definition of Probability – సంభావ్యత నిర్వచనం

Probability – సంభావ్యత అనేది ఒక event జరగడానికి ఎంత అవకాశం ఉందో numerically measure చేసే పద్ధతి. Probability value ఎల్లప్పుడూ 0 మరియు 1 మధ్య ఉంటుంది.

APPSC Focus:

Experiment & Outcome
Sample Space
Event
Simple & Compound Events
Equally Likely Outcomes
Classical Definition of Probability
Sure & Impossible Events
Complementary Events
Mutually Exclusive Events
Addition Rule
Conditional Probability
Independent & Dependent Events
Multiplication Rule
Important Probability Inequalities
Numerical Problems & Shortcuts

1. Meaning of Probability – సంభావ్యత

ఒక particular event జరిగే likelihood or chanceను numerical formలో తెలియజేసే measureను Probability అంటారు.

Definition:

Probability is a numerical measure of the likelihood that a specific event will occur.

Simple Meaning:

Impossible → Probability = 0

Possible → Probability lies between 0 and 1

Certain → Probability = 1

0 ≤ P(A) ≤ 1

2. Random Experiment

Possible outcomes ముందుగానే తెలిసినా, ఒక particular trialలో ఏ outcome వస్తుందో certaintyతో చెప్పలేని experimentను Random Experiment అంటారు.

Examples:

Coin toss → Head or Tail
Die throw → 1, 2, 3, 4, 5 or 6
Card selection → Any one card from the deck

3. Outcome

Random experimentను perform చేసినప్పుడు వచ్చే individual resultను Outcome అంటారు.

Example:

Die throwలో “4” రావడం → One Outcome

4. Sample Space

Random experimentకు సంబంధించిన all possible outcomes collectionను Sample Space అంటారు. దీనిని సాధారణంగా Sతో సూచిస్తారు.

Coin Toss:

S = {H, T}

Die Throw:

S = {1, 2, 3, 4, 5, 6}

Two Coins:

S = {HH, HT, TH, TT}

⚡ APPSC Shortcut:

One coin → 2 outcomes
Two coins → 2² = 4 outcomes
Three coins → 2³ = 8 outcomes

Therefore:

n coins → 2ⁿ outcomes

5. Event

Sample spaceలోని one or more outcomes యొక్క collectionను Event అంటారు.

Example:

Die throwలో even number రావడం:

S = {1,2,3,4,5,6}
A = {2,4,6}

Therefore A is an event.

6. Simple Event and Compound Event

Simple Event	Compound Event
Contains exactly one outcome	Contains more than one outcome
Example: Die shows 5 → {5}	Example: Even number → {2,4,6}
Also called Elementary Event	Also called Composite Event

7. Equally Likely Outcomes

Two or more outcomesకు occurrence probability equalగా ఉంటే వాటిని Equally Likely Outcomes అంటారు.

Fair Coin:

P(H) = P(T) = 1/2

Fair Die:

P(1) = P(2) = ... = P(6) = 1/6

8. Classical Definition of Probability

ఒక random experimentలో అన్ని outcomes equally likely అయితే, event A యొక్క probability:

P(A) = Number of Favourable Outcomes / Total Number of Equally Likely Outcomes

అంటే:

P(A) = n(A) / n(S)

APPSC Exam Trap:

Classical definitionను directly apply చేయాలంటే outcomes equally likelyగా ఉండాలి.

Numerical 1 – Coin

A fair coin is tossed once. Find the probability of getting a Head.

S = {H,T}

Total outcomes = 2
Favourable outcomes = 1

P(H) = 1/2

Answer: 1/2

Numerical 2 – Die

A fair die is thrown once. Find the probability of getting an even number.

S = {1,2,3,4,5,6}

Even numbers = {2,4,6}

P(Even) = 3/6 = 1/2

Answer: 1/2

⚡ Die Shortcut:

Prime → {2,3,5} → 3/6 = 1/2
Even → {2,4,6} → 1/2
Odd → {1,3,5} → 1/2
Greater than 4 → {5,6} → 1/3

9. Sure Event

తప్పకుండా జరిగే eventను Sure Event లేదా Certain Event అంటారు.

P(S) = 1

Example:

Ordinary die throwలో number ≤ 6 రావడం is certain.

Probability = 1

10. Impossible Event

జరగడానికి అవకాశం లేని eventను Impossible Event అంటారు.

P(∅) = 0

Example:

Ordinary six-sided dieలో 8 రావడం impossible.

Probability = 0

11. Complementary Events

Event A జరగకపోవడాన్ని complement of A అంటారు. దీనిని A′ లేదా A^cతో సూచించవచ్చు.

P(A) + P(A′) = 1

P(A′) = 1 − P(A)

⚡ Golden Shortcut:

Questionలో “at least one” కనిపిస్తే చాలా సందర్భాల్లో complement method వేగంగా ఉంటుంది.

P(At least one) = 1 − P(None)

Numerical 3 – Complement

If P(A) = 0.72, find P(A′).

P(A′) = 1 − 0.72 = 0.28

Answer: 0.28

Numerical 4 – At Least One Head

Two fair coins are tossed. Find the probability of getting at least one Head.

Instead of counting HH, HT and TH separately:

P(At least one H) = 1 − P(No Head)

No Head means TT.

= 1 − 1/4 = 3/4

Answer: 3/4

12. Mutually Exclusive Events

రెండు events ఒకే trialలో simultaneously occur కాలేకపోతే వాటిని Mutually Exclusive Events అంటారు.

A ∩ B = ∅

P(A ∩ B) = 0

Example:

One coin tossలో:

A = Head
B = Tail

Head మరియు Tail simultaneously రావు.

13. Addition Rule of Probability

Any two events A and Bకు:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

A and B mutually exclusive అయితే:

P(A ∪ B) = P(A) + P(B)

⚡ Memory:

OR → Think Addition

A or B → P(A ∪ B)

Numerical 5 – Addition Rule

If P(A)=0.5, P(B)=0.4 and P(A∩B)=0.2, find P(A∪B).

P(A∪B) = 0.5 + 0.4 − 0.2

= 0.7

Answer: 0.7

14. Conditional Probability

Event B already occurred అనే conditionలో event A యొక్క probabilityను Conditional Probability అంటారు.

P(A|B) = P(A ∩ B) / P(B)

where P(B) > 0.

⚡ Formula Memory:

A given B
→ Intersection on top
→ Given event B at bottom

Numerical 6 – Conditional Probability

P(A∩B)=0.24 and P(B)=0.60.

Find P(A|B).

P(A|B) = 0.24 / 0.60 = 0.4

Answer: 0.4

15. Independent Events

Event A occurrence వల్ల Event B probability మారకపోతే A and B are Independent Events.

P(A ∩ B) = P(A) × P(B)

Equivalent condition:

P(A|B) = P(A)

Very Important Trap:

Independent ≠ Mutually Exclusive

Independent → One event does not affect another.

Mutually Exclusive → Both cannot occur together.

16. Dependent Events

ఒక event occurrence మరొక event probabilityను మార్చితే అవి Dependent Events.

P(A ∩ B) = P(A) × P(B|A)

17. Multiplication Rule

General Rule:

P(A ∩ B) = P(A) × P(B|A)

If A and B are Independent:

P(A ∩ B) = P(A) × P(B)

⚡ Memory:

AND → Think Multiplication

OR → Think Addition

Numerical 7 – Independent Events

Two fair coins are tossed. Find probability of obtaining two Heads.

P(HH) = 1/2 × 1/2 = 1/4

Answer: 1/4

18. Repeated Independent Trials

Independent trialsలో success probability p అయితే failure probability:

q = 1 − p

First success exactly on nth trial కావాలంటే:

P = qⁿ⁻¹p

⚡ Shortcut:

First success on nth attempt

= Fail × Fail × ... × Fail × Success

= qⁿ⁻¹p

PYQ Pattern Numerical – Repeated Attempts

A person repeatedly attempts to destroy a target. Attempts are independent. Probability of destroying the target in any attempt is 0.7. Find the probability that the target is destroyed in the 7th attempt.

Success probability:

p = 0.7

Failure probability:

q = 1 − 0.7 = 0.3

To succeed for the first time on attempt 7:

P = (0.3)⁶(0.7)

= 0.0005103

Answer: 0.0005103

Exam Trap:

“Success in 7th attempt” and
“First success in 7th attempt” must be interpreted carefully.

For first success on 7th:
First 6 must fail + 7th must succeed.

19. Important Probability Inequalities

For any two events A and B:

P(A ∩ B) ≤ P(A)

P(A ∩ B) ≤ P(B)

Therefore:

P(A ∩ B) ≤ min[P(A), P(B)]

Also:

P(A ∪ B) ≤ P(A) + P(B)

Since P(A∪B) ≤ 1:

P(A ∩ B) ≥ P(A) + P(B) − 1

APPSC High-Yield Result:

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

PYQ Concept – Probability Inequality

For any two events A and B, which relation is always true?

P(A∩B) ≥ P(A) + P(B) − 1

This is an important APPSC PYQ-type result.

20. Useful Conditional Probability Result

If P(E)+P(F) ≥ 1, then:

P(E|F) ≥ P(E) + P(F) − 1

But APPSC may test such relations through transformed inequalities rather than asking only the basic definition. Therefore formula manipulation is important.

21. Probability of Two Dice

Two fair dice → Total ordered outcomes = 6 × 6 = 36

Required Sum	No. of Ways	Probability
2	1	1/36
3	2	2/36
4	3	3/36
5	4	4/36
6	5	5/36
7	6	6/36
8	5	5/36
9	4	4/36
10	3	3/36
11	2	2/36
12	1	1/36

⚡ Two-Dice Shortcut:

Number of ways for sums 2 to 12:

1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1

Numerical 8 – Two Dice

Two fair dice are thrown. Find probability that the sum is 7.

Favourable outcomes:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

P(Sum 7) = 6/36 = 1/6

Answer: 1/6

22. Basic Playing Card Probability

Standard Deck = 52 Cards

4 Suits = Hearts, Diamonds, Clubs, Spades

Each Suit = 13 cards

Red Cards = 26
Black Cards = 26

Kings = 4
Queens = 4
Jacks = 4

Face Cards = 12
Aces = 4

Numerical 9 – Cards

One card is drawn from a standard pack of 52 cards. Find the probability of drawing a King.

P(King) = 4/52 = 1/13

Answer: 1/13

23. Total Probability – Basic Idea

If B₁, B₂, ..., B_n divide the sample space into mutually exclusive and exhaustive events, then:

P(A) = Σ P(B_i) P(A|B_i)

This idea is useful in factory, machine, defective-item and population-group questions.

24. Bayes-Type Probability

When we know an observed result and want to find the probability of the source/cause, Bayes' rule is useful.

P(B_i|A) = [P(B_i)P(A|B_i)] / Σ[P(B_j)P(A|B_j)]

⚡ Bayes Shortcut:

Required Source Weight
÷
Total Weight

where

Weight = Source Probability × Conditional Probability

PYQ Pattern – Factory Problem

Three factories F₁, F₂ and F₃ produce 30%, 30% and 40% of total production. Their defective rates are 4%, 3% and 2%, respectively. If a randomly selected item is defective, find the probability that it came from F₃.

Weights:

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

⚡ Fast APPSC Method:

Ignore decimals initially:

30×4 = 120
30×3 = 90
40×2 = 80

Required = 80/(120+90+80)
= 80/290
= 8/29

25. Mutually Exclusive vs Independent

Mutually Exclusive	Independent
Cannot occur together	Occurrence of one does not affect other
P(A∩B)=0	P(A∩B)=P(A)P(B)
Joint occurrence impossible	Joint occurrence may be possible
Example: Head and Tail in one toss	Example: Heads in two separate coin tosses

Important Trap:

If A and B are mutually exclusive and both have positive probabilities, they cannot be independent.

APPSC Objective Questions

1. Probability of an impossible event is:

A) 1
B) 0
C) −1
D) ∞

Answer: B – 0

2. Probability of a sure event is:

A) 0
B) 1/2
C) 1
D) −1

Answer: C – 1

3. Which is always true?

A) P(A)>1
B) P(A)<0
C) 0≤P(A)≤1
D) P(A)=1/2

Answer: C

4. Collection of all possible outcomes is called:

A) Event
B) Sample Space
C) Trial Error
D) Frequency

Answer: B – Sample Space

5. If P(A)=0.35, then P(A′) is:

A) 0.35
B) 0.65
C) 1.35
D) 0

Answer: B – 0.65

6. For mutually exclusive events:

A) P(A∩B)=1
B) P(A∩B)=0
C) P(A)=P(B)
D) P(A|B)=1

Answer: B

7. For independent events:

A) P(A∩B)=0 always
B) P(A∩B)=P(A)P(B)
C) P(A)+P(B)=0
D) P(A)=0

Answer: B

8. If P(A∩B)=0.18 and P(B)=0.6, P(A|B) equals:

0.18/0.6 = 0.3

Answer: 0.3

9. Two fair dice are thrown. Probability of sum 11 is:

A) 1/6
B) 1/12
C) 1/18
D) 1/36

Sum 11 has 2 ways → (5,6),(6,5)

2/36 = 1/18

Answer: C – 1/18

10. Which is always true for any events A and B?

A) P(A∩B) ≥ P(A)
B) P(A∩B) ≤ P(A)
C) P(A∩B) = P(A)
D) P(A∩B) > 1

Answer: B

11. Which relation is always valid?

A) P(A∩B) ≥ P(A)+P(B)−1
B) P(A∩B) > P(A)+P(B)
C) P(A∩B) > 1
D) P(A∩B) < 0

Answer: A

12. If A and B are independent, then:

A) P(A|B)=P(A)
B) P(A|B)=0 always
C) P(A∩B)=0 always
D) A and B cannot occur together

Answer: A

APPSC Exam Traps

Trap 1:
Mutually Exclusive ≠ Independent

Trap 2:
P(A∪B) = P(A)+P(B) only when A and B are mutually exclusive.

Trap 3:
P(A∩B)=P(A)P(B) only when A and B are independent.

Trap 4:
“At least one” → Complement method often fastest.

Trap 5:
Classical probability requires equally likely outcomes.

Trap 6:
Conditional probability changes the effective sample space.

Trap 7:
A∩B is always a subset of A and B. Therefore:
P(A∩B)≤P(A) and P(A∩B)≤P(B).

One-Minute Formula Table

Concept	Formula / Result
Classical Probability	P(A)=n(A)/n(S)
Probability Range	0≤P(A)≤1
Sure Event	P(S)=1
Impossible Event	P(∅)=0
Complement	P(A′)=1−P(A)
Addition Rule	P(A∪B)=P(A)+P(B)−P(A∩B)
Mutually Exclusive	P(A∩B)=0
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	1−P(None)
First Success on nth Trial	qn−1p
Intersection Upper Bound	P(A∩B)≤min[P(A),P(B)]
Intersection Lower Bound	P(A∩B)≥P(A)+P(B)−1

చివరి నిమిషం పునశ్చరణ

Probability → Likelihood of occurrence

Range → 0 to 1

Sample Space → All possible outcomes

Event → One or more outcomes

Classical Probability → Favourable / Total equally likely outcomes

Sure Event → 1

Impossible Event → 0

Complement → 1 − P(A)

OR → Addition Rule

AND → Multiplication Rule

Mutually Exclusive → P(A∩B)=0

Independent → P(A∩B)=P(A)P(B)

Conditional Probability → P(A∩B)/P(B)

At Least One → 1 − None

First Success on nth Trial → qⁿ⁻¹p

Two Coins → 4 outcomes
Three Coins → 8 outcomes
Two Dice → 36 outcomes
Deck of Cards → 52 cards

Golden Inequality:

P(A)+P(B)−1 ≤ P(A∩B) ≤ min[P(A),P(B)]

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