Class 11 Maths · Chapter 14
Short answer:
Class 11 Maths Chapter 14 Probability random experiments, sample space, events (simple, compound, mutually exclusive, exhaustive), aur axiomatic approach to probability sikhaata hai — jisme addition theorem P(A∪B) = P(A)+P(B)−P(A∩B) sabse important formula hai. Rationalised 2026-27 NCERT me is chapter me 2 exercises (14.1: 7 questions, 14.2: 21 questions) aur 1 Miscellaneous Exercise (10 questions) hain — total 38 questions, poore class 11 maths ncert solutions set ka sabse scoring chapter.
Chapter 14 Probability Class 11 Maths ka aakhri aur sabse practical chapter hai — jo real-life uncertainty (dice, coins, cards, lottery) ko maths ki language me measure karna sikhaata hai. Yeh chapter Class 9-10 ki basic empirical probability se aage badhkar axiomatic approach introduce karta hai — jahan probability ko set theory (Chapter 1 Sets) ke through define kiya jaata hai. Isiliye jo students class 11 maths chapter 1 sets ncert solutions achhe se samajhte hain, unhe events, union, intersection aur complement ke concepts yahan bahut aasan lagte hain.
Yeh page poore chapter ke class 11 maths ncert solutions — dono exercises (14.1, 14.2) aur Miscellaneous Exercise — step-by-step working ke saath cover karta hai, taaki board exam aur school test dono ke liye ready ho sako.
Chapter 14 Summary — 5 Minute Revision
Probability chapter set theory (Sets, Chapter 1) ki foundation pe khada hai — sample space S, events as subsets of S, aur unke union/intersection/complement operations. Axiomatic approach me har event ki probability 0 aur 1 ke beech hoti hai, poore sample space ki probability 1 hoti hai, aur mutually exclusive events ke liye probabilities simply add ho jaati hain. General addition theorem P(A∪B)=P(A)+P(B)−P(A∩B) is chapter ka core tool hai jo dice, coins, cards aur real-life selection problems (lottery, committee selection) solve karne me use hota hai. Exercise 14.1 events ki classification (mutually exclusive/simple/compound) pe focus karta hai, Exercise 14.2 numeric probability calculation aur addition theorem application pe, aur Miscellaneous Exercise combinatorics (permutations/combinations) ke saath probability ko jodta hai — jo isse Chapter 6 (Permutations and Combinations) se bhi connect karta hai.
In-Text Questions — Solutions
Exercise 14.1, Q1. A die is rolled. Let E be the event 'die shows 4' and F be the event 'die shows an even number'. Are E and F mutually exclusive?
S = {1,2,3,4,5,6}. E = {4}. F = {2,4,6}.
E ∩ F = {4} ≠ φ
Chunki E aur F ka common outcome (4) hai, ye ek saath ho sakte hain. Isliye E aur F mutually exclusive NAHI hain.
Exercise 14.1, Q2. A die is thrown. Describe the following events: (i) A: a number less than 7 (ii) B: a number greater than 7 (iii) C: a multiple of 3 (iv) D: a number less than 4 (v) E: an even number greater than 4 (vi) F: a number not less than 3. Also find A∪B, A∩B, B∪C, E∩F, D∩E, A−C, D−E, E∩F′, F′.
S = {1,2,3,4,5,6}
A = {1,2,3,4,5,6} = S (sab numbers 7 se kam hain)
B = φ (koi bhi number 7 se zyada nahi)
C = {3,6}
D = {1,2,3}
E = {6} (4 se bada aur even, sirf 6)
F = {3,4,5,6}
Ab combinations nikalte hain:
A∪B = S = {1,2,3,4,5,6}
A∩B = φ
B∪C = {3,6}
E∩F = {6}∩{3,4,5,6} = {6}
D∩E = {1,2,3}∩{6} = φ
A−C = {1,2,4,5}
D−E = {1,2,3} (D me E ka koi element nahi)
F′ = S−F = {1,2}, isliye E∩F′ = {6}∩{1,2} = φ
Exercise 14.1, Q3. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: A: the sum is greater than 8, B: 2 occurs on either die, C: the sum is at least 7 and a multiple of 3. Which pairs of these events are mutually exclusive?
Pair of dice → n(S) = 36 ordered outcomes (i,j).
A (sum > 8) = {(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}
B (2 on either die) = {(1,2),(2,1),(2,2),(2,3),(3,2),(2,4),(4,2),(2,5),(5,2),(2,6),(6,2)}
C (sum ≥7 aur multiple of 3, matlab sum = 9 ya 12) = {(3,6),(4,5),(5,4),(6,3),(6,6)}
Check karte hain:
A∩B: A ke sum >8 hain, lekin agar ek die pe 2 hai to max sum 2+6=8, >8 nahi ho sakta → A∩B = φ → mutually exclusive
A∩C: C ⊂ A (C ke sab sums 9,12 already >8 hain) → A∩C = C ≠ φ → mutually exclusive NAHI
B∩C: C ke kisi bhi pair me 2 nahi hai → B∩C = φ → mutually exclusive
Answer: (A,B) aur (B,C) mutually exclusive hain; (A,C) nahi.
Exercise 14.1, Q4. Three coins are tossed once. Let A denote the event 'three heads show', B denote 'two heads and one tail show', C denote 'three tails show', and D denote 'a head shows on the first coin'. Which events are (i) Mutually exclusive? (ii) Simple? (iii) Compound?
S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
A = {HHH}, B = {HHT,HTH,THH}, C = {TTT}, D = {HHH,HHT,HTH,HTT}
(i) Pairwise intersections check karte hain: A∩B=φ, A∩C=φ, B∩C=φ, C∩D=φ. Lekin A∩D=A≠φ aur B∩D={HHT,HTH}≠φ.
Mutually exclusive pairs: (A,B), (A,C), (B,C), (C,D)
(ii) Simple events (ek hi outcome): A = {HHH} aur C = {TTT}
(iii) Compound events (ek se zyada outcome): B aur D
Exercise 14.1, Q5. Three coins are tossed. Describe (i) two events which are mutually exclusive (ii) three events which are mutually exclusive and exhaustive (iii) two events which are not mutually exclusive (iv) two events which are mutually exclusive but not exhaustive (v) three events which are mutually exclusive but not exhaustive.
S = 8 outcomes of three coins.
(i) A: exactly 1 head = {HTT,THT,TTH}; B: exactly 2 heads = {HHT,HTH,THH}. A∩B=φ → mutually exclusive.
(ii) A: no head={TTT}; B: exactly 1 head; C: at least 2 heads={HHT,HTH,THH,HHH}. Pairwise disjoint AND A∪B∪C=S → mutually exclusive and exhaustive.
(iii) A: at least 1 head (sab except TTT); B: at least 1 tail (sab except HHH). A∩B ≠ φ (jaise HHT dono me hai) → NOT mutually exclusive.
(iv) A: exactly 1 head; B: exactly 2 heads. A∩B=φ (mutually exclusive) lekin A∪B ≠ S (HHH, TTT missing) → exhaustive nahi.
(v) A: no head={TTT}; B: exactly 1 head; C: exactly 2 heads. Pairwise disjoint (mutually exclusive) lekin union me HHH missing hai → exhaustive nahi.
Exercise 14.1, Q6. Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice ≤ 5. Describe the events (i) A′ (ii) not B (iii) A or B (iv) A and B (v) A but not C (vi) B or C (vii) B and C (viii) A∩B′∩C′.
n(S)=36. A = {i even, i=first die}=18 outcomes. B={i odd}=18 outcomes (B=A′ hi hai). C (sum≤5) = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)} = 10 outcomes.
(i) A′ = B (first die odd)
(ii) not B = B′ = A (first die even)
(iii) A or B = A∪B = S (poora sample space, kyunki first die ya to even ya odd hoga hi)
(iv) A and B = A∩B = φ (first die dono even aur odd ek saath nahi ho sakta)
(v) A but not C = A∩C′. A ke jo elements C me bhi hain: (2,1),(2,2),(2,3),(4,1) — sirf ye 4. Baaki A∩C′ = 18−4 = 14 outcomes.
(vi) B or C = B∪C
(vii) B and C = B∩C: C ke jo pairs first-die-odd hain: (1,1),(1,2),(1,3),(1,4),(3,1),(3,2) = 6 outcomes
(viii) A∩B′∩C′ = A∩A∩C′ (kyunki B′=A) = A∩C′ = wahi 14 outcomes jo (v) me nikale
Exercise 14.1, Q7. Two dice are thrown. The events A, B, C, D, E and F are: A = even number on first die, B = odd number on first die, C = sum ≤ 5, D = sum > 5, E = at least one die shows 4, F = no die shows 4. State true or false (with reason): (i) A and B are mutually exclusive. (ii) A and B are mutually exclusive and exhaustive. (iii) A = B′. (iv) A and C are mutually exclusive. (v) A and D are mutually exclusive. (vi) A′, B′, C, D are mutually exclusive and exhaustive events. (vii) A, C and D are mutually exclusive and exhaustive events.
Same setup jaise Q6 — A∩B=φ, A∩C={(2,1),(2,2),(2,3),(4,1)}, D=C′ (sum>5 hai complement of sum≤5).
(i) A∩B=φ → TRUE (first die ek saath even+odd nahi ho sakta)
(ii) A∩B=φ aur A∪B=S → TRUE
(iii) B′ = complement of odd = even = A → TRUE
(iv) A∩C = {(2,1),(2,2),(2,3),(4,1)} ≠ φ → FALSE
(v) D = C′; A∩D = A∩C′ = 14 outcomes ≠ φ (Q6(v) se) → FALSE
(vi) A′=B, B′=A. B∩A=φ theek hai, lekin A′∩C = B∩C = 6 outcomes ≠ φ (Q6(vii) se) → mutually exclusive nahi → FALSE
(vii) C∪D = S (theek hai, kyunki D=C′) lekin A∩C ≠ φ (part iv se) → A aur C mutually exclusive nahi → FALSE
Poore Class 11 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–Q31)
Exercise 14.2, Q1. Which of the following cannot be valid assignment of probabilities for outcomes of sample space S = {ω1, ω2, ω3, ω4, ω5, ω6, ω7}?Assignment ω₁ ω₂ ω₃ ω₄ ω₅ ω₆ ω₇ (a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6 (b) 1/7 1/7 1/7 1/7 1/7 1/7 1/7 (c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (d) -0.1 0.2 0.3 0.4 -0.2 0.1 0.3 (e) 1/14 2/14 3/14 4/14 5/14 6/14 15/14
↔ Table ko side me swipe karein
| Assignment | ω₁ | ω₂ | ω₃ | ω₄ | ω₅ | ω₆ | ω₇ |
|---|---|---|---|---|---|---|---|
| (a) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
| (b) | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 |
| (c) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
| (d) | -0.1 | 0.2 | 0.3 | 0.4 | -0.2 | 0.1 | 0.3 |
| (e) | 1/14 | 2/14 | 3/14 | 4/14 | 5/14 | 6/14 | 15/14 |
Ek valid probability assignment ke liye do conditions honi chahiye: (1) har P(ωi) ≥ 0, aur (2) sabka sum exactly 1 ho.
(a) Sum = 0.1+0.01+0.05+0.03+0.01+0.2+0.6 = 1.00, sab non-negative → VALID.
(b) Sum = 7×(1/7) = 1, sab non-negative → VALID.
(c) Sum = 0.1+0.2+0.3+0.4+0.5+0.6+0.7 = 2.8 ≠ 1 → NOT VALID.
(d) ω₁ = -0.1 aur ω₅ = -0.2 negative hain, probability kabhi negative nahi ho sakti → NOT VALID.
(e) ω₇ = 15/14 > 1 hai, koi bhi single probability 1 se zyada nahi ho sakti → NOT VALID.
Exercise 14.2, Q2. A coin is tossed twice, what is the probability that at least one tail occurs?
S = {HH, HT, TH, TT}, n(S) = 4, sab equally likely.
At least one tail = {HT, TH, TT}, n(E) = 3
P(E) = n(E)/n(S) = 3/4
Exercise 14.2, Q3. A die is thrown, find the probability of following events:
(i) A prime number will appear
(ii) A number greater than or equal to 3 will appear
(iii) A number less than or equal to one will appear
(iv) A number more than 6 will appear
(v) A number less than 6 will appear
S = {1,2,3,4,5,6}, n(S) = 6, har outcome equally likely.
(i) Prime numbers = {2,3,5} → P = 3/6 = 1/2
(ii) ≥3 = {3,4,5,6} → P = 4/6 = 2/3
(iii) ≤1 = {1} → P = 1/6
(iv) >6 = φ (koi outcome nahi) → P = 0
(v) <6 = {1,2,3,4,5} → P = 5/6
Exercise 14.2, Q4. A card is selected from a pack of 52 cards. (i) How many points are there in the sample space? (ii) Calculate the probability that the card is an ace of spades. (iii) Calculate the probability that the card is (a) an ace (b) black card.
(i) Sample space points = 52 (har card ek equally likely outcome hai)
(ii) Ace of spades sirf 1 card hai → P = 1/52
(iii)(a) Total 4 aces (♠♥♦♣) → P = 4/52 = 1/13
(iii)(b) Black cards (♠+♣) = 26 → P = 26/52 = 1/2
Exercise 14.2, Q5. A fair coin with 1 marked on one face and 6 on the other, and a fair die, are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.
Coin outcomes {1,6}, die outcomes {1,2,3,4,5,6} → total equally likely outcomes n(S) = 2×6 = 12.
(i) Sum = 3: sirf (coin=1, die=2) kaam karta hai [(coin=6, die=-3) possible nahi] → 1 outcome → P = 1/12
(ii) Sum = 12: sirf (coin=6, die=6) → 1 outcome → P = 1/12
Exercise 14.2, Q6. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Total members = 4+6 = 10, sab equally likely candidates hain committee member banne ke liye.
P(woman) = 6/10 = 3/5
Exercise 14.2, Q7. A fair coin is tossed four times, and a person wins Re 1 for each head and loses Rs 1.50 for each tail that turns up. From the sample space, calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
4 tosses → n(S) = 2⁴ = 16, sab equally likely. Agar h heads aayein to (4−h) tails, net amount = h(1) − (4−h)(1.5).
h=4 (1 way — HHHH): amount = ₹4, P = 1/16
h=3 (4 ways): amount = 3 − 1.5 = ₹1.50, P = 4/16 = 1/4
h=2 (6 ways): amount = 2 − 3 = −₹1 (loss ₹1), P = 6/16 = 3/8
h=1 (4 ways): amount = 1 − 4.5 = −₹3.50, P = 4/16 = 1/4
h=0 (1 way — TTTT): amount = −₹6, P = 1/16
5 different amounts possible: ₹4, ₹1.50, −₹1, −₹3.50, −₹6 with probabilities 1/16, 1/4, 3/8, 1/4, 1/16 respectively (sum = 1, check ✓).
Exercise 14.2, Q8. Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails.
Teen coins tossed hone se S ke 8 equally likely outcomes bante hain: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
(i) 3 heads {HHH} → P = 1/8
(ii) exactly 2 heads {HHT,HTH,THH} → P = 3/8
(iii) at least 2 heads (2 ya 3 heads) = 3+1 = 4 → P = 4/8 = 1/2
(iv) at most 2 heads = 8 − P(3 heads outcome) = 7 → P = 7/8
(v) no head {TTT} → P = 1/8
(vi) 3 tails {TTT} → P = 1/8
(vii) exactly two tails {HTT,THT,TTH} → P = 3/8
(viii) no tail {HHH} → P = 1/8
(ix) at most two tails = 8 − 1(TTT) = 7 → P = 7/8
Exercise 14.2, Q9. If 2/11 is the probability of an event A, what is the probability of the event 'not A'?
Complement rule seedha laga dete hain — P(not A) hamesha 1 minus P(A) hota hai.
P(not A) = P(A′) = 1 − P(A) = 1 − 2/11 = 9/11
Exercise 14.2, Q10. A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that letter is (i) a vowel (ii) a consonant.
'ASSASSINATION' me total 13 letters hain: A,S,S,A,S,S,I,N,A,T,I,O,N
Count: A=3, S=4, I=2, N=2, T=1, O=1 → total = 3+4+2+2+1+1 = 13 ✓
(i) Vowels (A, I, O) = 3+2+1 = 6 → P(vowel) = 6/13
(ii) Consonants (S, N, T) = 4+2+1 = 7 → P(consonant) = 7/13
Exercise 14.2, Q11. In a lottery, a person chooses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? (Order of numbers is not important.)
Order matter nahi karta, isliye combinations use karenge.
Total ways to choose 6 numbers from 20 = C(20,6) = 38760
Winning way = sirf 1 (jo committee ne fix kiya)
P(winning) = 1/38760
Exercise 14.2, Q12. Check whether the following probabilities P(A) and P(B) are consistently defined:
(i) P(A) = 0.5, P(B) = 0.7, P(A∩B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A∪B) = 0.8
Rule: P(A∩B) hamesha min(P(A), P(B)) se kam ya barabar hona chahiye; aur addition theorem consistent hona chahiye.
(i) P(A∩B) = 0.6 > P(A) = 0.5 — yeh possible nahi kyunki intersection, A se bada nahi ho sakta → NOT consistently defined.
(ii) P(A∪B) = P(A)+P(B)−P(A∩B) → 0.8 = 0.5+0.4−P(A∩B) → P(A∩B) = 0.1
0.1, 0 aur min(0.5,0.4)=0.4 ke beech hai → consistently defined.
Exercise 14.2, Q13. Fill in the blanks in the following table:P(A) P(B) P(A∩B) P(A∪B) (i) 1/3 1/5 1/15 ... (ii) 0.35 ... 0.25 0.6 (iii) 0.5 0.35 ... 0.7
↔ Table ko side me swipe karein
| P(A) | P(B) | P(A∩B) | P(A∪B) | |
|---|---|---|---|---|
| (i) | 1/3 | 1/5 | 1/15 | ... |
| (ii) | 0.35 | ... | 0.25 | 0.6 |
| (iii) | 0.5 | 0.35 | ... | 0.7 |
Har blank ke liye wahi ek formula use hoga:
Formula: P(A∪B) = P(A) + P(B) − P(A∩B)
(i) P(A∪B) = 1/3 + 1/5 − 1/15 = 5/15+3/15−1/15 = 7/15
(ii) 0.6 = 0.35 + P(B) − 0.25 → P(B) = 0.6−0.35+0.25 = 0.5
(iii) 0.7 = 0.5 + 0.35 − P(A∩B) → P(A∩B) = 0.85−0.7 = 0.15
Exercise 14.2, Q14. Given P(A) = 3/5 and P(B) = 1/5. Find P(A or B), if A and B are mutually exclusive events.
Mutually exclusive hone se P(A∩B) = 0.
P(A or B) = P(A) + P(B) = 3/5 + 1/5 = 4/5
Exercise 14.2, Q15. If E and F are events such that P(E) = 1/4, P(F) = 1/2 and P(E and F) = 1/8, find (i) P(E or F) (ii) P(not E and not F).
Addition theorem aur complement rule dono yahan use honge.
(i) P(E or F) = P(E)+P(F)−P(E∩F) = 1/4+1/2−1/8 = 2/8+4/8−1/8 = 5/8
(ii) P(not E and not F) = P(E′∩F′) = 1 − P(E∪F) = 1 − 5/8 = 3/8
Exercise 14.2, Q16. Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.
De Morgan's law se: E′∪F′ = (E∩F)′
P((E∩F)′) = 0.25 → P(E∩F) = 1 − 0.25 = 0.75
Agar E, F mutually exclusive hote to P(E∩F) = 0 hona chahiye tha, but yahan P(E∩F) = 0.75 ≠ 0.
Isliye E aur F mutually exclusive NAHI hain.
Exercise 14.2, Q17. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A) (ii) P(not B) (iii) P(A or B).
Teeno parts me seedha complement aur addition theorem laga dete hain.
(i) P(A′) = 1 − 0.42 = 0.58
(ii) P(B′) = 1 − 0.48 = 0.52
(iii) P(A∪B) = 0.42 + 0.48 − 0.16 = 0.74
Exercise 14.2, Q18. In Class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Given values seedha addition theorem me daal dete hain.
P(M) = 0.40, P(B) = 0.30, P(M∩B) = 0.10
P(M∪B) = P(M)+P(B)−P(M∩B) = 0.40+0.30−0.10 = 0.60
Exercise 14.2, Q19. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?
Yahan addition theorem ko ulta ghumaakar P(A∩B) nikalna hai.
P(A) = 0.8, P(B) = 0.7, P(A∪B) = 0.95
P(A∩B) = P(A)+P(B)−P(A∪B) = 0.8+0.7−0.95 = 0.55
Exercise 14.2, Q20. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?
Neither ka matlab hai dono me se koi bhi pass nahi — isse P(E∪H) milta hai, phir addition theorem se P(H) nikaalte hain.
P(E∩H) = 0.5, P(neither) = P((E∪H)′) = 0.1 → P(E∪H) = 1−0.1 = 0.9. P(E) = 0.75.
P(E∪H) = P(E)+P(H)−P(E∩H) → 0.9 = 0.75+P(H)−0.5
P(H) = 0.9−0.75+0.5 = 0.65
Exercise 14.2, Q21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS, and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) the student opted for NCC or NSS (ii) the student has opted neither NCC nor NSS (iii) the student has opted NSS but not NCC.
Pehle teeno probabilities nikaal lete hain, phir addition theorem se aage badhte hain.
P(NCC) = 30/60 = 1/2, P(NSS) = 32/60 = 8/15, P(both) = 24/60 = 2/5.
(i) P(NCC∪NSS) = 1/2+8/15−2/5 = 15/30+16/30−12/30 = 19/30
(ii) P(neither) = 1 − 19/30 = 11/30
(iii) P(NSS but not NCC) = P(NSS)−P(both) = 8/15−2/5 = 8/15−6/15 = 2/15
Miscellaneous Exercise, Q1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box. What is the probability that (i) all will be blue? (ii) at least one will be green?
Total marbles = 10+20+30 = 60. 5 marbles draw karna hai, order matter nahi karta → combinations use karenge. Total ways = C(60,5).
(i) All blue: C(20,5) ways / C(60,5) → P = C(20,5)/C(60,5)
(ii) At least one green = 1 − P(no green) = 1 − C(30,5)/C(60,5) [C(30,5) = red+blue = 30 marbles se 5 chunna] → P = 1 − C(30,5)/C(60,5)
Miscellaneous Exercise, Q2. 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?
Combinations ka seedha use hoga — favourable divided by total.
Total ways to draw 4 cards from 52 = C(52,4).
Favourable = 3 diamonds (out of 13) × 1 spade (out of 13) = C(13,3) × C(13,1)
P = [C(13,3) × C(13,1)] / C(52,4)
Miscellaneous Exercise, Q3. A die has two faces each with number '1', three faces each with number '2' and one face with number '3'. If die is rolled once, determine (i) P(2) (ii) P(1 or 3) (iii) P(not 3).
Har face ek equally likely outcome maan kar seedha count kar lete hain.
Total faces = 2+3+1 = 6.
(i) P(2) = 3/6 = 1/2
(ii) P(1 or 3) = (2+1)/6 = 3/6 = 1/2
(iii) P(not 3) = 1 − 1/6 = 5/6
Miscellaneous Exercise, Q4. In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket? (b) two tickets? (c) 10 tickets?
Teeno parts me same logic hai — sirf tickets ki sankhya badalti hai.
Total tickets = 10000, prize-winning tickets = 10, no-prize tickets = 9990.
(a) One ticket: P(no prize) = 9990/10000 = 999/1000
(b) Two tickets: P(no prize) = C(9990,2)/C(10000,2)
(c) Ten tickets: P(no prize) = C(9990,10)/C(10000,10)
Miscellaneous Exercise, Q5. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that (a) you both enter the same section? (b) you both enter different sections?
Total ways 100 students ko divide karne ka relevant hai, lekin sirf tum aur friend ke section pe focus karte hain.
P(dono 40-wale section me) = (40×39)/(100×99) = 1560/9900 = 26/165
P(dono 60-wale section me) = (60×59)/(100×99) = 3540/9900 = 59/165
(a) P(same section) = 26/165 + 59/165 = 85/165 = 17/33
(b) P(different section) = 1 − 17/33 = 16/33
Miscellaneous Exercise, Q6. Three letters are dictated to three persons and an envelope is addressed to each of them; the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
Total arrangements of 3 letters in 3 envelopes = 3! = 6.
Derangements (koi bhi letter apne sahi envelope me nahi) for n=3: D₃ = 2 [123→231, 312 hi valid derangements hain]
P(no letter correct) = D₃/3! = 2/6 = 1/3
P(at least one correct) = 1 − 1/3 = 2/3
Miscellaneous Exercise, Q7. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A∩B) = 0.35. Find (i) P(A∪B) (ii) P(A′∩B′) (iii) P(A∩B′) (iv) P(B∩A′).
Chaaron parts me alag-alag set-formula seedha laga dete hain.
(i) P(A∪B) = 0.54+0.69−0.35 = 0.88
(ii) P(A′∩B′) = 1 − P(A∪B) = 1−0.88 = 0.12
(iii) P(A∩B′) = P(A) − P(A∩B) = 0.54−0.35 = 0.19
(iv) P(B∩A′) = P(B) − P(A∩B) = 0.69−0.35 = 0.34
Miscellaneous Exercise, Q8. From the employees of a company, 5 persons are selected to represent them in the managing committee. Particulars of five persons are: Harish (M, 30), Rohan (M, 33), Sheetal (F, 46), Alis (F, 28), Salim (M, 41). A person is selected at random from this group to act as spokesperson. What is the probability that the spokesperson will be either male or over 35 years?
Pehle A aur B events define karte hain, phir addition theorem lagate hain.
Total persons n(S) = 5. A = male = {Harish, Rohan, Salim}, n(A)=3, P(A)=3/5.
B = over 35 years = {Sheetal(46), Salim(41)}, n(B)=2, P(B)=2/5.
A∩B = male AND over 35 = {Salim}, n(A∩B)=1, P(A∩B)=1/5.
P(A∪B) = P(A)+P(B)−P(A∩B) = 3/5+2/5−1/5 = 4/5
Miscellaneous Exercise, Q9. If four-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when (i) the digits are repeated? (ii) repetition of digits is not allowed?
(i) Repetition allowed: 4-digit numbers > 5000 banane ke liye thousands digit ∈ {5, 7} lena hoga — kyunki 5 ya usse bada digit thousands place par ho to number automatically > 5000 ho jaata hai.
Thousands = 5: units divisible-by-5 digit ∈{0,5} → 2 choices; hundreds & tens: 5×5=25 each free → 2×25 = 50
Thousands = 7: units ∈{0,5} → 2 choices; hundreds & tens: 5×5=25 → 2×25 = 50
Favourable = 50+50 = 100. Total 4-digit numbers (thousands digit ≠0, repetition allowed) = 4×5×5×5 = 500
P = 100/500 = 1/5
(ii) Repetition not allowed:
Thousands=5: units must be 0 (5 already used) → hundreds,tens from remaining 3 digits arrange 2 = 3×2=6 → total 6
Thousands=7, units=0: remaining {1,3,5} arrange 2 = 6; units=5: remaining {0,1,3} arrange 2 = 6 → total 12
Favourable = 6+12 = 18. Total numbers (no repetition, thousands≠0) = 4×4×3×2 = 96
P = 18/96 = 3/16
Miscellaneous Exercise, Q10. The number lock of a suitcase has 4 wheels, each labelled with ten digits from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?
Total possible sequences with no repeats = 10×9×8×7 = 5040.
Correct sequence sirf 1 hai → P = 1/5040
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Sample Space & Event | S = set of all possible outcomes of a random experiment; Event E ⊆ S |
| Classical Probability (equally likely outcomes) | P(E) = n(E) / n(S) |
| Range of Probability | 0 ≤ P(E) ≤ 1 |
| Sure & Impossible Event | P(S) = 1, P(φ) = 0 |
| Complement Rule | P(E′) = 1 − P(E) |
| Mutually Exclusive Events | P(A∩B) = 0 ⇒ P(A∪B) = P(A) + P(B) |
| General Addition Theorem | P(A∪B) = P(A) + P(B) − P(A∩B) |
| 'A but not B' | P(A∩B′) = P(A) − P(A∩B) |
| Neither A nor B | P(A′∩B′) = 1 − P(A∪B) |
| Not (A and B) — De Morgan | P(A′∪B′) = 1 − P(A∩B) |
| Exhaustive + Mutually Exclusive Set {E₁,…,Eₙ} | P(E₁) + P(E₂) + … + P(Eₙ) = 1 |
| Valid Probability Assignment Check | Har P(ωᵢ) ≥ 0 AND Σ P(ωᵢ) = 1 |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- 'Mutually exclusive' ko 'independent' samajh lena — yeh dono alag concepts hain; independence Class 12 me detail me aata hai, Class 11 me sirf mutually exclusive/exhaustive tak focus rakho.
- Addition theorem P(A∪B) = P(A)+P(B)−P(A∩B) me common part (A∩B) subtract karna bhool jaana — is se answer double-count ho jaata hai.
- P(E) = n(E)/n(S) formula har jagah laga dena, chahe outcomes equally likely na hon (jaise biased die ya asymmetric coin ke sawalon me).
- Sample space likhte waqt outcomes miss karna ya repeat karna — jaise do dice ke 36 outcomes me se kuch pairs chhod dena ya (i,j) aur (j,i) ko same maan lena.
- 'At least' aur 'at most' ko ulta samajh lena — 'at least 2 heads' ka matlab '2 ya usse zyada' hai, na ki '2 ya usse kam'.
- Valid probability assignment check karte waqt sirf sum = 1 dekh lena, individual values ka negative ya 1 se zyada hona check na karna — dono conditions zaroori hain.
Board-Style Important Questions
- A die is thrown once. What is the probability of getting a number greater than 4?
- If P(A) = 0.4, find P(not A).
- A card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is either a king or a queen.
- If A and B are events such that P(A) = 0.5, P(B) = 0.6 and P(A∩B) = 0.2, find P(A∪B) and P(A′∩B′).
- Two dice are thrown simultaneously. Find the probability of getting a sum greater than 9.
- Three coins are tossed once. Describe (a) two events which are mutually exclusive and exhaustive, and (b) two events which are mutually exclusive but not exhaustive, giving reasons for each.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 14 Probability me kitne exercises hain aur kitne questions hain?
Rationalised (2026-27) NCERT chapter me 2 exercises + 1 Miscellaneous Exercise hain: Exercise 14.1 me 7 questions (events — mutually exclusive, simple, compound), Exercise 14.2 me 21 questions (axiomatic probability, addition theorem), aur Miscellaneous Exercise me 10 questions. Yeh chapter class 11 maths ncert solutions ke poore set ka last aur sabse scoring chapter hai.
Probability chapter ke important formulas kahan milenge?
Is page ke 'formulas' section me poore chapter ka ek hi table hai — sample space se lekar addition theorem tak. Agar poore syllabus ka class 11 maths formulas pdf download chahiye (Sets se Probability tak), woh alag consolidated PDF resource se milega.
Class 11 Maths Probability important questions kaunse type ke bar-bar puchhe jaate hain?
Sabse common class 11 maths probability important questions hain: addition theorem based (P(A∪B) nikalna), mutually exclusive vs exhaustive events identify karna, card/dice/coin ke classical probability sums, aur 'consistency check' type questions (Q12, Q16 jaisa) jo concept-testing hote hain.
Kya Class 11 Probability chapter me Bayes' theorem ya conditional probability aata hai?
Nahi. Class 11 Chapter 14 sirf axiomatic approach tak limited hai — random experiment, sample space, events, aur addition theorem. Conditional probability, Bayes' theorem, aur random variables Class 12 Probability chapter me aate hain.
Poore Class 11 Maths syllabus ke NCERT solutions kahan milenge, sirf Probability nahi?
Yeh page sirf Chapter 14 cover karta hai. Poore syllabus ke class 11 maths ncert solutions ke liye Chapter 1 Sets, class 11 maths chapter 3 trigonometric functions ncert solutions, class 11 maths straight lines ncert solutions, class 11 maths limits and derivatives ncert solutions, aur class 11 maths chapter 8 sequences and series important questions jaise dedicated chapter-pages dekho.
2026-27 session ke liye Class 11 Maths syllabus aur NCERT book PDF kahan se download karein?
Sabse authentic source ncert.nic.in hai — wahan se class 11 maths syllabus 2026-27 pdf aur class 11 maths ncert book pdf download dono official milte hain. Extra practice ke liye class 11 maths exemplar solutions aur class 11 maths conic sections notes jaise resources bhi dekh sakte ho.
Class 11 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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