Class 10 Maths · Chapter 14
Short answer:
Chapter 14 Probability me sirf ek exercise hai — Exercise 14.1, jisme 25 questions hain. Ye poora chapter classical (theoretical) probability pe based hai — P(E) = number of favourable outcomes / number of all possible equally likely outcomes. Coverage: coin tossing, single die aur do dice ek saath (36 outcomes), playing cards ka poora deck (52 cards, 4 suits), bag/box me marbles ya balls, aur real-life scenarios jaise defective bulbs, piggy bank coins, aur do dice ka combined sum table. Is chapter me koi calculus ya heavy formula nahi hai — sirf outcomes ko sahi tarah count karna aata hai.
Probability ka formula seedha hai — favourable outcomes divide by total outcomes. Mushkil part sirf outcomes ko sahi se count karna hota hai — khaas kar cards aur do dice waale questions me, jahan ek galat count poora answer ulta kar deta hai. Is chapter me Class 9 waali experimental probability (baar baar trial karke) chhod ke ab hum theoretical ya classical probability pe aa gaye hain, jisme outcomes 'equally likely' maane jaate hain — jaise coin ka head/tail, ya die ka 1-6.
Chapter 14 Summary — 5 Minute Revision
1. Probability kya hai
Kisi event E ke liye:
| Term | Matlab |
|---|---|
| Trial | Experiment jo perform kiya jaata hai — coin toss, die throw, card draw |
| Outcome | Trial ka ek possible result — jaise die pe 4 aana |
| Equally likely outcomes | Har outcome ke aane ke chances barabar hain (fair coin, unbiased die) |
| Event (E) | Ek ya zyada outcomes ka set jisme hum interested hain |
| Sure event | Jo hamesha hota hai — P = 1 |
| Impossible event | Jo kabhi nahi hota — P = 0 |
↔ Table ko side me swipe karein
Classical formula:
| P(E) = (number of outcomes favourable to E) / (number of all possible outcomes) |
↔ Table ko side me swipe karein
2. Coin tossing
- 1 coin: 2 outcomes — H, T
- 2 coins together (ya 1 coin 2 baar): 4 outcomes — HH, HT, TH, TT
- 3 coins together (ya 1 coin 3 baar): 8 outcomes — HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
3. Dice — single aur double
- Ek die: 6 outcomes — 1, 2, 3, 4, 5, 6
- Do dice ek saath (ya ek die 2 baar) = 36 outcomes — (1,1) se (6,6) tak, 6×6 grid. Ye sabse common mistake ki jagah hai — students isse 12 outcomes maan lete hain, jo galat hai.
4. Playing cards — poora deck yaad rakho
| Group | Count | Detail |
|---|---|---|
| Total cards | 52 | 4 suits × 13 cards |
| Suits | 4 | Hearts, Diamonds (red), Spades, Clubs (black) |
| Cards per suit | 13 | Ace, 2–10, Jack, Queen, King |
| Red cards | 26 | Hearts + Diamonds |
| Black cards | 26 | Spades + Clubs |
| Face cards | 12 | Jack, Queen, King × 4 suits (Ace face card NAHI hai!) |
| Red face cards | 6 | 3 face cards × 2 red suits |
| Aces | 4 | 1 per suit |
| Kings/Queens/Jacks | 4 each | 1 per suit |
↔ Table ko side me swipe karein
5. Complementary events
Har event E ke liye ek "not E" event hota hai jo E ke alawa sab kuch cover karta hai:
| P(E) + P(not E) = 1 → P(not E) = 1 − P(E) |
↔ Table ko side me swipe karein
Ye shortcut bahut kaam aata hai jab "at least one" ya "not" waala event poochha jaaye.
6. Bag/box problems ka approach
- Total items count karo (sab colours/types add karo)
- Favourable items count karo (jo event me chahiye)
- P(E) = favourable / total likho, fir simplify karo
Poore Class 10 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–Q25)
Q1. Complete the following statements: (i) Probability of an event E + Probability of the event 'not E' = ___. (ii) The probability of an event that cannot happen is ___. Such an event is called ___. (iii) The probability of an event that is certain to happen is ___. Such an event is called ___. (iv) The sum of the probabilities of all the elementary events of an experiment is ___. (v) The probability of an event is greater than or equal to ___ and less than or equal to ___.
(i) 1 (ii) 0 — impossible event (iii) 1 — sure (certain) event (iv) 1 (v) 0 aur 1
P(E) + P(not E) = 1 hamesha
Q2. Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car — it starts or does not start. (ii) A player attempts to shoot a basketball — she/he shoots or misses. (iii) A true-false question — answer right or wrong. (iv) A baby is born — boy or girl.
(i) NOT equally likely — car ka start hona bahut factors (battery, fuel, engine condition) pe depend karta hai, dono outcomes ki chances barabar nahi.
(ii) NOT equally likely — shot lagana player ki skill pe depend karta hai.
(iii) Equally likely — random guess me right ya wrong dono ke chances barabar hain.
(iv) Equally likely — boy ya girl hone ke chances (ideal case me) barabar maane jaate hain.
Q3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Kyunki coin toss ke sirf 2 possible outcomes hote hain — head ya tail — aur dono ke aane ke chances exactly barabar (equally likely) hote hain. Isliye kisi bhi team ko unfair advantage nahi milta.
P(Head) = P(Tail) = 1/2
Q4. Which of the following cannot be the probability of an event? (A) 2/3 (B) −1.5 (C) 15% (D) 0.7
(B) −1.5 nahi ho sakta.
Probability hamesha 0 ≤ P(E) ≤ 1 ke beech hoti hai. Negative value kabhi possible nahi.
Q5. If P(E) = 0.05, what is the probability of 'not E'?
P(not E) = 0.95
P(not E) = 1 − P(E) = 1 − 0.05 = 0.95
Q6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy?
(i) P(orange) = 0 — impossible event, kyunki bag me orange candy hai hi nahi.
(ii) P(lemon) = 1 — sure event, kyunki bag me sirf lemon candies hain.
Q7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
P(same birthday) = 0.008
P(same) = 1 − P(not same) = 1 − 0.992 = 0.008
Q8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
(i) P(red) = 3/8
(ii) P(not red) = 5/8
Total balls = 3 + 5 = 8. P(red) = 3/8. P(not red) = 1 − 3/8 = 5/8
Q9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
(i) P(red) = 5/17
(ii) P(white) = 8/17
(iii) P(not green) = 13/17
Total = 5 + 8 + 4 = 17. P(not green) = 1 − 4/17 = 13/17
Q10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out, what is the probability that the coin (i) will be a 50p coin? (ii) will not be a ₹5 coin?
(i) P(50p) = 5/9
(ii) P(not ₹5) = 17/18
Total coins = 100 + 50 + 20 + 10 = 180. P(50p) = 100/180 = 5/9. P(not ₹5) = 1 − 10/180 = 170/180 = 17/18
Q11. Gopi buys a fish for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
P(male fish) = 5/13
Total fish = 5 + 8 = 13. P(male) = 5/13
Q12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,8, and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
(i) P(8) = 1/8
(ii) P(odd) = 1/2
(iii) P(>2) = 3/4
(iv) P(<9) = 1
Total = 8. Odd numbers {1,3,5,7} = 4/8 = 1/2. Numbers >2: {3,4,5,6,7,8} = 6/8 = 3/4. All numbers are <9, so P = 8/8 = 1
Q13. A die is thrown once. Find the probability of getting (i) a prime number (ii) a number lying between 2 and 6 (iii) an odd number.
(i) P(prime) = 1/2
(ii) P(between 2 and 6) = 1/2
(iii) P(odd) = 1/2
Total = 6. Prime {2,3,5} = 3/6 = 1/2. Between 2 and 6 (exclusive) {3,4,5} = 3/6 = 1/2. Odd {1,3,5} = 3/6 = 1/2
Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds.
(i) P = 1/26
(ii) P = 3/13
(iii) P = 3/26
(iv) P = 1/52
(v) P = 1/4
(vi) P = 1/52
Total = 52. Red kings = 2 → 2/52 = 1/26. Face cards = 12 (J,Q,K × 4) → 12/52 = 3/13. Red face cards = 6 → 6/52 = 3/26. Jack of hearts = 1 → 1/52. Spades = 13 → 13/52 = 1/4. Queen of diamonds = 1 → 1/52
Q15. Five cards — the ten, jack, queen, king and ace of diamonds are well shuffled with their face downwards. One card is picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
(i) P(queen) = 1/5
(ii)(a) P(ace) = 1/4
(ii)(b) P(queen) = 0
Total = 5 cards, so P(queen) = 1/5. Queen removed, 4 cards left → P(ace) = 1/4. Queen already removed, so P(queen again) = 0/4 = 0
Q16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is good.
P(good pen) = 11/12
Total pens = 132 + 12 = 144. P(good) = 132/144 = 11/12
Q17. (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
(i) P(defective) = 1/5
(ii) P(not defective) = 15/19
(i) Total = 20, defective = 4 → 4/20 = 1/5. (ii) After removing one good bulb: total left = 19, good bulbs left = 132... yaha 16-4=... total good originally 16, ek nikal gaya to 15 bache. P = 15/19
Q18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
(i) P(two-digit) = 9/10
(ii) P(perfect square) = 1/10
(iii) P(divisible by 5) = 1/5
Total = 90. Two-digit numbers 10–90 = 81 → 81/90 = 9/10. Perfect squares 1–90: 1,4,9,16,25,36,49,64,81 = 9 → 9/90 = 1/10. Divisible by 5: 5,10,...,90 = 18 → 18/90 = 1/5
Q19. A child has a die whose six faces show the letters as given below: A, B, C, D, E, A. The die is thrown once. What is the probability of getting (i) A? (ii) D?
(i) P(A) = 1/3
(ii) P(D) = 1/6
Total faces = 6. 'A' 2 baar hai → P(A) = 2/6 = 1/3. 'D' 1 baar hai → P(D) = 1/6
Q20. Suppose you drop a die at random on a rectangular region of size 3 m × 2 m in which a circle of diameter 1 m is inscribed. What is the probability that it will land inside the circle?
P(inside circle) = π/24 ≈ 0.131
Area of rectangle = 3 × 2 = 6 m². Radius of circle = 0.5 m → Area = π(0.5)² = 0.25π. P = 0.25π / 6 = π/24 ≈ 0.131
Q21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) she will buy it? (ii) she will not buy it?
(i) P(buy) = 31/36
(ii) P(not buy) = 5/36
Total = 144, defective = 20, good = 124. P(buy = good) = 124/144 = 31/36. P(not buy = defective) = 20/144 = 5/36
Q22. Two dice are thrown together. (i) Complete the table showing probability of each possible sum from 2 to 12. (ii) A student argues that there are 11 possible outcomes 2,3,4,...,12 and therefore each has probability 1/11. Do you agree with this argument? Justify your answer.
Sum table (out of 36 total outcomes):
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| P | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
↔ Table ko side me swipe karein
Nahi, student ka argument galat hai — ye 11 outcomes equally likely NAHI hain. Sum=7 banane ke 6 tareeke hain lekin sum=2 sirf 1 tareeke se banta hai, isliye har sum ki probability barabar (1/11) nahi ho sakti.
Q23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result (three heads or three tails) and loses otherwise. Calculate the probability that Hanif will lose the game.
P(Hanif loses) = 3/4
Total outcomes (3 coins) = 2³ = 8: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Winning outcomes = HHH, TTT = 2. P(win) = 2/8 = 1/4. P(lose) = 1 − 1/4 = 3/4
Q24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once? [Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
(i) P(no 5 either time) = 25/36
(ii) P(at least one 5) = 11/36
Total outcomes = 6 × 6 = 36. Outcomes with no 5 on either die = 5 × 5 = 25 → 25/36. P(at least one 5) = 1 − 25/36 = 11/36
Q25. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
(i) P(same day) = 1/5
(ii) P(consecutive days) = 8/25
(iii) P(different days) = 4/5
Total outcomes = 5 × 5 = 25 (5 days each). Same day = 5 favourable → 5/25 = 1/5. Consecutive-day pairs = 8 → 8/25. Different days = 1 − 1/5 = 4/5
Important Equations — Ek Nazar Me
| Formula / Fact | Note |
|---|---|
| P(E) = favourable outcomes / total outcomes | Classical (theoretical) probability — outcomes equally likely hone chahiye |
| 0 ≤ P(E) ≤ 1 | Probability kabhi negative ya 1 se zyada nahi ho sakti |
| P(E) + P(not E) = 1 | Complementary event shortcut — P(not E) = 1 − P(E) |
| P(sure event) = 1, P(impossible event) = 0 | Extreme cases |
| 1 die = 6 outcomes | 1, 2, 3, 4, 5, 6 |
| 2 dice together = 36 outcomes | 6 × 6 grid — sabse common galti yahi hai |
| 1 coin = 2 outcomes, 2 coins = 4, 3 coins = 8 | 2ⁿ outcomes for n coins |
| Deck of cards = 52 | 4 suits × 13 cards; face cards = 12 (J,Q,K only, Ace nahi); red = 26, black = 26 |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Face cards ko 16 samajhna. Face cards sirf Jack, Queen, King hote hain — 3 × 4 suits = 12. Ace ko face card mat gino.
- Do dice ke total outcomes 12 maan lena. Do dice ek saath fenkne pe outcomes 6+6=12 NAHI hote — ye 6×6=36 hote hain, kyunki har die ka har face doosre die ke har face ke saath pair banata hai.
- P(not E) = 1 − P(E) shortcut bhool jaana. 'At least one' ya 'not' waale questions me seedha complement nikalne ke bajaye poori list gin lete hain — time waste aur galti dono hoti hai.
- Final fraction simplify na karna. Jaise 12/36 ko simplify karke 1/3 likhna zaroori hai — bina simplify kiye chhodna marks kaat sakta hai.
- 'At least one' ko 'exactly one' samajh lena. Dice problems me 'at least one 5' ka matlab hai 5 ek ya do baar aaye — sirf 'exactly one baar 5 aana' nahi. In dono ka answer alag hota hai.
- Without replacement wale questions me total outcomes update na karna. Jab ek item bina replace kiye nikal liya jaata hai (jaise Q15, Q17), to agla probability naye (chhote) total aur naye favourable count se nikalta hai — purana total use karna galat answer dega.
Board-Style Important Questions
- 1 mark: A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn is a face card.
- 1 mark: Two coins are tossed simultaneously. Find the probability of getting exactly one head.
- 2 marks: A bag contains 4 red, 5 blue and 3 green balls. A ball is drawn at random. Find the probability that it is (i) not blue (ii) red or green.
- 3 marks: Two dice are thrown together. Find the probability that the sum of the numbers on the two dice is (i) 8 (ii) a multiple of 3 (iii) greater than 10.
- 3 marks: A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Quick Quiz — Score Check Karein
Q1. Do dice ek saath fenke jaate hain. Total kitne possible equally likely outcomes hote hain?
Q2. Ek well-shuffled deck of 52 cards me face cards kitne hote hain?
Q3. Agar P(E) = 0.05 hai, to P(not E) kya hoga?
Q4. Kisi event E ke liye probability P(E) ki range kya hai?
Q5. Ek bag me 3 red aur 5 black balls hain. Ek ball random draw ki jaati hai. P(red) kya hoga?
Q6. Ek die ek baar throw kiya jaata hai. Prime number aane ki probability kya hai?
Q7. Ek student kehta hai ki 2 dice ke sum 2 se 12 tak 11 possible outcomes hain, isliye har sum ki probability 1/11 hai. Ye statement sahi hai ya galat?
Q8. Ek piggy bank me 100 fifty-paise coins, 50 ₹1 coins, 20 ₹2 coins aur 10 ₹5 coins hain. Random coin nikalne par woh ₹5 coin NA hone ki probability kya hai?
Q9. Ek coin 3 baar toss kiya jaata hai. Hanif jeetega agar teeno outcomes same hon (HHH ya TTT). Hanif ke haarne ki probability kya hai?
Q10. Class 9 ki experimental probability aur Class 10 ki classical (theoretical) probability me main farak kya hai?
Aksar Poochhe Jaane Wale Sawaal
Class 10 Probability me kitne exercises hain?
Rationalised NCERT syllabus me sirf ek exercise hai — Exercise 14.1, jisme 25 questions hain. Poora chapter classical (theoretical) probability pe based hai.
Do dice fenkne pe total kitne outcomes hote hain?
36 outcomes hote hain, na ki 12. Har die ke 6 faces hote hain, aur dono dice independent hote hain, isliye 6 × 6 = 36 combinations banti hain.
Deck of cards me face cards kitne hote hain?
12 face cards hote hain — Jack, Queen, King, har ek 4 suits me, to 3 × 4 = 12. Ace ko face card nahi maana jaata.
Probability kabhi 1 se zyada ya negative ho sakti hai kya?
Nahi. Probability hamesha 0 aur 1 ke beech hoti hai (0 ≤ P(E) ≤ 1). Agar koi calculation isse bahar aaye to kahin galti hui hai.
P(not E) nikalne ka shortcut kya hai?
P(not E) = 1 − P(E). Ye tab bahut kaam aata hai jab 'at least one' ya 'is event nahi hoga' type ka sawaal poocha jaaye — seedha subtract karke answer mil jaata hai.
Classical probability aur experimental probability me kya farak hai?
Experimental probability (Class 9) me hum baar baar trial karke observed frequency se probability nikalte hain. Classical/theoretical probability (Class 10) me hum outcomes ko equally likely maan kar seedha favourable/total se probability nikalte hain, bina koi experiment kiye.
Class 10 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
Class 10 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
NCERT Kaksha ek swatantra shaikshik platform hai aur NCERT ya CBSE se aadhikarik roop se sambaddh (officially affiliated) nahi hai. Kisi galti ki jaankari dene ke liye contact kijiye.