Important Questions Class 10 Maths Chapter 14 Probability

Probability is the measure of how likely an event is to occur in an experiment. For equally likely outcomes, probability equals favourable outcomes divided by total possible outcomes.

Probability becomes accurate only when the sample space is counted correctly. Important Questions Class 10 Maths Chapter 14 help students practise theoretical probability, equally likely outcomes, elementary events, complementary events, impossible events, sure events, cards, coins, dice, marbles, defective items, and two-dice cases. The 2026 NCERT Chapter 14, Probability, focuses on the classical approach, where all outcomes are assumed equally likely.

Key Takeaways

  • Probability Formula: P(E) = Number of favourable outcomes / Number of total outcomes.
  • Complementary Events: P(not E) = 1 − P(E).
  • Probability Range: For any event E, 0 ≤ P(E) ≤ 1.
  • Two Dice: Two dice have 36 equally likely ordered outcomes.

Important Questions Class 10 Maths Chapter 14 Structure 2026

Concept Formula Or Rule Key Variables
Theoretical probability P(E) = Favourable outcomes / Total outcomes Event E, sample space
Complement P(not E) = 1 − P(E) Event E, not E
Range 0 ≤ P(E) ≤ 1 Impossible event, sure event

Probability Class 10 Chapter Overview

Probability Class 10 studies the theoretical chance of an event under equally likely outcomes. NCERT contrasts this with experimental probability from Class 9.

The chapter uses coins, dice, cards, marbles, birthday cases, defective items, and two-dice tables. It also explains impossible events, sure events, elementary events, and complementary events.

Class 10 Maths Chapter 14 Important Questions On Theoretical Probability

Theoretical probability avoids repeated trials when outcomes can be counted logically. It works only when all outcomes are equally likely.

Students should always identify total outcomes before counting favourable outcomes.

Q1. What Is Theoretical Probability Class 10?

Theoretical probability is the probability calculated by counting equally likely outcomes.

It predicts the chance of an event without repeating the experiment many times. NCERT also calls it classical probability.

Formula:
P(E) = Number of outcomes favourable to E / Number of all possible outcomes

Q2. What Is The Probability Formula Class 10?

The probability formula is P(E) = Favourable outcomes / Total possible outcomes.

Here, E is the event whose probability is required. The formula applies when outcomes are equally likely.

Example: For one coin toss, P(head) = 1/2.

Q3. What Are Equally Likely Outcomes Class 10?

Equally likely outcomes are outcomes that have the same chance of occurring.

A fair coin gives head and tail with equal chance. A fair die gives 1, 2, 3, 4, 5, and 6 with equal chance.

Example: Each face of a fair die has probability 1/6.

Q4. Are All Experiments Based On Equally Likely Outcomes?

No, all experiments do not have equally likely outcomes.

A bag with 4 red balls and 1 blue ball gives red a higher chance. The outcomes red and blue are not equally likely.

Example: P(red) = 4/5, while P(blue) = 1/5.

Probability Formula Class 10 Questions With Answers

The formula becomes simple when students count outcomes carefully. Most errors happen when students treat unequal outcomes as equal.

Use fractions in simplest form unless the answer asks for decimals.

Q5. Find The Probability Of Getting A Head When A Coin Is Tossed Once.

The probability of getting a head is 1/2.

  1. Given Data:
    Possible outcomes = Head, Tail
  2. Total outcomes:
    2
  3. Favourable outcomes for head:
    1
  4. Formula Used:
    P(E) = Favourable outcomes / Total outcomes
  5. Calculation:
    P(head) = 1/2

Final Result: P(head) = 1/2

Q6. Find The Probability Of Getting A Tail When A Coin Is Tossed Once.

The probability of getting a tail is 1/2.

  1. Given Data:
    Possible outcomes = Head, Tail
  2. Total outcomes:
    2
  3. Favourable outcomes for tail:
    1
  4. Calculation:
    P(tail) = 1/2

Final Result: P(tail) = 1/2

Q7. A Bag Has One Red, One Blue, And One Yellow Ball. Find P(yellow).

The probability of drawing a yellow ball is 1/3.

  1. Given Data:
    Balls = red, blue, yellow
  2. Total outcomes:
    3
  3. Favourable outcomes for yellow:
    1
  4. Calculation:
    P(yellow) = 1/3

Final Result: P(yellow) = 1/3

Q8. A Box Contains 3 Blue, 2 White, And 4 Red Marbles. Find P(white), P(blue), And P(red).

P(white) = 2/9, P(blue) = 1/3, and P(red) = 4/9.

  1. Given Data:
    Blue marbles = 3
    White marbles = 2
    Red marbles = 4
  2. Total marbles:
    3 + 2 + 4 = 9
  3. Calculations:
    P(white) = 2/9
    P(blue) = 3/9 = 1/3
    P(red) = 4/9

Final Result: 2/9, 1/3, 4/9

Elementary Event Class 10 And Probability Range

An elementary event has exactly one outcome. The sum of all elementary event probabilities equals 1.

Probability can never be negative and can never be greater than 1.

Q9. What Is An Elementary Event Class 10?

An elementary event is an event with only one outcome.

Getting a head in one coin toss is an elementary event. Getting a 4 in one die throw is also elementary.

Example: In a die throw, event “getting 6” has one favourable outcome.

Q10. What Is The Sum Of Probabilities Of All Elementary Events?

The sum of probabilities of all elementary events is 1.

All elementary outcomes together cover the complete sample space. One of them must occur in the experiment.

Example: For a coin, P(head) + P(tail) = 1/2 + 1/2 = 1.

Q11. What Is The Probability Range For Any Event?

The probability of any event lies between 0 and 1.

The number of favourable outcomes cannot exceed total outcomes. It also cannot be less than zero.

Rule:
0 ≤ P(E) ≤ 1

Q12. Which Of These Cannot Be The Probability Of An Event: 2/3, −1.5, 15%, 0.7?

−1.5 cannot be the probability of an event.

  1. Probability must lie between 0 and 1.
  2. 2/3 lies between 0 and 1.
  3. 15% = 0.15 lies between 0 and 1.
  4. 0.7 lies between 0 and 1.
  5. −1.5 is less than 0.

Final Result: −1.5 cannot be a probability.

Complementary Events Class 10 Questions

Complementary events cover “event happens” and “event does not happen”. Their probabilities always add to 1.

This idea saves time in “not” and “at least once” questions.

Q13. What Are Complementary Events Class 10?

Complementary events are an event E and the event not E.

If E happens, not E cannot happen. If E does not happen, not E happens.

Formula:
P(not E) = 1 − P(E)

Q14. If P(E) = 0.05, Find P(not E).

P(not E) = 0.95.

  1. Given Data:
    P(E) = 0.05
  2. Formula Used:
    P(not E) = 1 − P(E)
  3. Calculation:
    P(not E) = 1 − 0.05
    P(not E) = 0.95

Final Result: P(not E) = 0.95

Q15. Sangeeta’s Probability Of Winning A Tennis Match Is 0.62. Find Reshma’s Probability Of Winning.

Reshma’s probability of winning is 0.38.

  1. Given Data:
    P(Sangeeta wins) = 0.62
  2. Since one player wins:
    P(Reshma wins) = 1 − P(Sangeeta wins)
  3. Calculation:
    P(Reshma wins) = 1 − 0.62
    P(Reshma wins) = 0.38

Final Result: 0.38

Q16. If Two Friends Have Different Birthdays With Probability 364/365, Find The Probability Of Same Birthday.

The probability of the same birthday is 1/365.

  1. Given Data:
    P(different birthdays) = 364/365
  2. Formula Used:
    P(same birthday) = 1 − P(different birthdays)
  3. Calculation:
    P(same birthday) = 1 − 364/365
    P(same birthday) = 1/365

Final Result: 1/365

Probability Of Impossible Event And Sure Event

An impossible event has no favourable outcome. A sure event has every possible outcome favourable.

These two cases create the boundary values 0 and 1.

Q17. What Is The Probability Of Impossible Event?

The probability of an impossible event is 0.

An impossible event has no favourable outcome. It cannot occur in the experiment.

Example: Getting 8 on a standard die is impossible, so P(8) = 0/6 = 0.

Q18. What Is The Probability Of Sure Event?

The probability of a sure event is 1.

A sure event includes all possible outcomes. It must occur in the experiment.

Example: Getting a number less than 7 on a die is sure, so P(number < 7) = 6/6 = 1.

Q19. A Bag Contains Lemon Candies Only. Find P(orange candy) And P(lemon candy).

P(orange candy) = 0 and P(lemon candy) = 1.

  1. Given Data:
    Bag contains lemon candies only.
  2. Orange candy:
    No orange candy exists in the bag.
    P(orange candy) = 0
  3. Lemon candy:
    Every candy is lemon flavoured.
    P(lemon candy) = 1

Final Result: P(orange) = 0, P(lemon) = 1

Q20. What Is P(getting A Number Less Than 9) On A Spinner Marked 1 To 8?

The probability is 1.

  1. Given Data:
    Spinner numbers = 1, 2, 3, 4, 5, 6, 7, 8
  2. Total outcomes:
    8
  3. Favourable outcomes for number less than 9:
    All 8 outcomes
  4. Calculation:
    P(number < 9) = 8/8 = 1

Final Result: 1

Probability Of Dice Class 10 Questions

A fair die has six equally likely outcomes. The possible outcomes are 1, 2, 3, 4, 5, and 6.

Prime, odd, even, and range-based questions are common in NCERT Class 10 Maths Chapter 14 questions.

Q21. A Die Is Thrown Once. Find P(number Greater Than 4).

The probability of getting a number greater than 4 is 1/3.

  1. Total outcomes:
    1, 2, 3, 4, 5, 6
  2. Favourable outcomes:
    5, 6
  3. Number of favourable outcomes:
    2
  4. Calculation:
    P(number > 4) = 2/6 = 1/3

Final Result: 1/3

Q22. A Die Is Thrown Once. Find P(number Less Than Or Equal To 4).

The probability is 2/3.

  1. Total outcomes:
    6
  2. Favourable outcomes:
    1, 2, 3, 4
  3. Number of favourable outcomes:
    4
  4. Calculation:
    P(number ≤ 4) = 4/6 = 2/3

Final Result: 2/3

Q23. A Die Is Thrown Once. Find P(prime number).

The probability of getting a prime number is 1/2.

  1. Total outcomes:
    1, 2, 3, 4, 5, 6
  2. Prime outcomes:
    2, 3, 5
  3. Number of favourable outcomes:
    3
  4. Calculation:
    P(prime number) = 3/6 = 1/2

Final Result: 1/2

Q24. A Die Is Thrown Once. Find P(number Between 2 And 6).

The probability is 1/2 if “between 2 and 6” means excluding 2 and 6.

  1. Total outcomes:
    1, 2, 3, 4, 5, 6
  2. Favourable outcomes:
    3, 4, 5
  3. Number of favourable outcomes:
    3
  4. Calculation:
    P(number between 2 and 6) = 3/6 = 1/2

Final Result: 1/2

Q25. A Die Is Thrown Once. Find P(odd number).

The probability of getting an odd number is 1/2.

  1. Total outcomes:
    6
  2. Odd outcomes:
    1, 3, 5
  3. Number of favourable outcomes:
    3
  4. Calculation:
    P(odd number) = 3/6 = 1/2

Final Result: 1/2

Probability Of Cards Class 10 Questions

A standard deck has 52 cards. It has 4 suits, 13 cards in each suit, 4 aces, and 12 face cards.

Spades and clubs are black. Hearts and diamonds are red.

Q26. One Card Is Drawn From A Well-Shuffled Deck. Find P(ace).

The probability of drawing an ace is 1/13.

  1. Total cards:
    52
  2. Number of aces:
    4
  3. Calculation:
    P(ace) = 4/52 = 1/13

Final Result: 1/13

Q27. One Card Is Drawn From A Deck. Find P(not an ace).

The probability of drawing a card that is not an ace is 12/13.

  1. Total cards:
    52
  2. Number of aces:
    4
  3. Cards that are not aces:
    52 − 4 = 48
  4. Calculation:
    P(not ace) = 48/52 = 12/13

Final Result: 12/13

Q28. One Card Is Drawn. Find P(face card).

The probability of drawing a face card is 3/13.

  1. Face cards:
    Kings, queens, jacks
  2. Face cards per suit:
    3
  3. Total face cards:
    3 × 4 = 12
  4. Total cards:
    52
  5. Calculation:
    P(face card) = 12/52 = 3/13

Final Result: 3/13

Q29. One Card Is Drawn. Find P(red face card).

The probability of drawing a red face card is 3/26.

  1. Red suits:
    Hearts and diamonds
  2. Face cards in each red suit:
    3
  3. Total red face cards:
    2 × 3 = 6
  4. Total cards:
    52
  5. Calculation:
    P(red face card) = 6/52 = 3/26

Final Result: 3/26

Q30. One Card Is Drawn. Find P(queen of diamonds).

The probability of drawing the queen of diamonds is 1/52.

  1. Total cards:
    52
  2. Queen of diamonds:
    1 card
  3. Calculation:
    P(queen of diamonds) = 1/52

Final Result: 1/52

Probability Of Coins Class 10 Questions

Coin questions become harder when two or three coins are tossed. Students must write the complete sample space.

For two different coins, HT and TH are different outcomes.

Q31. Two Different Coins Are Tossed Together. Find P(at least one head).

The probability of at least one head is 3/4.

  1. Sample space:
    HH, HT, TH, TT
  2. Total outcomes:
    4
  3. Favourable outcomes:
    HH, HT, TH
  4. Calculation:
    P(at least one head) = 3/4

Final Result: 3/4

Q32. Two Coins Are Tossed Together. Find P(no head).

The probability of no head is 1/4.

  1. Sample space:
    HH, HT, TH, TT
  2. Favourable outcome for no head:
    TT
  3. Calculation:
    P(no head) = 1/4

Final Result: 1/4

Q33. A Coin Is Tossed Three Times. Find The Probability That Hanif Loses If He Wins Only On HHH Or TTT.

The probability that Hanif loses is 3/4.

  1. Sample space:
    HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
  2. Total outcomes:
    8
  3. Winning outcomes:
    HHH, TTT
  4. Losing outcomes:
    8 − 2 = 6
  5. Calculation:
    P(losing) = 6/8 = 3/4

Final Result: 3/4

Two Dice Probability Class 10 Board Pattern Questions

Two dice create ordered pairs. The pair (1, 4) differs from (4, 1), so total outcomes are 36.

The NCERT two-dice table is essential for sum-based probability questions.

Q34. Two Dice Are Thrown Together. How Many Outcomes Are Possible?

Two dice have 36 possible outcomes.

  1. First die outcomes:
    6
  2. Second die outcomes:
    6
  3. Total ordered outcomes:
    6 × 6 = 36

Example: (1, 4) and (4, 1) are different outcomes.

Final Result: 36 outcomes

Q35. Two Dice Are Thrown Together. Find P(sum = 8).

The probability of getting sum 8 is 5/36.

  1. Total outcomes:
    36
  2. Favourable outcomes:
    (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
  3. Number of favourable outcomes:
    5
  4. Calculation:
    P(sum = 8) = 5/36

Final Result: 5/36

Q36. Two Dice Are Thrown Together. Find P(sum = 13).

The probability of getting sum 13 is 0.

  1. Maximum sum with two dice:
    6 + 6 = 12
  2. Sum 13 cannot occur.
  3. Favourable outcomes:
    0
  4. Calculation:
    P(sum = 13) = 0/36 = 0

Final Result: 0

Q37. Two Dice Are Thrown Together. Find P(sum ≤ 12).

The probability of getting sum less than or equal to 12 is 1.

  1. Minimum possible sum:
    1 + 1 = 2
  2. Maximum possible sum:
    6 + 6 = 12
  3. Every outcome gives sum ≤ 12.
  4. Calculation:
    P(sum ≤ 12) = 36/36 = 1

Final Result: 1

Q38. A Die Is Thrown Twice. Find P(5 Will Not Come Up Either Time).

The probability that 5 will not come up either time is 25/36.

  1. Throwing a die twice gives:
    6 × 6 = 36 outcomes
  2. Outcomes without 5 in one throw:
    1, 2, 3, 4, 6
  3. Outcomes without 5 in both throws:
    5 × 5 = 25
  4. Calculation:
    P(no 5 both times) = 25/36

Final Result: 25/36

Q39. A Die Is Thrown Twice. Find P(5 Will Come Up At Least Once).

The probability that 5 comes up at least once is 11/36.

  1. Use complement:
    P(at least one 5) = 1 − P(no 5)
  2. From Q38:
    P(no 5) = 25/36
  3. Calculation:
    P(at least one 5) = 1 − 25/36
    = 11/36

Final Result: 11/36

NCERT Class 10 Maths Chapter 14 Questions On Mixed Cases

Mixed probability questions use coins, marbles, pens, discs, fish, and number cards. The method stays the same, but counting changes.

Students must read whether the question asks for an event or its complement.

Q40. A Bag Contains 3 Red Balls And 5 Black Balls. Find P(red) And P(not red).

P(red) = 3/8 and P(not red) = 5/8.

  1. Total balls:
    3 + 5 = 8
  2. Favourable outcomes for red:
    3
  3. P(red):
    P(red) = 3/8
  4. P(not red):
    P(not red) = 1 − 3/8 = 5/8

Final Result: P(red) = 3/8, P(not red) = 5/8

Q41. A Box Has 5 Red, 8 White, And 4 Green Marbles. Find P(not green).

The probability of not drawing a green marble is 13/17.

  1. Total marbles:
    5 + 8 + 4 = 17
  2. Green marbles:
    4
  3. Not green marbles:
    5 + 8 = 13
  4. Calculation:
    P(not green) = 13/17

Final Result: 13/17

Q42. A Piggy Bank Has 100 Fifty-Paise Coins, 50 One-Rupee Coins, 20 Two-Rupee Coins, And 10 Five-Rupee Coins. Find P(50p coin).

The probability of getting a 50p coin is 5/9.

  1. Total coins:
    100 + 50 + 20 + 10 = 180
  2. Fifty-paise coins:
    100
  3. Calculation:
    P(50p coin) = 100/180 = 5/9

Final Result: 5/9

Q43. In The Same Piggy Bank, Find P(not a ₹5 coin).

The probability of not getting a ₹5 coin is 17/18.

  1. Total coins:
    180
  2. ₹5 coins:
    10
  3. Coins that are not ₹5:
    180 − 10 = 170
  4. Calculation:
    P(not ₹5 coin) = 170/180 = 17/18

Final Result: 17/18

Q44. A Tank Has 5 Male Fish And 8 Female Fish. Find P(male fish).

The probability of drawing a male fish is 5/13.

  1. Total fish:
    5 + 8 = 13
  2. Male fish:
    5
  3. Calculation:
    P(male fish) = 5/13

Final Result: 5/13

Q45. A Box Contains Discs Numbered 1 To 90. Find P(two-digit number).

The probability of drawing a two-digit number is 9/10.

  1. Total discs:
    90
  2. Two-digit numbers from 1 to 90:
    10 to 90
  3. Count:
    90 − 10 + 1 = 81
  4. Calculation:
    P(two-digit number) = 81/90 = 9/10

Final Result: 9/10

Q46. A Box Contains Discs Numbered 1 To 90. Find P(perfect square).

The probability of drawing a perfect square number is 1/10.

  1. Perfect squares from 1 to 90:
    1, 4, 9, 16, 25, 36, 49, 64, 81
  2. Number of favourable outcomes:
    9
  3. Total outcomes:
    90
  4. Calculation:
    P(perfect square) = 9/90 = 1/10

Final Result: 1/10

Q47. A Box Contains Discs Numbered 1 To 90. Find P(number Divisible By 5).

The probability of drawing a number divisible by 5 is 1/5.

  1. Multiples of 5 from 1 to 90:
    5, 10, 15, ..., 90
  2. Number of multiples:
    90 ÷ 5 = 18
  3. Total outcomes:
    90
  4. Calculation:
    P(divisible by 5) = 18/90 = 1/5

Final Result: 1/5

Probability Questions Class 10 With Answers On Defective Items

Defective-item questions need careful counting of good and defective pieces. “Acceptable” may mean different things for different people.

NCERT uses shirts, pens, bulbs, and ball pens to test this idea.

Q48. A Carton Has 100 Shirts: 88 Good, 8 Minor Defects, 4 Major Defects. Find P(acceptable To Jimmy).

The probability that the shirt is acceptable to Jimmy is 0.88.

  1. Jimmy accepts only good shirts.
  2. Good shirts = 88
  3. Total shirts = 100
  4. Calculation:
    P(acceptable to Jimmy) = 88/100 = 0.88

Final Result: 0.88

Q49. For The Same Carton, Find P(acceptable To Sujatha).

The probability that the shirt is acceptable to Sujatha is 0.96.

  1. Sujatha rejects only major defects.
  2. Acceptable shirts = good shirts + minor defects
  3. Acceptable shirts = 88 + 8 = 96
  4. Total shirts = 100
  5. Calculation:
    P(acceptable to Sujatha) = 96/100 = 0.96

Final Result: 0.96

Q50. 12 Defective Pens Are Mixed With 132 Good Pens. Find P(good pen).

The probability of drawing a good pen is 11/12.

  1. Defective pens = 12
  2. Good pens = 132
  3. Total pens = 12 + 132 = 144
  4. Calculation:
    P(good pen) = 132/144 = 11/12

Final Result: 11/12

Important Questions Class 10 Maths All Chapters

Chapter No Chapter Name
Chapter 1 Real Numbers
Chapter 2 Polynomials
Chapter 3 Pair of Linear Equations in Two Variables
Chapter 4 Quadratic Equations
Chapter 5 Arithmetic Progressions
Chapter 6 Triangles
Chapter 7 Coordinate Geometry
Chapter 8 Introduction to Trigonometry
Chapter 9 Some Applications of Trigonometry
Chapter 10 Circles
Chapter 11 Areas Related to Circles
Chapter 12 Surface Areas and Volumes
Chapter 13 Statistics
Chapter 14 Probability

Q.1 Find the mean of the following given data:

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 4 10 18 12 6

Marks:2
Ans

The frequency table is

Class interval fi Class Mark

xi

 Product

fixi
0-10

10-20

20-30

30-40

40-50

 4

10

18

12

6

 5

15

25

35

45

 20

150

450

420

270
fi = 50 fixi = 1310

Therefore,Meanx=fixifi=131050=26.2

Q.2 A survey conducted on 15 houses in a locality by a group resulted in the following frequency table for the number of family members in a house.

Family Size 1-3 3-5 5-7 7-9 9-11
No. of families 4 6 2 2 1

Find the median family size.

Marks:3
Ans

Family size (Class-intervals) Number of families (f) Cumulative frequency

1-3

3-5

5-7

7-9

9-11

4

6

2

2

2

4

10

12

14

16

n = 16

Here,n2=8,So,medianclass=35therefore,l=3cf=cumulativefrequencyofclassprecedingthemedianclass=4f=frequencyofthemedianclass=10h=classsize=2So,median=3+8410—2=3.8Therefore,themedianfamilysizeis3.8

Q.3 Find the mode for the following data:

Classes 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 4 8 10 12 10 4 2

Marks:2
Ans

Classes Frequency
fi
10 – 20 4
20 – 30 8
30 – 40 10 = f0
40 – 50 12 = f1
50 – 60 10 = f2
60 – 70 4
70 – 80 2
Total 50

Sincemaximumfrequencyis12,therefore,themodalclassis4050.Thus,thelowerlimitlofthemodalclassis40.mode=l+f1f02f1f0f2—h=40+12102—121010—10=40+24—10=45.Thus,themodeis45.

Q.4 During the medical check-up of 35 students of a class, their weights were recorded as follows

Weight (in kg) Number of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

Draw a less than type graph for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.

Marks:4
Ans

By locating upper limits on X-axis and number of students on Y-axis, we get the following graph.

From the graph, we see that median is 46.5.
The frequency distribution table with the given cumulative frequencies becomes:

Class Intervals Frequency Cumulative Frequency
Below 38 0 0
38-40 3 3
40-42 2 5
42-44 4 9
44-46 5 14
46-48 14 28
48-50 4 32
50-52 3 35

Heren2=352=17.5.Now4648istheclasswhosecumulativefrequencyis28isgreaterthann2i.e.17.5.So,4648isthemedianclass.Fromthetable:f=14,cf=14,h=2Median=l+n2cff—h=46+17.51414—2=46+3.57=46+0.5=46.5

Q.5 If the mean of the following distribution is 27, find the value of p:

Class-Interval 0-10 10-20 20-30 30-40 40-50
No. of Workers 8 p 12 13 10

Marks:4
Ans

Class Int. (xi) (fi) di = xi A Ui = xi A/h fiUi
0-10 5 8 -20 -2 -16
10-20 15 p -10 -1 -p
20-30 25 12 0 0 0
30-40 35 13 10 1 13
40-50 45 10 20 2 20
fi = 43 + p fiUi = 3

mean=A+hfiUifiMean=25+1017p43+p27=25+1017p43+p2=1017p43+p15=17p43+p43+p=855p6p=42p=7

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FAQs (Frequently Asked Questions)

Probability is always between 0 and 1 because favourable outcomes cannot be less than 0 or more than total outcomes.
If an event is impossible, P(E) = 0. If it is certain, P(E) = 1.

Outcomes are equally likely when each outcome has the same chance of occurring. A fair coin gives head and tail equally likely outcomes, while a bag with 4 red balls and 1 blue ball does not make red and blue equally likely.

(1, 4) and (4, 1) are different outcomes because the first number belongs to the first die and the second number belongs to the second die. For two dice, total outcomes = 6 × 6 = 36.

“Not” probability questions are solved using P(not E) = 1 − P(E).
For example, if P(E) = 0.05, then P(not E) = 1 − 0.05 = 0.95.

“At least one head” is easier using complement because its opposite is only “no head”.
For two coins, P(no head) = 1/4, so P(at least one head) = 1 − 1/4 = 3/4.