Sampling Methods and Probability based problems

Sampling & Probability – Combined Numerical Problems, Shortcuts & APPSC Practice MCQs

Sampling మరియు Probability ఒకదానితో ఒకటి closely related topics. Random sample selectionలో ప్రతి unit selectionకు probability ఉంటుంది. అందువల్ల APPSC objective examinationలో sampling conceptను probability calculationతో combine చేసి questions అడగవచ్చు.

ఈ Practice Postలో:

✓ Probability of Selecting a Unit
✓ With Replacement
✓ Without Replacement
✓ At Least One Problems
✓ Exactly One Problems
✓ Systematic Sampling Numericals
✓ Stratified Sampling Numericals
✓ Sampling Fraction
✓ Sampling Error
✓ Conditional Probability
✓ Independent & Dependent Events
✓ Bayes' Theorem Problems
✓ Repeated Independent Trials
✓ Mixed APPSC Numerical Problems
✓ Fast Calculation Shortcuts

PART–A: Basic Random Sampling & Probability

Problem 1 – Probability of Selecting One Particular Unit

A population consists of 50 persons. One person is selected at random. What is the probability that a particular person is selected?

P = 1 / 50

Answer: 1/50

⚡ Simple Random Selection:

If one unit is selected from N equally likely units,

P(particular unit) = 1/N

Problem 2 – Selecting a Category

A population contains 60 men and 40 women. One person is selected at random. Find the probability that the selected person is a woman.

Total population:

N = 60 + 40 = 100

P(Woman) = 40/100 = 2/5

Answer: 2/5 = 0.40

Problem 3 – Complement Method

From the same population of 60 men and 40 women, find the probability of selecting a person who is not a woman.

P(Not Woman) = 1 − P(Woman)

= 1 − 2/5 = 3/5

Answer: 3/5

PART–B: Without Replacement Problems

Without Replacement:

Selected item is NOT returned.
Therefore population size changes after each selection.

Problem 4 – Both Selected are Women

A group consists of 6 men and 4 women. Two persons are selected randomly without replacement. Find the probability that both are women.

First woman:

4/10

After selecting one woman, 3 women remain among 9 persons.

P(Both Women) = 4/10 × 3/9

= 12/90 = 2/15

Answer: 2/15

Problem 5 – Both Selected are Men

From the same group of 6 men and 4 women, two persons are selected without replacement. Find P(Both Men).

P = 6/10 × 5/9

= 30/90 = 1/3

Answer: 1/3

Problem 6 – One Man and One Woman

From 6 men and 4 women, two persons are selected without replacement. Find the probability of selecting exactly one man and one woman.

There are two possible orders:

Man → Woman

OR

Woman → Man

P = (6/10 × 4/9) + (4/10 × 6/9)

= 24/90 + 24/90 = 48/90 = 8/15

Answer: 8/15

Important Trap:

For "one man and one woman", do not calculate only Man → Woman.

You must also include Woman → Man.

Problem 7 – At Least One Woman

From 6 men and 4 women, two persons are selected without replacement. Find the probability that at least one is a woman.

At least one Woman
= 1 − P(No Woman)
= 1 − P(Both Men)

= 1 − 1/3

Answer: 2/3

⚡ Golden Rule:

"At least one success"
= 1 − P(No Success)

PART–C: With Replacement Problems

With Replacement:

After each selection, selected item is returned.

Therefore probabilities remain unchanged.

Problem 8 – Both Women With Replacement

A group contains 6 men and 4 women. Two persons are selected with replacement. Find P(Both Women).

P = 4/10 × 4/10

= 16/100 = 4/25

Answer: 4/25

Problem 9 – At Least One Woman With Replacement

From the same group, two selections are made with replacement. Find the probability of at least one woman.

Probability of a man:

P(M) = 6/10 = 3/5

Probability of no woman in two selections:

P(MM) = 3/5 × 3/5 = 9/25

P(At Least One Woman) = 1 − 9/25

Answer: 16/25

PART–D: With vs Without Replacement – Exam Trap

With Replacement	Without Replacement
Population size remains same	Population size decreases
Probability remains unchanged	Probability generally changes
Selections can be independent	Successive selections are generally dependent
Example: 4/10 × 4/10	Example: 4/10 × 3/9

PART–E: Combination Method in Sampling Probability

Problem 10 – Both Women Using Combinations

From 6 men and 4 women, two persons are selected randomly. Find P(Both Women).

Total ways:

10C2 = 45

Favourable ways:

4C2 = 6

P = 6/45 = 2/15

Answer: 2/15

Problem 11 – Exactly One Woman Using Combinations

Favourable Ways = 4C1 × 6C1

= 4 × 6 = 24

P = 24/45 = 8/15

Answer: 8/15

⚡ Combination Method:

Probability = Favourable Combinations / Total Combinations

PART–F: Sampling Fraction Numericals

Sampling Fraction = n/N

where:
n = Sample Size
N = Population Size

Problem 12

Population size = 2000.
Sample size = 200.
Find the sampling fraction.

n/N = 200/2000 = 1/10

Answer: 0.10 or 10%

Problem 13 – Find Sample Size

Population = 5000.
Sampling fraction = 8%.
Find sample size.

n = 8/100 × 5000

Answer: 400

PART–G: Systematic Sampling Numericals

Sampling Interval k = N/n

Problem 14

A population contains 2000 units. A sample of 200 is required. Find the systematic sampling interval.

k = 2000/200 = 10

Answer: k = 10

Problem 15 – Identify Sample Units

N = 1000, n = 100.
Random starting unit = 6.
Find the first five selected units.

k = 1000/100 = 10

Selected units:

6,
6 + 10 = 16,
26,
36,
46

Answer: 6, 16, 26, 36, 46

Problem 16 – Textbook-Type Systematic Sample

There are 45,000 households and a sample of 150 households is required. Find the sampling interval.

k = 45000/150 = 300

Answer: Every 300th household after a random start.

APPSC Trap:

Systematic sampling does NOT mean simply selecting units 1, k+1, 2k+1... automatically.

A random starting unit is first selected.

PART–H: Stratified Sampling Numericals

nₕ = (Nₕ / N) × n

Problem 17 – Proportional Allocation

A population contains:

Rural = 700
Urban = 300

Required sample size = 200.

Find proportional sample sizes.

Total:

N = 700 + 300 = 1000

Rural:

700/1000 × 200 = 140

Urban:

300/1000 × 200 = 60

Answer: Rural = 140, Urban = 60

Problem 18 – Three Strata

Three strata contain 1000, 600 and 400 units respectively. A proportional sample of 200 is required.

Total:

N = 1000 + 600 + 400 = 2000

Stratum 1:

1000/2000 × 200 = 100

Stratum 2:

600/2000 × 200 = 60

Stratum 3:

400/2000 × 200 = 40

Answer: 100, 60, 40

Problem 19 – Missing Stratum Size

In proportional stratified sampling, a stratum containing 400 units receives a sample of 80. Total population is 2000. Find total sample size.

80 = (400/2000)n

80 = n/5

Answer: n = 400

PART–I: Sampling Error Numericals

Sampling Error = Sample Statistic − Population Parameter

Problem 20

Population mean = 80.60.
Sample mean = 81.67.
Find sampling error.

81.67 − 80.60 = +1.07

Answer: +1.07

Problem 21 – Negative Sampling Error

Population mean = 75.
Sample mean = 72.
Find sampling error.

72 − 75 = −3

Answer: −3

Important:

Sampling error can be positive or negative when expressed as a signed difference.

Problem 22 – Sampling and Non-Sampling Components

Population mean = 80.60.
Correct sample mean = 81.67.
Because of a recording mistake, calculated sample mean = 82.33.

Find:
1. Sampling error
2. Total observed difference
3. Error attributable to the recording mistake

Sampling error:

81.67 − 80.60 = 1.07

Total observed difference:

82.33 − 80.60 = 1.73

Recording-error component:

82.33 − 81.67 = 0.66

Sampling Error = 1.07
Total Difference = 1.73
Recording Error Component = 0.66

PART–J: Conditional Probability + Sampling

P(A|B) = P(A ∩ B) / P(B)

Problem 23

A group contains 6 men and 4 women. Two persons are selected without replacement. Given that the first selected person is a woman, find the probability that the second is also a woman.

After one woman has already been selected:

Remaining persons = 9
Remaining women = 3

P(Second Woman | First Woman) = 3/9

Answer: 1/3

Problem 24

A box contains 5 defective and 15 good items. Two items are selected without replacement. Given that the first item is defective, find the probability that the second is good.

After selecting one defective item:

Total remaining = 19
Good items remaining = 15

P(G₂ | D₁) = 15/19

Answer: 15/19

PART–K: Independent vs Dependent Selection

Situation	Nature
Repeated selection with replacement	Independent in the standard random-selection setup
Repeated selection without replacement	Dependent
Independent trials	P(A ∩ B) = P(A)P(B)
Dependent trials	P(A ∩ B) = P(A)P(B|A)

Problem 25 – Independent Events

A fair coin is tossed twice. Find the probability of obtaining two heads.

P(HH) = 1/2 × 1/2 = 1/4

Answer: 1/4

PART–L: At Least One – High-Yield Probability Pattern

Problem 26

Probability of success in one independent trial is 0.7. Three trials are conducted. Find the probability of at least one success.

Failure probability:

q = 1 − 0.7 = 0.3

Probability of no success:

0.3³ = 0.027

P(At Least One Success) = 1 − 0.027

Answer: 0.973

⚡ At Least One Success in n Independent Trials:

1 − qⁿ

Problem 27 – No Success

Success probability = 0.8. Four independent attempts are made. Find the probability of no success.

q = 1 − 0.8 = 0.2

P(No Success) = 0.2⁴ = 0.0016

Answer: 0.0016

PART–M: First Success on nth Trial

P(First Success on nth Trial) = qⁿ⁻¹p

Problem 28

Probability of success in each independent attempt is 0.7. Find the probability that the first success occurs on the 4th attempt.

For first success on attempt 4:

Failure, Failure, Failure, Success

P = (0.3)³(0.7)

= 0.0189

Answer: 0.0189

Problem 29 – First Success on 7th Attempt

p = 0.7. Find the probability that the first success occurs on the 7th independent attempt.

P = (0.3)⁶(0.7)

= 0.0005103

Answer: 0.0005103

Exam Trap:

"Success on 7th trial" and
"First success on 7th trial"

are not automatically the same statement.

First success requires the first six attempts to fail.

PART–N: Exactly One Success

Problem 30

Probability of success = 0.6. Two independent trials are performed. Find the probability of exactly one success.

Possible sequences:

Success–Failure
Failure–Success

P = (0.6)(0.4) + (0.4)(0.6)

Answer: 0.48

⚡ Exactly One Success in Two Independent Trials:

2pq

PART–O: Bayes' Theorem + Sampling

P(Fᵢ|D) = [P(Fᵢ) × P(D|Fᵢ)] / Σ[P(Fⱼ) × P(D|Fⱼ)]

Problem 31 – Factory Defect Problem

Three factories F₁, F₂ and F₃ produce respectively 30%, 30% and 40% of total items.

Their defective rates are respectively:

F₁ = 4%
F₂ = 3%
F₃ = 2%

A randomly selected item is found defective. Find the probability that it came from F₃.

Defective contribution of F₁:

0.30 × 0.04 = 0.012

F₂:

0.30 × 0.03 = 0.009

F₃:

0.40 × 0.02 = 0.008

Total defective probability:

0.012 + 0.009 + 0.008 = 0.029

P(F₃|D) = 0.008 / 0.029 = 8/29

Answer: 8/29 ≈ 0.276

⚡ Bayes Shortcut:

Source Share × Defect Rate
→ calculate for every source
→ add all contributions
→ required contribution / total

PART–P: Mixed Stratified Sampling + Probability

Problem 32

A population consists of:

Rural = 600
Urban = 400

A proportional stratified sample of 200 is selected. Find the number selected from each stratum.

Rural = 600/1000 × 200 = 120

Urban = 400/1000 × 200 = 80

Answer: Rural = 120, Urban = 80

Suppose 30 of the 120 sampled rural respondents have characteristic A. If one rural sampled respondent is selected randomly, find P(A).

P(A | Rural Sample) = 30/120

Answer: 1/4 = 0.25

PART–Q: Mixed Systematic Sampling + Probability

Problem 33

N = 1200 and n = 100. A systematic random sample is required.

Q1. Find sampling interval.

k = 1200/100 = 12

Answer: 12

Q2. If random start is 7, find first four sample units.

7, 19, 31, 43

Q3. If the random start is chosen uniformly from the first 12 units, what is the probability that 7 is selected as the starting point?

P = 1/12

Answer: 1/12

PART–R: Probability Bounds – APPSC Tricky Area

For any two events A and B:

P(A ∩ B) ≤ P(A)

P(A ∩ B) ≤ P(B)

P(A ∩ B) ≥ P(A) + P(B) − 1

Problem 34

P(A) = 0.7 and P(B) = 0.6. What is the minimum possible value of P(A ∩ B)?

P(A ∩ B) ≥ 0.7 + 0.6 − 1

Answer: Minimum = 0.3

Problem 35

P(A)=0.4 and P(B)=0.7. What is the maximum possible value of P(A ∩ B)?

P(A ∩ B) ≤ min[P(A),P(B)]

Answer: Maximum = 0.4

PART–S: APPSC Mixed Practice MCQs

Question 36

A population has 500 units. One unit is selected randomly. Probability that a specified unit is selected is:

A) 1/50
B) 1/100
C) 1/500
D) 500

Answer: C – 1/500

Question 37

A population contains 80 men and 20 women. One person is selected randomly. Probability of selecting a woman is:

A) 0.20
B) 0.25
C) 0.40
D) 0.80

Answer: A – 0.20

Question 38

If N = 3000 and n = 150, systematic sampling interval is:

3000/150 = 20

Answer: 20

Question 39

A population has two strata containing 800 and 200 units. For a proportional sample of 100, the second stratum should contribute:

200/1000 × 100 = 20

Answer: 20

Question 40

Population mean = 50 and sample mean = 53. Using Sample Statistic − Population Parameter, sampling error is:

Answer: +3

Question 41

Probability of success = 0.8. Probability of failure is:

1 − 0.8 = 0.2

Answer: 0.2

Question 42

For independent events A and B: P(A)=0.5 and P(B)=0.4. Find P(A ∩ B).

0.5 × 0.4 = 0.20

Answer: 0.20

Question 43

P(A)=0.6, P(B)=0.5 and P(A ∩ B)=0.3. Find P(A ∪ B).

P(A ∪ B) = P(A)+P(B)−P(A ∩ B)

= 0.6 + 0.5 − 0.3 = 0.8

Answer: 0.8

Question 44

If two events are mutually exclusive and P(A)=0.3, P(B)=0.4, find P(A ∪ B).

0.3 + 0.4 = 0.7

Answer: 0.7

Question 45

If P(A)=0.7, find P(Aᶜ).

1 − 0.7 = 0.3

Answer: 0.3

PART–T: APPSC Statement Traps

1. A random sample is always perfectly representative of the population.
→ FALSE

2. In simple random sampling, all samples of the same size have equal chance of selection.
→ TRUE

3. In systematic sampling, k = N/n.
→ TRUE

4. Stratified sampling selects a sample from each stratum.
→ TRUE

5. Cluster sampling necessarily selects some units from every cluster.
→ FALSE

6. Sampling error occurs in a census because all population units are observed.
→ FALSE

7. Non-sampling errors may occur in both census and sample surveys.
→ TRUE

8. Increasing sample size automatically removes response bias.
→ FALSE

9. Mutually exclusive events must always be independent.
→ FALSE

10. If A and B are independent, P(A ∩ B)=P(A)P(B).
→ TRUE

11. Without replacement, successive selection probabilities generally change.
→ TRUE

12. At least one success is conveniently calculated using the complement of no success.
→ TRUE

PART–U: 10-Second Question Identification

Words in Question	Immediate Approach
One particular unit from N	1/N
With replacement	Probabilities remain same
Without replacement	Update numerator and denominator
At least one	1 − P(None)
Exactly one of two	Success-Failure + Failure-Success
First success on nth trial	qⁿ⁻¹p
Every kth unit	Systematic Sampling
N and n given	k=N/n or n/N
Proportion from each group	Stratified Sampling
Sample mean and population mean	Sampling Error
Given that...	Conditional Probability
Source known after observing outcome	Bayes' Theorem
Independent events	Multiply probabilities
Mutually exclusive events	Intersection = 0

PART–V: One-Minute Formula Box

Probability
P(A) = Favourable Outcomes / Total Outcomes

Complement
P(Aᶜ) = 1 − P(A)

Addition Rule
P(A ∪ B) = P(A)+P(B)−P(A ∩ B)

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

Independent Events
P(A ∩ B)=P(A)P(B)

Dependent Events
P(A ∩ B)=P(A)P(B|A)

At Least One Success
1 − qⁿ

First Success on nth Trial
qⁿ⁻¹p

Sampling Fraction
n/N

Systematic Sampling Interval
k=N/n

Proportional Stratified Allocation
nₕ=(Nₕ/N)n

Sampling Error
Sample Statistic − Population Parameter

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

చివరి నిమిషం పునశ్చరణ

Random Selection
→ Probability-based selection

One particular unit among N
→ 1/N

With Replacement
→ Probability remains unchanged

Without Replacement
→ Probability changes

At Least One
→ 1 − None

Systematic Sampling
→ k=N/n + Random Start

Stratified Sampling
→ Sample from each stratum

Proportional Allocation
→ nₕ=(Nₕ/N)n

Sampling Fraction
→ n/N

Sampling Error
→ Sample Statistic − Population Parameter

Conditional Probability
→ P(A∩B)/P(B)

Independent Events
→ Multiply probabilities

Mutually Exclusive
→ P(A∩B)=0

First Success on nth Trial
→ qⁿ⁻¹p

Bayes' Theorem
→ Reverse probability: Outcome → Source

Most Important APPSC Numerical Areas

With vs Without Replacement
At Least One
Systematic Sampling Interval
Proportional Stratified Allocation
Sampling Error
Conditional Probability
Independent/Dependent Events
Bayes' Theorem
Repeated Trials
Probability Bounds

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