	Yes
Examples	Sample result differs from census result	Selection, nonresponse, response, recording errors

Example 18 – Sampling Error

True population mean = 80.
Sample mean = 83.
Assuming no other errors, find the signed sampling error.

83 − 80 = +3

Answer: +3

Example 19 – Nonresponse Error

A random sample of 1000 households is selected. Only 400 households answer the survey, and the nonrespondents systematically differ from respondents. Which error should be suspected?

Answer: Nonresponse Error.

Example 20 – Response Error

A respondent intentionally reports an incorrect income. What type of error is this?

Answer: Response Error.

Example 21 – Voluntary Response Error

A newspaper publishes an online question and asks readers to voluntarily vote. Which error/bias is particularly likely?

Answer: Voluntary Response Error/Bias.

Most Important Trap:

Increasing sample size may reduce random sampling variability, but it does not automatically remove systematic bias or non-sampling errors.

PART–9: Two-Way Table + Probability

500 employees are classified as follows:

	Retirement Benefit – Yes	Retirement Benefit – No	Total
Men	225	75	300
Women	150	50	200
Total	375	125	500

Example 22 – Marginal Probability

One employee is selected randomly. Find P(Woman).

P(Woman)=200/500

Answer: 0.4

Example 23 – Joint Probability

Find P(Woman ∩ Retirement Benefit).

P = 150/500

Answer: 0.30

Example 24 – Conditional Probability

Given that the selected employee is a woman, find the probability that she has retirement benefits.

P(Benefit | Woman)=150/200

Answer: 0.75

Example 25 – Reverse Conditional Probability

Given that an employee has retirement benefits, find the probability that the employee is a woman.

P(Woman | Benefit)=150/375

Answer: 0.40

APPSC Trap:

P(Benefit | Woman) = 150/200

but

P(Woman | Benefit) = 150/375

Conditional probability changes the denominator.

PART–10: Independence from a Two-Way Table

Example 26

Using the previous employee table:

P(Woman)=200/500=0.4

P(Benefit)=375/500=0.75

P(Woman∩Benefit)=150/500=0.30

Now:

P(Woman)P(Benefit)=0.4×0.75=0.30

Since:

P(Woman∩Benefit)=P(Woman)P(Benefit)

For this table, the two events are independent.

⚡ Independence Test:

Calculate:

P(A∩B)

and

P(A) × P(B)

Equal → Independent
Not Equal → Dependent

PART–11: Mutually Exclusive vs Independent – Major Trap

Mutually Exclusive	Independent
Cannot occur together	Occurrence of one does not change probability of other
P(A∩B)=0	P(A∩B)=P(A)P(B)
For non-zero probabilities, generally dependent	Can occur together

Example 27

A and B are mutually exclusive events with P(A)=0.4 and P(B)=0.3. Are they independent?

Since mutually exclusive:

P(A∩B)=0

But if independent:

P(A)P(B)=0.4×0.3=0.12

Since:

0 ≠ 0.12

Answer: No. They are dependent.

PART–12: Bayes' Theorem – Source Identification Problems

Example 28 – Factory Defect Problem

Factories F₁, F₂ and F₃ produce 30%, 30% and 40% of total production. Their defective rates are 4%, 3% and 2% respectively. A randomly selected product is defective. Find the probability that it came from F₃.

Factory	Production Share	Defect Rate	Joint Contribution
F₁	0.30	0.04	0.012
F₂	0.30	0.03	0.009
F₃	0.40	0.02	0.008

P(D)=0.012+0.009+0.008=0.029

P(F₃|D)=0.008/0.029

Answer: 8/29

Example 29 – Another Bayes Pattern

Machines A and B produce 60% and 40% of a factory's output. Their defect rates are 2% and 5% respectively. A defective product is selected. Find P(B | Defective).

A contribution = 0.60×0.02 = 0.012

B contribution = 0.40×0.05 = 0.020

Total Defective = 0.032

P(B|D)=0.020/0.032

Answer: 5/8 = 0.625

⚡ Bayes Table Shortcut:

Step 1: Source Probability
Step 2: Multiply by conditional outcome probability
Step 3: Add all joint contributions
Step 4: Required contribution ÷ Total contribution

PART–13: Repeated Independent Attempts

Let:

P(Success)=p     and     P(Failure)=q=1−p

Example 30 – No Success

Probability of success in one independent attempt is 0.7. Find the probability of no success in 4 attempts.

q=0.3

P(No Success)=0.3⁴=0.0081

Answer: 0.0081

Example 31 – At Least One Success

P(At Least One)=1−q⁴

=1−0.0081

Answer: 0.9919

Example 32 – First Success on 4th Attempt

Required sequence:

Failure → Failure → Failure → Success

P=q³p

=(0.3)³(0.7)

Answer: 0.0189

Example 33 – Exactly One Success in Four Attempts

p=0.7, q=0.3. Find probability of exactly one success in four independent attempts.

The one success can occupy any one of four positions.

P = 4 × (0.7)(0.3)³

= 4 × 0.0189

Answer: 0.0756

Do Not Confuse:

Success on 4th attempt → wording must be interpreted carefully.

First success on 4th attempt → q³p

At least one success in 4 attempts → 1−q⁴

Exactly one success in 4 attempts → 4pq³

PART–14: Identify the Sampling Method

Example 34

A university gives every student a number and randomly selects 100 numbers.

Simple Random Sampling

Example 35

After a random start, every 50th household is selected.

Systematic Random Sampling

Example 36

Population is divided into rural and urban groups and random samples are taken from both groups.

Stratified Random Sampling

Example 37

A state is divided into geographical clusters. A few clusters are randomly chosen for the survey.

Cluster Sampling

Example 38

A researcher interviews people who are easiest to contact.

Convenience Sampling

Example 39

An expert personally selects units considered most suitable for the study.

Judgment Sampling

Example 40

A population consists of 48% men and 52% women. The researcher intentionally chooses a sample containing exactly the same percentages.

Quota Sampling

PART–15: Stratified vs Quota – Favourite Trap

Feature	Stratified Random Sampling	Quota Sampling
Main category	Random/Probability Sampling	Non-Random Sampling
Population divided into groups?	Yes	Yes
Representation of groups	Sample drawn from each stratum	Quota fixed for groups
Random selection within groups	Yes, in stratified random sampling	Not required

Exam Trap:

“Groups are proportionately represented” alone does not prove that the method is stratified random sampling.

Check whether random selection is actually used.

PART–16: Stratified vs Cluster Sampling

Stratified Sampling	Cluster Sampling
Population → Strata	Population → Clusters
Sample from every stratum	One or a few clusters selected
Strata internally similar for stratification characteristic	Clusters ideally represent population
Example: income groups	Example: geographical areas

PART–17: Advanced APPSC Practice MCQs

Question 41

If P(A)=0.65 and P(B)=0.55, the minimum possible value of P(A∩B) is:

A) 0.10
B) 0.20
C) 0.55
D) 0.65

0.65+0.55−1=0.20

Answer: B – 0.20

Question 42

If P(A)=0.35 and P(B)=0.60, the maximum possible value of P(A∩B) is:

A) 0
B) 0.21
C) 0.35
D) 0.60

Answer: C – 0.35

Question 43

For N=101, n=20 and σ=12, SE(X̄) under SRSWR is:

A) 1.08√5
B) 1.2√5
C) 2.4√5
D) 12√5

Answer: B – 1.2√5

Question 44

For the same data, SE(X̄) under SRSWOR is:

A) 1.08√5
B) 1.2√5
C) 0
D) 6√5

Answer: A – 1.08√5

Question 45

Which error can occur only because a sample rather than the entire population is observed?

A) Response Error
B) Nonresponse Error
C) Sampling Error
D) Recording Error

Answer: C – Sampling Error

Question 46

Which can occur even in a complete census?

A) Sampling Error only
B) Non-Sampling Error
C) Finite Population Correction
D) Sample-to-sample variation

Answer: B – Non-Sampling Error

Question 47

A sample is selected from every income group after dividing a population into low, middle and high income groups. The method is:

A) Cluster Sampling
B) Convenience Sampling
C) Stratified Random Sampling
D) Judgment Sampling

Answer: C – Stratified Random Sampling

Question 48

A sample includes people who voluntarily respond to a newspaper poll. The principal concern is:

A) Sampling interval
B) Voluntary Response Bias
C) Finite Population Correction
D) Stratification

Answer: B – Voluntary Response Bias

Question 49

If A and B are mutually exclusive with P(A)>0 and P(B)>0, then:

A) They must be independent
B) P(A∩B)=P(A)P(B)
C) They are not independent
D) P(A∪B)=0

Answer: C – They are not independent

Question 50

If P(A)=0.4, P(B)=0.5 and P(A∩B)=0.20, A and B are:

A) Mutually exclusive
B) Independent
C) Impossible events
D) Complementary events

P(A)P(B)=0.4×0.5=0.20

Answer: B – Independent

PART–18: Assertion–Reason Practice

Question 51

Assertion (A): Sampling error does not occur in a census.

Reason (R): A census includes every element of the population.

Answer: Both A and R are true, and R explains A.

Question 52

Assertion (A): Non-sampling errors can occur in a census.

Reason (R): Recording and response errors are not eliminated merely by observing every population unit.

Answer: Both A and R are true, and R explains A.

Question 53

Assertion (A): A random sample must always exactly reproduce all population characteristics.

Reason (R): Random selection eliminates every possible sample-to-sample difference.

Answer: Both A and R are false.

Question 54

Assertion (A): Under SRSWOR, the standard error is generally smaller than under SRSWR for the same N, n and σ.

Reason (R): The finite population correction is at most 1.

Answer: Both A and R are true, and R explains A.

PART–19: APPSC 10-Second Shortcuts

Question Clue	Immediate Thought
Every kth unit	Systematic Sampling
List ends and selection continues from beginning	Circular Systematic Sampling
Each same-size sample equally likely	Simple Random Sampling
Sample from every subgroup	Stratified Sampling
One/few geographical groups selected	Cluster Sampling
Easiest people selected	Convenience Sampling
Expert chooses	Judgment Sampling
Exact population proportions imposed without random selection	Quota Sampling
Incomplete list used to select sample	Selection Error / Sampling Frame problem
Selected people do not reply	Nonresponse Error
Incorrect answer supplied	Response Error
People choose themselves to respond	Voluntary Response Error
At least one	1 − None
First success on nth trial	qⁿ⁻¹p
Minimum intersection	max[0,P(A)+P(B)−1]
Maximum intersection	min[P(A),P(B)]
Without replacement SE	Apply FPC

PART–20: One-Minute Formula Sheet

Systematic Sampling Interval
k = N/n

Sampling Fraction
f = n/N

SRSWR Standard Error
SE(X̄)=σ/√n

Finite Population Correction
FPC=√[(N−n)/(N−1)]

SRSWOR Standard Error
SE(X̄)=(σ/√n)√[(N−n)/(N−1)]

Conditional Probability
P(A|B)=P(A∩B)/P(B)

Independent Events
P(A∩B)=P(A)P(B)

Addition Rule
P(A∪B)=P(A)+P(B)−P(A∩B)

Minimum Intersection
max[0,P(A)+P(B)−1]

Maximum Intersection
min[P(A),P(B)]

At Least One Success
1−qⁿ

First Success on nth Attempt
qⁿ⁻¹p

Exactly One Success in n Independent Trials
npqⁿ⁻¹

PART–21: Final APPSC Exam Traps

Trap 1:
Random sample ≠ guaranteed perfectly representative sample.

Trap 2:
Random sampling and non-random sampling are not distinguished merely by sample size.

Trap 3:
Stratified Sampling ≠ Quota Sampling.

Trap 4:
Stratified Sampling ≠ Cluster Sampling.

Trap 5:
Sampling Error can occur in sample surveys, not as sampling error in a complete census.

Trap 6:
Non-Sampling Errors can occur in both sample surveys and censuses.

Trap 7:
Larger sample size does not automatically remove response or selection bias.

Trap 8:
Mutually Exclusive ≠ Independent.

Trap 9:
P(A|B) ≠ P(B|A) in general.

Trap 10:
With Replacement ≠ Without Replacement.

Trap 11:
Without replacement from a finite population → check whether FPC is required.

Trap 12:
“At least one” → complement is usually fastest.

Trap 13:
“First success on nth attempt” requires all previous attempts to fail.

Trap 14:
In probability-bound questions, P(A∩B) can never exceed the smaller marginal probability.

Trap 15:
In circular systematic sampling, crossing the last unit does not end selection; continue from the beginning.

చివరి నిమిషం పునశ్చరణ

Circular Systematic Sampling
→ Fixed interval + wrap around population list

SRSWR
→ With Replacement
→ SE = σ/√n

SRSWOR
→ Without Replacement
→ Apply Finite Population Correction

FPC
→ √[(N−n)/(N−1)]

Minimum P(A∩B)
→ max[0,P(A)+P(B)−1]

Maximum P(A∩B)
→ min[P(A),P(B)]

Sampling Frame Problem
→ Selection Error

No Reply
→ Nonresponse Error

Incorrect Reply
→ Response Error

Self-selected Respondents
→ Voluntary Response Error

Sampling Error
→ Sample Survey only

Non-Sampling Error
→ Sample Survey + Census

Two-Way Table
→ Joint + Marginal + Conditional Probability

Independence Check
→ P(A∩B)=P(A)P(B)

Bayes
→ Observed Outcome → Find Source

At Least One
→ 1−qⁿ

First Success on nth Attempt
→ qⁿ⁻¹p

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