Class 12 Maths · Chapter 13
Short answer:
NCERT Class 12 Maths Chapter 13 — Probability me total 22 practice questions yahan cover hote hain (Exercise 13.1 se 13.5 tak, miscellaneous included in rationalised syllabus). Is chapter me conditional probability, multiplication theorem, independent events, Bayes' theorem, random variables ki probability distribution, mean-variance, aur Bernoulli trials + binomial distribution cover hota hai. Yeh Class 12 Maths ka final aur last chapter hai (13 of 13) — CBSE board exam me is chapter se Bayes' theorem aur binomial distribution ke word problems bohot common hain.
Probability chapter mein hum 'condition' laga kar probability nikalna seekhte hain — matlab kisi event ka pehle se ho jana, doosre event ki probability ko kaise badal deta hai. Class 11 me humne basic probability P(A) padha tha — is chapter me hum aage badhte hain: agar ek event B already ho chuka hai, toh event A ki probability kya hogi (conditional probability), do events independent hain ya nahi, aur sabse important — Bayes' theorem, jisse hum 'effect' dekh kar 'cause' ki probability nikalte hain. Last me random variables aur binomial distribution aata hai jo statistics aur real-life prediction problems (jaise coin tosses, defective items, disease testing) me use hota hai.
Chapter 13 Summary — 5 Minute Revision
1. Conditional Probability
Agar event B ho chuka hai (P(B) > 0), toh event A ki probability, B ke given hone par:
| Formula | P(A|B) = P(A∩B) / P(B), jahan P(B) ≠ 0 |
| Reading | "Probability of A given B" — B pehle se ho chuka hai, ab A ki chance kya hai |
↔ Table ko side me swipe karein
Isi tarah P(B|A) = P(A∩B) / P(A), jahan P(A) ≠ 0. Dhyan rahe P(A|B) aur P(B|A) alag hote hain — dono ka numerator same hai (P(A∩B)) par denominator different hai.
2. Properties of Conditional Probability
- 0 ≤ P(A|B) ≤ 1
- P(S|B) = P(B|B) = 1 (S = sample space)
- P((A∪C)|B) = P(A|B) + P(C|B) − P((A∩C)|B), agar A aur C mutually exclusive nahi hain
- P(A'|B) = 1 − P(A|B)
3. Multiplication Theorem on Probability
Conditional probability ke formula ko rearrange karke:
| Two events | P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B), jahan P(A) ≠ 0, P(B) ≠ 0 |
| Three events | P(A∩B∩C) = P(A)·P(B|A)·P(C|A∩B) |
↔ Table ko side me swipe karein
Ye "without replacement" wale problems me bohot kaam aata hai — jaise bina replace kiye 2 balls nikalna.
4. Independent Events
Do events A aur B independent kehlate hain agar ek ka hona doosre ki probability ko affect nahi karta:
| Test | P(A∩B) = P(A) · P(B) |
| Equivalent | P(A|B) = P(A) aur P(B|A) = P(B) |
↔ Table ko side me swipe karein
Independent ≠ Mutually Exclusive. Mutually exclusive events me P(A∩B) = 0 hota hai — agar dono ki probability non-zero hai, toh mutually exclusive events kabhi independent nahi ho sakte.
5. Bayes' Theorem
Bayes' theorem tab use karte hain jab humein 'effect' pata ho aur 'cause' ki probability nikalni ho — jaise diagnosis test positive aaya, ab actual disease hone ki probability kya hai. Agar E₁, E₂, ..., Eₙ sample space ki partition hain (mutually exclusive, exhaustive, sab ki probability > 0), aur A koi event hai:
| Formula | P(Eᵢ|A) = [P(Eᵢ)·P(A|Eᵢ)] / [Σ P(Eⱼ)·P(A|Eⱼ)] for j = 1 to n |
↔ Table ko side me swipe karein
Yahan P(Eᵢ) ko prior probability kehte hain (event A hone se pehle ka gyaan), aur P(Eᵢ|A) ko posterior probability kehte hain (A ho chuka hai, ab uske baad ka updated gyaan). Denominator me saare cases ka total probability P(A) aata hai — is step ko miss karna sabse badi galti hoti hai.
6. Random Variable aur Probability Distribution
Random variable X ek function hai jo sample space ke har outcome ko ek real number assign karta hai. Iski probability distribution table isi tarah likhi jaati hai:
| X | x₁ | x₂ | ... | xₙ |
|---|---|---|---|---|
| P(X) | p₁ | p₂ | ... | pₙ |
↔ Table ko side me swipe karein
Condition: pᵢ > 0 for all i, aur Σ pᵢ = 1 (sab probabilities ka sum hamesha 1 hona chahiye — agar nahi ho raha toh galti hai).
7. Mean aur Variance of a Random Variable
| Mean (Expectation) | E(X) = μ = Σ xᵢ·pᵢ |
| E(X²) | Σ xᵢ²·pᵢ |
| Variance | Var(X) = E(X²) − [E(X)]² |
| Standard Deviation | SD(X) = √Var(X) |
↔ Table ko side me swipe karein
8. Bernoulli Trials
Ek trial Bernoulli trial hai agar: (i) finite number of independent trials hain, (ii) har trial ka result sirf "success" ya "failure" ho sakta hai, (iii) success ki probability p har trial me constant rehti hai.
9. Binomial Distribution
Agar n independent Bernoulli trials hain jisme success ki probability p hai (aur failure ki q = 1−p), toh exactly r successes ki probability:
| Formula | P(X = r) = ⁿCᵣ · pʳ · qⁿ⁻ʳ, jahan r = 0, 1, 2, ..., n |
| Mean | E(X) = np |
| Variance | Var(X) = npq |
↔ Table ko side me swipe karein
Binomial distribution wale problems me sabse pehle n (total trials), p (success probability), q = 1−p, aur r (jo required outcome hai) identify karo — phir formula lagao.
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q22)
Q1. If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find P(A∩B).
Multiplication theorem se:
P(A∩B) = P(A) · P(B|A) = 0.8 × 0.4 = 0.32
Q2. If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find P(A|B).
Pehle P(A∩B) nikala tha = 0.32. Ab:
P(A|B) = P(A∩B) / P(B) = 0.32 / 0.5 = 0.64
Q3. A die is thrown. If E is the event 'number appearing is a multiple of 3' and F is the event 'number appearing is even', find P(E|F).
Sample space S = {1,2,3,4,5,6}, so n(S) = 6
E = {3, 6} → P(E) = 2/6
F = {2, 4, 6} → P(F) = 3/6
E∩F = {6} → P(E∩F) = 1/6
P(E|F) = P(E∩F)/P(F) = (1/6)/(3/6) = 1/3
Q4. A card is drawn from a well-shuffled deck of 52 cards. Find P(the card drawn is a king | the card drawn is a face card).
Face cards = 12 (4 kings, 4 queens, 4 jacks)
Let A = king, B = face card
P(A∩B) = P(king, which is also a face card) = 4/52
P(B) = 12/52
P(A|B) = (4/52)/(12/52) = 4/12 = 1/3
Q5. Two coins are tossed once. Find P(A|B) if A: 'no tail appears', B: 'no head appears'.
Sample space S = {HH, HT, TH, TT}
A = {HH} → P(A) = 1/4
B = {TT} → P(B) = 1/4
A∩B = {} (empty set) → P(A∩B) = 0
P(A|B) = P(A∩B)/P(B) = 0/(1/4) = 0
Q6. A fair die is rolled. Consider events E = {1,3,5}, F = {2,3}, G = {2,3,4,5}. Find P(E|F) and P(F|E).
P(E) = 3/6, P(F) = 2/6, P(E∩F) = P({3}) = 1/6
P(E|F) = P(E∩F)/P(F) = (1/6)/(2/6) = 1/2
P(F|E) = P(E∩F)/P(E) = (1/6)/(3/6) = 1/3
Q7. A black and a red die are rolled together. Find the conditional probability of getting a sum greater than 9, given that the black die resulted in a 5.
Let A = black die shows 5, B = sum > 9
A = {(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)} → P(A) = 6/36
A∩B = sum > 9 AND black = 5 → (5,5), (5,6) → P(A∩B) = 2/36
P(B|A) = (2/36)/(6/36) = 2/6 = 1/3
Q8. A pair of dice is thrown. Find P(exactly one die shows 4 | sum shown is less than 5) — i.e. E: at least one die is 4, F: sum of numbers is less than 5.
F (sum < 5) = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} → P(F) = 6/36
E∩F = outcomes with a 4 AND sum < 5 = none → P(E∩F) = 0
P(E|F) = 0/(6/36) = 0
Q9. Given that the two numbers appearing on throwing two dice are different, find P(the sum of numbers is 4).
Let A = sum is 4, B = numbers are different
A = {(1,3),(3,1),(2,2)} but (2,2) has same numbers so excluded when intersecting with B
A∩B = {(1,3),(3,1)} → P(A∩B) = 2/36
Total outcomes with different numbers: 36 − 6 = 30 → P(B) = 30/36
P(A|B) = (2/36)/(30/36) = 2/30 = 1/15
Q10. A die is thrown three times. Events A: 4 appears on the third throw, B: 6 and 5 appear on first two throws. Check whether A and B are independent.
P(A) = 1/6 (probability of 4 on 3rd throw, independent of other throws)
P(B) = P(6 on 1st) × P(5 on 2nd) = 1/6 × 1/6 = 1/36
P(A∩B) = P(6,5,4) = 1/6 × 1/6 × 1/6 = 1/216
Check: P(A)·P(B) = (1/6)(1/36) = 1/216 = P(A∩B)
Since P(A∩B) = P(A)·P(B), events A and B are independent.
Q11. A fair coin and an unbiased die are tossed together. Let A = 'head appears on the coin' and B = 'a 3 appears on the die'. Check independence.
P(A) = 1/2, P(B) = 1/6
P(A∩B) = P(head, 3) = 1/2 × 1/6 = 1/12 (since coin and die are physically independent)
P(A)·P(B) = (1/2)(1/6) = 1/12 = P(A∩B)
A and B are independent events.
Q12. Probability of solving a specific problem independently by A and B are 1/2 and 1/3 respectively. If both try independently, find the probability that the problem is solved.
P(A) = 1/2, P(B) = 1/3, A and B independent
P(problem solved) = P(A∪B) = P(A) + P(B) − P(A)P(B)
= 1/2 + 1/3 − (1/2)(1/3) = 1/2 + 1/3 − 1/6
= 3/6 + 2/6 − 1/6 = 4/6 = 2/3
Q13. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn. Find the probability that the ball drawn is red.
Let E₁ = bag 1 selected, E₂ = bag 2 selected, A = red ball drawn
P(E₁) = P(E₂) = 1/2
P(A|E₁) = 4/8 = 1/2, P(A|E₂) = 2/8 = 1/4
By theorem of total probability:
P(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) = (1/2)(1/2) + (1/2)(1/4)
= 1/4 + 1/8 = 2/8 + 1/8 = 3/8
Q14. Two urns: Urn I has 3 white and 2 red balls, Urn II has 2 white and 3 red balls. An urn is chosen at random and a ball drawn is found to be white. Find the probability it was drawn from Urn I (Bayes' theorem).
Let E₁ = Urn I chosen, E₂ = Urn II chosen, A = white ball drawn
P(E₁) = P(E₂) = 1/2
P(A|E₁) = 3/5, P(A|E₂) = 2/5
By Bayes' theorem:
P(E₁|A) = [P(E₁)·P(A|E₁)] / [P(E₁)P(A|E₁) + P(E₂)P(A|E₂)]
= [(1/2)(3/5)] / [(1/2)(3/5) + (1/2)(2/5)]
= (3/10) / (3/10 + 2/10) = (3/10)/(5/10) = 3/5
Q15. Bag I contains 3 red, 4 black balls; Bag II contains 5 red, 6 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. It turns out to be red. Find the probability that the transferred ball was red.
Let E₁ = red ball transferred, E₂ = black ball transferred, A = ball drawn from Bag II is red
P(E₁) = 3/7, P(E₂) = 4/7
After transfer, Bag II has 12 balls:
If red transferred: Bag II has 6 red, 6 black → P(A|E₁) = 6/12 = 1/2
If black transferred: Bag II has 5 red, 7 black → P(A|E₂) = 5/12
By Bayes' theorem:
P(E₁|A) = [(3/7)(1/2)] / [(3/7)(1/2) + (4/7)(5/12)]
= (3/14) / (3/14 + 20/84) = (18/84) / (18/84 + 20/84) = 18/38 = 9/19
Q16. A factory has three machines A, B, C producing 25%, 35%, 40% of total output. Their defective rates are 5%, 4%, 2% respectively. A bulb is drawn at random and found defective. Find the probability it was made by machine A.
Let E₁, E₂, E₃ = machine A, B, C produced the item; D = item is defective
P(E₁) = 0.25, P(E₂) = 0.35, P(E₃) = 0.40
P(D|E₁) = 0.05, P(D|E₂) = 0.04, P(D|E₃) = 0.02
Total probability of defective:
P(D) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02)
= 0.0125 + 0.014 + 0.008 = 0.0345
By Bayes' theorem:
P(E₁|D) = (0.0125)/(0.0345) = 125/345 = 25/69 ≈ 0.362
Q17. A test for a disease correctly detects the disease 99% of the time (true positive) and gives a false positive 0.5% of the time. 0.1% of the population has the disease. If a person tests positive, find the probability they actually have the disease (classic medical Bayes' problem).
Let D = has disease, D' = does not have disease, T = tests positive
P(D) = 0.001, P(D') = 0.999
P(T|D) = 0.99 (true positive rate)
P(T|D') = 0.005 (false positive rate)
Total probability of positive test:
P(T) = P(D)P(T|D) + P(D')P(T|D') = (0.001)(0.99) + (0.999)(0.005)
= 0.00099 + 0.004995 = 0.005985
By Bayes' theorem:
P(D|T) = 0.00099 / 0.005985 ≈ 0.1654 ≈ 16.5%
Yahi Bayes' theorem ka sabse important real-life takeaway hai — bahut accurate test hone ke bawajood, agar disease rare hai toh positive test ka matlab actual disease hona sirf ~16.5% hi hai, kyunki false positives ki sankhya (rare disease ke case me) bohot zyada dominate karti hai.
Q18. Two cards are drawn without replacement from a pack of 52 cards. Find the probability distribution of the number of aces.
Let X = number of aces drawn, X can be 0, 1, or 2
P(X=0) = (48C2)/(52C2) = (48×47/2)/(52×51/2) = 1128/1326 = 188/221
P(X=1) = (4C1 × 48C1)/(52C2) = (4×48)/1326 = 192/1326 = 32/221
P(X=2) = (4C2)/(52C2) = 6/1326 = 1/221
Check: 188/221 + 32/221 + 1/221 = 221/221 = 1 ✓
| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | 188/221 | 32/221 | 1/221 |
↔ Table ko side me swipe karein
Q19. Find the mean and variance of the number of tails in the simultaneous tosses of three fair coins.
Let X = number of tails, X can be 0, 1, 2, 3. Each with p = 1/2 for tail, q = 1/2 for head, n = 3 (binomial with p=1/2)
P(X=0) = ³C₀(1/2)⁰(1/2)³ = 1/8
P(X=1) = ³C₁(1/2)¹(1/2)² = 3/8
P(X=2) = ³C₂(1/2)²(1/2)¹ = 3/8
P(X=3) = ³C₃(1/2)³(1/2)⁰ = 1/8
Check: 1/8 + 3/8 + 3/8 + 1/8 = 8/8 = 1 ✓
Mean E(X) = np = 3 × 1/2 = 3/2
Variance Var(X) = npq = 3 × 1/2 × 1/2 = 3/4
Q20. If a fair coin is tossed 10 times, find the probability of getting exactly 6 heads.
This is a binomial distribution problem: n = 10, p = 1/2 (head), q = 1/2, r = 6
P(X=6) = ¹⁰C₆ (1/2)⁶ (1/2)⁴ = ¹⁰C₆ (1/2)¹⁰
¹⁰C₆ = ¹⁰C₄ = (10×9×8×7)/(4×3×2×1) = 210
P(X=6) = 210 × (1/1024) = 210/1024 = 105/512
Q21. The probability of a bulb being defective is 0.1. Find the probability that out of a sample of 5 bulbs, none is defective (binomial distribution).
n = 5, p = 0.1 (defective), q = 0.9, r = 0 (none defective)
P(X=0) = ⁵C₀ (0.1)⁰ (0.9)⁵ = 1 × 1 × (0.9)⁵
(0.9)⁵ = 0.59049
P(none defective) ≈ 0.59049
Q22. On a multiple-choice exam with 4 possible answers for each of 5 questions, find the probability that a student guessing at random gets exactly 4 or more correct answers (binomial distribution).
n = 5, p = 1/4 (correct guess), q = 3/4
P(X≥4) = P(X=4) + P(X=5)
P(X=4) = ⁵C₄ (1/4)⁴ (3/4)¹ = 5 × (1/256) × (3/4) = 15/1024
P(X=5) = ⁵C₅ (1/4)⁵ (3/4)⁰ = 1 × (1/1024) = 1/1024
P(X≥4) = 15/1024 + 1/1024 = 16/1024 = 1/64
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Conditional probability | P(A|B) = P(A∩B) / P(B), P(B) ≠ 0 |
| Multiplication theorem (2 events) | P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B) |
| Multiplication theorem (3 events) | P(A∩B∩C) = P(A)·P(B|A)·P(C|A∩B) |
| Independence test | A, B independent ⇔ P(A∩B) = P(A)·P(B) |
| Theorem of total probability | P(A) = Σ P(Eᵢ)·P(A|Eᵢ), Eᵢ partition of sample space |
| Bayes' theorem | P(Eᵢ|A) = [P(Eᵢ)·P(A|Eᵢ)] / [Σ P(Eⱼ)·P(A|Eⱼ)] |
| Probability distribution condition | Σ pᵢ = 1, pᵢ > 0 |
| Mean / Expectation | E(X) = Σ xᵢ·pᵢ |
| E(X²) | Σ xᵢ²·pᵢ |
| Variance | Var(X) = E(X²) − [E(X)]² |
| Standard deviation | SD(X) = √Var(X) |
| Binomial probability | P(X=r) = ⁿCᵣ·pʳ·qⁿ⁻ʳ, q = 1−p, r = 0,1,...,n |
| Binomial mean | E(X) = np |
| Binomial variance | Var(X) = npq |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- P(A|B) ko P(B|A) samajh lena. Dono ka numerator P(A∩B) same hota hai lekin denominator different hai — P(A|B) me denominator P(B) hota hai, P(B|A) me P(A). Question dhyan se padho ki 'given' kaun sa event hai.
- Bayes' theorem me denominator bhool jaana. Sabse common galti — sirf numerator P(Eᵢ)·P(A|Eᵢ) likh kar answer de dena, poore total probability (sab cases ka sum) se divide karna bhool jaana. Denominator hamesha saare Eⱼ cases ka sum hona chahiye.
- Independence check karte waqt wrong intersection nikalna. P(A∩B) = P(A)·P(B) check karne se pehle A∩B ka actual sample space se sahi calculation karo, sirf formula assume mat karo ki independent honge.
- Probability distribution table ka sum 1 na hona check na karna. Har probability distribution problem ke end me Σpᵢ = 1 verify zaroor karo — agar nahi ho raha toh calculation me galti hai.
- Binomial distribution me n, p, q, r ko galat identify karna. n = total trials, p = 'success' ki probability (jo question me define hui hai), q = 1−p, r = required successes. 'Defective' ya 'failure' ko galti se p bana dena common mistake hai.
- Independent events aur mutually exclusive events ko confuse karna. Agar dono events ki probability non-zero hai, toh wo kabhi bhi mutually exclusive AND independent dono nahi ho sakte — ye do alag concepts hain jo overlap nahi karte.
Board-Style Important Questions
- 1 mark: If P(A) = 0.6, P(B) = 0.3 and A, B are independent events, find P(A∩B).
- 2 marks: A die is thrown twice. Find the probability that the sum of the numbers is even, given that the first throw shows an even number.
- 3 marks: Find the probability distribution of the number of heads when a fair coin is tossed 3 times, and find its mean.
- 4 marks: A bag contains 5 red and 3 black balls. Three balls are drawn one by one without replacement. Find the probability distribution of the number of red balls drawn.
- 5 marks: There are two bags: Bag I has 4 white and 4 red balls, Bag II has 2 white and 6 red balls. One bag is selected at random and a ball is drawn from it, which turns out to be red. Find the probability that the ball was drawn from Bag II (Bayes' theorem).
- 5 marks: In a group of students, the probability that a randomly chosen student passes a test is 0.7. If 5 students are chosen at random, find the probability that exactly 4 of them pass (binomial distribution).
Aksar Poochhe Jaane Wale Sawaal
Bayes' theorem kab use karte hain?
Jab hume kisi 'effect' ya outcome ka pata ho (jaise test positive aaya) aur humein uske 'cause' ki probability nikalni ho (jaise actual disease hone ki probability). Yeh hamesha reverse conditional probability hoti hai.
Conditional probability aur multiplication theorem me kya farak hai?
Conditional probability formula P(A|B) = P(A∩B)/P(B) hume P(A∩B) nikalne pe use hota hai jab P(A) aur P(B) pehle se pata ho — ismein hum formula ko rearrange karke P(A∩B) = P(A)·P(B|A) nikalte hain, jo multiplication theorem kehlata hai.
Independent events aur mutually exclusive events same hain kya?
Bilkul nahi. Independent events ka matlab hai ek event ka hona doosre ko affect nahi karta. Mutually exclusive ka matlab hai dono events ek saath ho hi nahi sakte (P(A∩B)=0). Agar dono events ki probability non-zero hai, toh wo kabhi independent nahi ho sakte agar mutually exclusive hain.
Probability distribution table me kya check karna zaroori hai?
Har table ke end me saari probabilities ka sum (Σpᵢ) exactly 1 hona chahiye. Agar 1 nahi aa raha, iska matlab calculation me galti hai — ye har question me verify karna chahiye.
Binomial distribution kab apply hoti hai?
Jab trials fixed number (n) me ho, har trial independent ho, aur har trial ka result sirf do outcomes me se ek ho (success/failure) jisme success ki probability p constant rahe har trial me — jaise coin toss, defective item check, ya MCQ guessing.
Mean aur variance nikalne ka shortcut kya hai binomial distribution me?
Agar problem binomial hai (n, p, q pata hai), toh poori probability distribution table banane ki zaroorat nahi — directly E(X) = np aur Var(X) = npq formula use kar sakte ho, jo bohot time bachata hai exam me.
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