Probability 🎲

Chance, luck, uncertainty — in sab ko mathematically measure karna yahi hai Probability! Class 9 se aage — ab Axiomatic approach seekhenge.

🌧️ Weather Forecast aur Probability

News par sunते ho: "Kal baarish ki 70% probability hai." Yeh kya matlab hai? 100 aisi hi days mein 70 times baarish hogi? Ya kuch aur? Class 9 mein humne basic probability seekhi thi — classical definition. Lekin real life mein events equal likely nahi hote. Ek loaded dice mein 6 aane ki probability 1/6 nahi ho sakti! Isliye Class 11 mein Probability ka Axiomatic Approach sikhate hain — jo zyada rigorous aur universal hai.

1. Kuch Important Terms — Seedha Clear Karo

🎯 Random Experiment

Woh experiment jiska result pehle se predict nahi kar sakte — lekin saare possible results pata hain. Example: Coin tossna, dice phenkna, card draw karna.

📋 Sample Space (S)

Random experiment ke saare possible outcomes ka set. Coin: S={H,T}. Dice: S={1,2,3,4,5,6}. Two coins: S={HH,HT,TH,TT}.

🎪 Event (E)

Sample space ka koi bhi subset. "Even number aana" on dice = E={2,4,6} — yeh ek event hai. P(E) = n(E)/n(S) is classical approach mein.

⚡ Elementary Event

Sirf ek outcome wala event. Dice par E={3} — sirf ek elementary event. P(elementary event) = 1/n(S) jab equally likely hoon.

Do dice phenko — Sample Space S (36 outcomes)

Die 2 → 1 2 3 4 5 6 1 2 3 4 5 6 (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) (4,1)...(4,6) (5,1)(5,6) (6,1)(6,6) Total = 6×6 = 36 outcomes

2. Axiomatic Approach — Class 11 ka New Way

Axiomatic approach mein probability ko 3 axioms (rules) ke basis par define karte hain. Yeh sirf "favourable/total" se zyada powerful hai — non-equal likely events bhi handle kar sakta hai!

Axiom 1

Har event ki probability 0 aur 1 ke beech (inclusive) hoti hai.

0 ≤ P(E) ≤ 1

Probability kabhi negative ya 1 se badi nahi hoti!

Axiom 2

Puri sample space ki probability = 1 (certain event).

P(S) = 1

Kuch toh result aayega — guarantee!

Axiom 3

Mutually exclusive events ke liye probabilities add hoti hain.

P(A∪B) = P(A)+P(B)
jab A∩B = ∅

Ek saath nahi ho sakte → add kar do.

3. Types of Events

Event TypeMatlabExample
Sure EventHamesha hoga — P=1Dice par 1-6 ke beech koi number aayega
Impossible EventKabhi nahi hoga — P=0Dice par 7 aana
Complementary A'A nahi hogaA = even, A' = odd (dice)
Mutually ExclusiveEk saath nahi ho sakte — A∩B=∅Even number aur odd number (same toss)
ExhaustiveMilake pura S cover karte hain{Even} ∪ {Odd} = S
IndependentEk ka outcome doosre par asar nahi2 alag coins tossna

4. Important Probability Formulas

P(A') = 1 − P(A)   (Complement)
P(A∪B) = P(A) + P(B) − P(A∩B)   (Addition Theorem)
P(A∩B) = 0 jab A,B mutually exclusive

Addition Theorem wahi hai jo Sets wala formula tha: n(A∪B) = n(A)+n(B)−n(A∩B) — probability mein probability version!

P(A∪B) = P(A) + P(B) − P(A∩B) — visually

A only A B only B A∩B both S (Sample Space) A∪B = orange + red + overlap

✅ Dice phenkne par: P(even), P(multiple of 3), P(even OR multiple of 3) nikalo

S = {1,2,3,4,5,6}, n(S) = 6
A = Even = {2,4,6}, P(A) = 3/6 = 1/2
B = Multiple of 3 = {3,6}, P(B) = 2/6 = 1/3
A∩B = {6} (even AND multiple of 3), P(A∩B) = 1/6
P(A∪B) = 1/2 + 1/3 − 1/6 = 3/6 + 2/6 − 1/6 = 4/6 = 2/3

5. Classical Probability — Quick Problems

✅ 52 cards ki deck mein se ek card draw karo. P(king or heart) nikalo.

P(King) = 4/52. P(Heart) = 13/52.
P(King AND Heart) = 1/52 (King of Hearts)
P(King or Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13

✅ 3 coins toss karo. P(atleast 2 heads) nikalo.

S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}, n(S) = 8
Atleast 2H = {HHH,HHT,HTH,THH} — 4 outcomes
P(atleast 2H) = 4/8 = 1/2

6. Practice Questions

Q1. Ek bag mein 5 red, 3 blue, 2 green balls hain. Ek ball nikalo. P(red), P(not green) nikalo.

Solution

Total=10. P(red)=5/10=1/2. P(not green)=P(red or blue)=8/10=4/5

Q2. Do dice phenko. P(sum = 7) nikalo.

Solution

Sum=7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 outcomes. P=6/36=1/6

Q3. P(A)=0.4, P(B)=0.3, P(A∩B)=0.1. P(A∪B) nikalo.

Solution

P(A∪B)=0.4+0.3−0.1=0.6

Q4. P(A)=2/5. P(A') nikalo.

Solution

P(A')=1−2/5=3/5

Q5. 52 cards se ek draw karo. P(face card) nikalo. (Face cards = J,Q,K)

Solution

Face cards = 3×4=12. P=12/52=3/13

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