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Probability – CBSE NCERT Study Resources

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9th

9th - Mathematics

Probability

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Chapter Overview: Probability

This chapter introduces the fundamental concepts of probability, which is a branch of mathematics that deals with calculating the likelihood of events occurring. Students will learn about experiments, outcomes, events, and how to compute probabilities using theoretical and experimental approaches.

Probability: Probability is a measure of the likelihood that an event will occur, expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

Key Concepts

  • Experiment: An action that results in one or more outcomes.
  • Outcome: A possible result of an experiment.
  • Event: A collection of one or more outcomes.
  • Sample Space: The set of all possible outcomes of an experiment.

Types of Probability

  • Theoretical Probability: Probability based on reasoning and theoretical principles.
  • Experimental Probability: Probability calculated based on actual experiments and observations.

Probability Formula

The probability of an event E is given by:

P(E) = Number of favorable outcomes / Total number of possible outcomes

Examples

Example 1: Tossing a fair coin. The probability of getting heads is 1/2.

Example 2: Rolling a die. The probability of getting an even number is 3/6 = 1/2.

Applications

Probability is widely used in various fields such as statistics, finance, science, and engineering to predict outcomes and make informed decisions.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define probability as per NCERT.
Answer:

Probability is the measure of the likelihood of an event.

Question 2:
What is the probability of getting a head when tossing a fair coin?
Answer:
Numeric answer:
0.5
Question 3:
If a die is rolled, what is the probability of getting an even number?
Answer:
Numeric answer:
0.5
Question 4:
A bag has 3 red and 2 blue balls. What is the probability of drawing a red ball?
Answer:
Numeric answer:
0.6
Question 5:
What is the probability of an impossible event?
Answer:
Numeric answer:
0
Question 6:
What is the probability of a certain event?
Answer:
Numeric answer:
1
Question 7:
A spinner has 4 equal sectors. What is the probability of landing on one specific sector?
Answer:
Numeric answer:
0.25
Question 8:
From a deck of 52 cards, what is the probability of drawing an ace?
Answer:
Numeric answer:
1/13
Question 9:
If P(E) = 0.3, what is P(not E)?
Answer:
Numeric answer:
0.7
Question 10:
A jar has 5 green and 7 yellow marbles. What is the probability of picking a green marble?
Answer:
Numeric answer:
5/12
Question 11:
What is the sum of probabilities of all possible outcomes of an experiment?
Answer:
Numeric answer:
1
Question 12:
A fair die is rolled. What is the probability of getting a number greater than 4?
Answer:
Numeric answer:
1/3
Question 13:
Define probability in simple terms.
Answer:

The probability of an event is a measure of the likelihood that the event will occur. It is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

Question 14:
If a die is rolled once, what is the probability of getting an even number?
Answer:

The probability is 1/2 because there are 3 favorable outcomes (2, 4, 6) out of 6 possible outcomes (1 to 6).

Question 15:
What is the range of possible values for probability?
Answer:

The probability of an event always lies between 0 (impossible event) and 1 (certain event), inclusive.

Question 16:
A bag contains 5 red and 3 blue balls. What is the probability of drawing a red ball?
Answer:

The probability is 5/8 because there are 5 favorable (red) balls out of a total of 8 balls.

Question 17:
If the probability of an event is 0.75, what is the probability of its complement?
Answer:

The probability of the complement event is 0.25 because
1 - 0.75 = 0.25.

Question 18:
A spinner has 4 equal sectors colored red, blue, green, and yellow. What is the probability of landing on blue?
Answer:

The probability is 1/4 because there is 1 favorable outcome (blue) out of 4 possible outcomes.

Question 19:
If a card is drawn from a well-shuffled deck of 52 cards, what is the probability of getting a king?
Answer:

The probability is 4/52 or 1/13 because there are 4 kings in a deck of 52 cards.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
If a coin is tossed once, what is the probability of getting a head?
Answer:

When a coin is tossed, there are two possible outcomes: head or tail.
The probability of getting a head is:
P(Head) = Number of favorable outcomes / Total outcomes = 1/2.

Question 2:
A die is rolled. Find the probability of getting an even number.
Answer:

Possible even numbers on a die: 2, 4, 6 (3 outcomes).
Total outcomes = 6.
P(Even) = Favorable outcomes / Total outcomes = 3/6 = 1/2.

Question 3:
If P(E) = 0.35, what is the probability of 'not E'?
Answer:

P(not E) = 1 - P(E)
= 1 - 0.35
= 0.65.

Question 4:
A card is drawn from a deck of 52 cards. What is the probability of drawing a king?
Answer:

Number of kings in a deck = 4.
Total cards = 52.
P(King) = 4/52 = 1/13.

Question 5:
A spinner has 8 equal sectors numbered 1 to 8. What is the probability of landing on an odd number?
Answer:

Odd numbers: 1, 3, 5, 7 (4 outcomes).
Total sectors = 8.
P(Odd) = 4/8 = 1/2.

Question 6:
If the probability of raining tomorrow is 0.75, what is the probability that it will not rain?
Answer:

P(Not Raining) = 1 - P(Raining)
= 1 - 0.75
= 0.25.

Question 7:
What is the probability of drawing a red card from a standard deck of 52 playing cards?
Answer:

A standard deck has 26 red cards (hearts and diamonds) out of 52 total cards.


Probability = Number of red cards / Total number of cards = 26/52 = 1/2.

Question 8:
A bag contains 5 red, 3 blue, and 2 green balls. What is the probability of drawing a blue ball?
Answer:

Total balls = 5 (red) + 3 (blue) + 2 (green) = 10.


Probability of drawing a blue ball = Number of blue balls / Total balls = 3/10.

Question 9:
A spinner has 8 equal sectors numbered 1 to 8. What is the probability of landing on a prime number?
Answer:

Prime numbers between 1 and 8 are 2, 3, 5, and 7 (4 outcomes).


Probability = Number of prime numbers / Total sectors = 4/8 = 1/2.

Question 10:
In a class of 40 students, 25 are girls. What is the probability of randomly selecting a boy?
Answer:

Number of boys = Total students - Number of girls = 40 - 25 = 15.


Probability of selecting a boy = Number of boys / Total students = 15/40 = 3/8.

Question 11:
What is the probability of getting a sum of 7 when two dice are rolled?
Answer:

Total possible outcomes when two dice are rolled = 6 × 6 = 36.


Favorable outcomes for sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes.


Probability = 6/36 = 1/6.

Question 12:
If P(E) = 0.35, what is P(not E)?
Answer:

P(not E) is the complement of P(E).


P(not E) = 1 - P(E) = 1 - 0.35 = 0.65.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
A bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. Find the probability that the ball drawn is not red.
Answer:

Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10.
Number of non-red balls = 3 (blue) + 2 (green) = 5.
Probability of drawing a non-red ball = Number of favorable outcomes / Total outcomes = 5/10 = 1/2.

Question 2:
A die is rolled once. What is the probability of getting a prime number?
Answer:

Possible outcomes when rolling a die: 1, 2, 3, 4, 5, 6 (Total outcomes = 6).
Prime numbers in this range: 2, 3, 5 (Favorable outcomes = 3).
Probability = 3/6 = 1/2.

Question 3:
In a class of 30 students, 12 are girls. If a student is selected at random, what is the probability that the student is a boy?
Answer:

Total students = 30.
Number of girls = 12, so number of boys = 30 - 12 = 18.
Probability of selecting a boy = Favorable outcomes / Total outcomes = 18/30 = 3/5.

Question 4:
A card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing a black queen.
Answer:

Total cards in a deck = 52.
Number of black queens (Queen of Spades and Queen of Clubs) = 2.
Probability = 2/52 = 1/26.

Question 5:
A box contains 20 discs numbered 1 to 20. A disc is drawn at random. What is the probability that the number on the disc is divisible by 5?
Answer:

Total discs = 20.
Numbers divisible by 5 between 1 and 20: 5, 10, 15, 20 (Favorable outcomes = 4).
Probability = 4/20 = 1/5.

Question 6:
A bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. What is the probability that the ball drawn is not red?
Answer:

Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10.
Number of non-red balls = 3 (blue) + 2 (green) = 5.
Probability of drawing a non-red ball = Number of non-red balls / Total balls = 5/10 = 1/2.
Thus, the probability is 0.5 or 50%.

Question 7:
A die is rolled once. Find the probability of getting a prime number.
Answer:

Total possible outcomes when rolling a die = 6 (1, 2, 3, 4, 5, 6).
Prime numbers between 1 and 6 are 2, 3, 5 (total of 3 outcomes).
Probability of getting a prime number = Number of favorable outcomes / Total outcomes = 3/6 = 1/2.
Thus, the probability is 0.5 or 50%.

Question 8:
In a class of 40 students, 25 are girls. If a student is selected at random, what is the probability that the student is a boy?
Answer:

Total number of students = 40.
Number of girls = 25, so number of boys = 40 - 25 = 15.
Probability of selecting a boy = Number of boys / Total students = 15/40 = 3/8.
Thus, the probability is 0.375 or 37.5%.

Question 9:
A spinner has 8 equal sectors numbered 1 to 8. What is the probability of landing on an even number?
Answer:

Total number of sectors = 8.
Even numbers between 1 and 8 are 2, 4, 6, 8 (total of 4 outcomes).
Probability of landing on an even number = Number of favorable outcomes / Total outcomes = 4/8 = 1/2.
Thus, the probability is 0.5 or 50%.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
A bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. Find the probability that it is (a) red (b) not green. Explain using probability formula and NCERT examples.
Answer:
Introduction

Probability measures the likelihood of an event. Our textbook shows similar problems with colored balls.


Argument 1
  • Total balls = 5 (red) + 3 (blue) + 2 (green) = 10.
  • P(red) = Favorable outcomes / Total outcomes = 5/10 = 1/2.

Argument 2
  • P(not green) = 1 - P(green) = 1 - (2/10) = 4/5.
  • This matches NCERT Example 5 where non-favorable outcomes are subtracted.

Conclusion

Probability helps predict real-life scenarios like drawing colored balls.

Question 2:
In a class of 40 students, 25 play cricket and 20 play basketball. If 10 play both, find the probability that a student plays neither sport. Use Venn diagram logic.
Answer:
Introduction

Probability of mutually exclusive events is common in NCERT. We studied Venn diagrams for such cases.


Argument 1
  • Total students = 40.
  • Students playing at least one sport = 25 (cricket) + 20 (basketball) - 10 (both) = 35.

Argument 2
  • Students playing neither = 40 - 35 = 5.
  • P(neither) = 5/40 = 1/8, similar to NCERT Exercise 15.1 Q5.

Conclusion

Venn diagrams simplify real-world problems like sports participation.

Question 3:
A die is rolled once. Find the probability of getting (a) a prime number (b) a number greater than 4. Relate to NCERT examples and step notation.
Answer:
Introduction

Probability of outcomes in a die roll is a basic concept. Our textbook uses similar examples.


Argument 1
  • Prime numbers on a die: 2, 3, 5 → 3 outcomes.
  • P(prime) = 3/6 = 1/2, as in NCERT Example 4.

Argument 2
  • Numbers > 4: 5, 6 → 2 outcomes.
  • P(>4) = 2/6 = 1/3, following step notation from Exercise 15.1.

Conclusion

Dice problems help understand probability in games and real-life decisions.

Question 4:
A bag contains 5 red, 3 blue, and 2 green marbles. Calculate the probability of drawing a blue marble and then a red marble without replacement. Show steps.
Answer:
Introduction

We studied probability in NCERT Chapter 15. Here, we find the probability of sequential events without replacement.


Argument 1
  • Total marbles = 5 (red) + 3 (blue) + 2 (green) = 10.
  • P(blue first) = 3/10.

Argument 2
  • After removing 1 blue marble, remaining marbles = 9.
  • P(red next) = 5/9.
  • Combined probability = (3/10) × (5/9) = 15/90 = 1/6.

Conclusion

Our textbook shows similar problems. The final probability is 1/6.

Question 5:
In a class of 30 students, 12 prefer cricket, 8 prefer football, and the rest prefer basketball. If a student is chosen randomly, what is the probability they prefer basketball? Explain using NCERT methods.
Answer:
Introduction

Probability helps predict outcomes. We use the formula P(E) = Number of favorable outcomes / Total outcomes.


Argument 1
  • Total students = 30.
  • Students preferring basketball = 30 - (12 + 8) = 10.

Argument 2
  • Favorable outcomes = 10.
  • Total outcomes = 30.
  • P(basketball) = 10/30 = 1/3.

Conclusion

Our NCERT examples confirm this method. The probability is 1/3.

Question 6:
A die is rolled twice. Find the probability that the sum of outcomes is 9. Use step notation and justify with NCERT concepts.
Answer:
Introduction

We learned about probability with dice in NCERT. Rolling a die twice gives 36 possible outcomes.


Argument 1
  • Favorable pairs for sum 9: (3,6), (4,5), (5,4), (6,3).
  • Total favorable outcomes = 4.

Argument 2
  • Total possible outcomes = 6 × 6 = 36.
  • P(sum 9) = 4/36 = 1/9.

Conclusion

Our textbook shows similar derivations. The probability is 1/9.

Question 7:
A bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. Find the probability that it is (a) red (b) not green. Explain using probability formula.
Answer:
Introduction

We studied probability as the likelihood of an event happening. Our textbook shows the formula: P(E) = Number of favorable outcomes / Total outcomes.


Argument 1
  • Total balls = 5 red + 3 blue + 2 green = 10.
  • P(red) = 5/10 = 1/2.

Argument 2
  • P(not green) = 1 - P(green) = 1 - (2/10) = 4/5.
  • This uses the complement rule.

Conclusion

Probability helps predict real-life events, like drawing colored balls.

Question 8:
In a class of 30 students, 12 like cricket, 8 like football, and the rest like basketball. If a student is chosen randomly, find the probability they like (a) cricket (b) basketball. Use step notation.
Answer:
Introduction

Probability measures chance. Our textbook explains it using ratios.


Argument 1
  • Total students = 30.
  • P(cricket) = 12/30 = 2/5.

Argument 2
  • Basketball lovers = 30 - (12 + 8) = 10.
  • P(basketball) = 10/30 = 1/3.

Conclusion

This method applies to surveys or preference analysis.

Question 9:
A die is rolled once. Find the probability of getting (a) an even number (b) a prime number. Relate to NCERT examples.
Answer:
Introduction

Probability for a die is a classic example. Our textbook uses similar problems.


Argument 1
  • Even numbers: 2, 4, 6 → 3 outcomes.
  • P(even) = 3/6 = 1/2.

Argument 2
  • Prime numbers: 2, 3, 5 → 3 outcomes.
  • P(prime) = 3/6 = 1/2.

Conclusion

Dice problems help understand basic probability concepts.

Question 10:
A bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. Find the probability of getting (a) a red ball (b) a non-blue ball. Explain using probability concepts.
Answer:
Introduction

Probability measures the likelihood of an event. Our textbook shows examples like this to understand basic probability.


Argument 1
  • Total balls = 5 (red) + 3 (blue) + 2 (green) = 10.
  • P(red) = Favourable outcomes / Total outcomes = 5/10 = 1/2.

Argument 2
  • Non-blue balls = 5 (red) + 2 (green) = 7.
  • P(non-blue) = 7/10.

Conclusion

We studied that probability ranges from 0 to 1. Here, P(red) is 0.5, and P(non-blue) is 0.7.

Question 11:
In a class of 40 students, 25 like cricket and 15 like football. If a student is chosen randomly, find the probability that they like (a) cricket (b) either cricket or football. Justify with probability rules.
Answer:
Introduction

Probability helps predict outcomes. Our textbook uses such examples to explain simple events.


Argument 1
  • Total students = 40.
  • P(cricket) = 25/40 = 5/8.

Argument 2
  • Since all students like at least one sport, P(cricket or football) = (25 + 15)/40 = 40/40 = 1.

Conclusion

We learned that P(cricket) is 5/8, and P(either) is 1, meaning it’s certain.

Question 12:
A die is rolled once. Calculate the probability of getting (a) an even number (b) a prime number. Use the probability formula and explain.
Answer:
Introduction

Probability of an event is calculated using outcomes. Our textbook explains this with dice examples.


Argument 1
  • Total outcomes = 6 (1-6).
  • Even numbers = {2, 4, 6} → P(even) = 3/6 = 1/2.

Argument 2
  • Prime numbers = {2, 3, 5} → P(prime) = 3/6 = 1/2.

Conclusion

We studied that P(even) and P(prime) are both 1/2, showing equal likelihood.

Question 13:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag.

(i) Find the probability that the ball drawn is red.
(ii) If the red ball drawn is not replaced, find the probability of drawing a blue ball next.

Show all steps clearly.
Answer:

To solve this problem, we will follow the basic principles of probability.



(i) Probability of drawing a red ball:
Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10.
Number of favorable outcomes (red balls) = 5.
Probability = Number of favorable outcomes / Total number of outcomes.
So, P(red) = 5/10 = 1/2 or 0.5.

(ii) Probability of drawing a blue ball after removing one red ball:
Since the red ball is not replaced, the total number of balls left = 10 - 1 = 9.
Number of blue balls remains unchanged = 3.
Probability = 3/9 = 1/3 or approximately 0.333.

Thus, the probabilities are:

  • P(red) = 1/2
  • P(blue after removing red) = 1/3
Question 14:
In a survey of 50 students, 20 students like cricket, 25 like football, and 5 like both.

(i) Represent this data in a Venn diagram.
(ii) Find the probability that a randomly selected student likes only cricket.

Explain your steps.
Answer:

This problem involves set theory and probability.



(i) Venn Diagram Representation:
Let:
- C = Students who like cricket (20)
- F = Students who like football (25)
- C ∩ F = Students who like both (5)

The Venn diagram will have:
- Only cricket: 20 - 5 = 15
- Only football: 25 - 5 = 20
- Both: 5

(ii) Probability of selecting a student who likes only cricket:
Total students = 50.
Students who like only cricket = 15.
Probability = Number of favorable outcomes / Total number of outcomes.
P(only cricket) = 15/50 = 3/10 or 0.3.

Key takeaways:

  • Venn diagram shows overlaps clearly.
  • Probability calculations rely on accurate counting of exclusive and shared preferences.
Question 15:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag.

(i) Find the probability that the ball drawn is red.
(ii) If the red ball drawn is not replaced, find the probability of drawing a blue ball next.
Answer:

Solution:

Total number of balls in the bag initially = 5 (red) + 3 (blue) + 2 (green) = 10 balls.


(i) Probability of drawing a red ball:

Number of favorable outcomes (red balls) = 5
Total possible outcomes = 10

Probability = Number of favorable outcomes / Total possible outcomes
= 5/10
= 1/2 or 0.5.


(ii) Probability of drawing a blue ball after removing one red ball:

After removing one red ball, total remaining balls = 10 - 1 = 9 balls.
Number of blue balls remains unchanged = 3.

Probability = Number of blue balls / Remaining total balls
= 3/9
= 1/3.


Final Answers:
(i) Probability of drawing a red ball = 1/2
(ii) Probability of drawing a blue ball next = 1/3.

Question 16:
A fair six-sided die is rolled once.

(i) Find the probability of getting an even number.
(ii) Find the probability of getting a prime number.
Answer:

Solution:

A fair die has 6 faces numbered from 1 to 6. Each outcome is equally likely.


(i) Probability of getting an even number:

Even numbers on a die: 2, 4, 6 → 3 favorable outcomes.
Total possible outcomes = 6.

Probability = Favorable outcomes / Total outcomes
= 3/6
= 1/2.


(ii) Probability of getting a prime number:

Prime numbers on a die: 2, 3, 5 → 3 favorable outcomes.
Total possible outcomes = 6.

Probability = Favorable outcomes / Total outcomes
= 3/6
= 1/2.


Final Answers:
(i) Probability of even number = 1/2
(ii) Probability of prime number = 1/2.

Question 17:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag.

(i) Find the probability that the ball drawn is red.
(ii) Find the probability that the ball drawn is not green.

Show all steps clearly and explain the concept of probability in your answer.
Answer:

The probability of an event is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes. It is a measure of how likely an event is to occur, ranging from 0 (impossible) to 1 (certain).


Given:
Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10 balls.


(i) Probability of drawing a red ball:
Number of favorable outcomes (red balls) = 5.
Total possible outcomes = 10.
Probability = Number of favorable outcomes / Total possible outcomes = 5/10 = 1/2.


(ii) Probability that the ball drawn is not green:
Number of favorable outcomes (not green) = Total balls - Green balls = 10 - 2 = 8.
Total possible outcomes = 10.
Probability = 8/10 = 4/5.


Explanation:
The probability of an event not happening is calculated by subtracting the probability of the event happening from 1. Here, the probability of drawing a green ball is 2/10 = 1/5. Thus, the probability of not drawing a green ball is 1 - 1/5 = 4/5, which matches our earlier calculation.

Question 18:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag.

(i) Find the probability that the ball drawn is red.
(ii) If one red ball is removed from the bag, find the new probability of drawing a red ball.
(iii) Compare the probabilities in (i) and (ii) and explain the change.
Answer:

Solution:

Total number of balls initially = 5 (red) + 3 (blue) + 2 (green) = 10 balls.


(i) Probability of drawing a red ball initially:

Probability = Number of favorable outcomes / Total number of outcomes

P(Red) = 5/10 = 1/2 or 0.5.


(ii) After removing one red ball:

New count of red balls = 5 - 1 = 4 red balls.

Total balls remaining = 10 - 1 = 9 balls.

New P(Red) = 4/9 ≈ 0.444.


(iii) Comparison and explanation:

  • Initial P(Red) = 0.5, New P(Red) ≈ 0.444.
  • The probability decreases because the number of red balls reduces, while the total number of balls also reduces but not proportionally.
  • This shows that probability depends on the ratio of favorable outcomes to total outcomes.

Conclusion: Removing a red ball reduces the chance of drawing a red ball, as the favorable outcomes decrease relative to the total outcomes.

Question 19:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag.

(i) Find the probability that the ball drawn is red.
(ii) Find the probability that the ball drawn is not blue.
(iii) If one more red ball is added to the bag, what will be the new probability of drawing a red ball?
Answer:

Given: The bag contains 5 red balls, 3 blue balls, and 2 green balls. Total number of balls = 5 + 3 + 2 = 10.


(i) Probability of drawing a red ball:


Probability = Number of favorable outcomes / Total number of outcomes


P(Red) = 5/10 = 1/2 or 0.5.


(ii) Probability that the ball drawn is not blue:


Number of non-blue balls = Total balls - Blue balls = 10 - 3 = 7.


P(Not Blue) = 7/10 = 0.7.


(iii) After adding one more red ball:


New number of red balls = 5 + 1 = 6.


Total balls now = 10 + 1 = 11.


New P(Red) = 6/11 ≈ 0.545 (approx).


Key Takeaways:

  • Probability is always between 0 and 1.
  • Adding more favorable outcomes increases the probability.
  • The sum of probabilities of all possible outcomes is 1.
Question 20:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag.

(i) Find the probability that the ball drawn is red.
(ii) Find the probability that the ball drawn is not green.
(iii) If one more red ball is added to the bag, what will be the new probability of drawing a red ball?
Answer:

Given: A bag contains 5 red balls, 3 blue balls, and 2 green balls. Total balls = 5 + 3 + 2 = 10.


(i) Probability of drawing a red ball:


Number of red balls = 5
Total number of balls = 10
Probability = Number of red balls / Total balls = 5/10 = 1/2


(ii) Probability that the ball drawn is not green:


Number of non-green balls = Red + Blue = 5 + 3 = 8
Total number of balls = 10
Probability = Number of non-green balls / Total balls = 8/10 = 4/5


(iii) New probability if one more red ball is added:


New number of red balls = 5 + 1 = 6
New total number of balls = 10 + 1 = 11
New probability = Number of red balls / Total balls = 6/11


Key Takeaways:

  • Probability is calculated as Favorable Outcomes / Total Outcomes.
  • For not events, subtract the probability of the event from 1 or count non-favorable outcomes.
  • Adding/removing items changes the sample space, affecting probability.
Question 21:
A bag contains 5 red, 8 blue, and 7 green marbles. A marble is drawn at random from the bag.
(i) Find the probability that the marble drawn is red.
(ii) If the red marble drawn is not replaced, find the probability of drawing a blue marble next.
Answer:

Given: A bag contains 5 red, 8 blue, and 7 green marbles. Total marbles = 5 + 8 + 7 = 20.

(i) Probability of drawing a red marble:


Probability = (Number of red marbles) / (Total marbles)
P(Red) = 5 / 20 = 1/4 or 0.25.

(ii) Probability of drawing a blue marble after removing one red marble:


After removing one red marble, remaining red marbles = 4.
Total marbles now = 20 - 1 = 19.
P(Blue) = (Number of blue marbles) / (Total marbles) = 8 / 19.

Note: Since the first marble is not replaced, the total number of marbles decreases, affecting the probability of the second draw.

Question 22:
Two dice are rolled simultaneously. Find the probability that:
(i) The sum of the numbers on the dice is 9.
(ii) The numbers on the two dice are different.
Answer:

Given: Two dice are rolled. Total possible outcomes = 6 × 6 = 36.

(i) Probability that the sum is 9:


Favorable outcomes: (3,6), (4,5), (5,4), (6,3) → 4 outcomes.
P(Sum = 9) = 4 / 36 = 1/9.

(ii) Probability that the numbers are different:


Favorable outcomes: Total outcomes - outcomes where numbers are same.
Outcomes with same numbers: (1,1), (2,2), ..., (6,6) → 6 outcomes.
Favorable outcomes = 36 - 6 = 30.
P(Numbers different) = 30 / 36 = 5/6.

Note: For probability questions involving dice, always list the possible outcomes systematically to avoid errors.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
In a class of 30 students, 12 prefer cricket, 8 prefer football, and the rest prefer badminton. If a student is chosen at random, find the probability that the student prefers (a) cricket (b) badminton.
Answer:
Problem Interpretation

We need to find the probability of a student preferring cricket or badminton from a class of 30.

Mathematical Modeling
  • Total students = 30
  • Cricket lovers = 12
  • Badminton lovers = 30 - (12 + 8) = 10
Solution

(a) P(Cricket) = 12/30 = 2/5

(b) P(Badminton) = 10/30 = 1/3

Question 2:
A bag contains 5 red, 3 blue, and 2 green marbles. A marble is drawn randomly. Find the probability of getting (a) a red marble (b) a green or blue marble.
Answer:
Problem Interpretation

We studied probability in NCERT. Here, we find the chance of drawing marbles of specific colors.

Mathematical Modeling
  • Total marbles = 5 + 3 + 2 = 10
  • Red marbles = 5
  • Green + Blue marbles = 2 + 3 = 5
Solution

(a) P(Red) = 5/10 = 1/2

(b) P(Green or Blue) = 5/10 = 1/2

Question 3:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. If one ball is drawn at random, find the probability that it is (a) red (b) not green.
Answer:
Problem Interpretation

We need to find the probability of drawing a red ball and a non-green ball from the bag.


Mathematical Modeling
  • Total balls = 5 (red) + 3 (blue) + 2 (green) = 10
  • (a) P(red) = Red balls / Total balls = 5/10 = 1/2
  • (b) P(not green) = (Total - Green) / Total = 8/10 = 4/5

Solution

Our textbook shows similar examples. The probabilities are (a) 1/2 and (b) 4/5.

Question 4:
In a class of 30 students, 12 like cricket, 8 like football, and the rest like basketball. If a student is chosen randomly, what is the probability they like (a) cricket (b) basketball?
Answer:
Problem Interpretation

We must calculate the probability of a student liking cricket or basketball.


Mathematical Modeling
  • Total students = 30
  • Basketball lovers = 30 - (12 + 8) = 10
  • (a) P(cricket) = 12/30 = 2/5
  • (b) P(basketball) = 10/30 = 1/3

Solution

We studied such problems in NCERT. The probabilities are (a) 2/5 and (b) 1/3.

Question 5:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. If a ball is drawn at random, find the probability of getting: (a) a red ball (b) a green ball.
Answer:
Problem Interpretation

We need to find the probability of drawing a red or green ball from a bag with given balls.


Mathematical Modeling
  • Total balls = 5 (red) + 3 (blue) + 2 (green) = 10
  • Probability = (Favorable outcomes) / (Total outcomes)

Solution
  • (a) P(red) = 5/10 = 1/2
  • (b) P(green) = 2/10 = 1/5
Question 6:
In a class of 30 students, 12 like cricket, 8 like football, and the rest like basketball. If a student is chosen randomly, what is the probability they like basketball?
Answer:
Problem Interpretation

We studied probability in our textbook. Here, we find the chance of a student liking basketball.


Mathematical Modeling
  • Total students = 30
  • Basketball lovers = 30 - (12 + 8) = 10

Solution

Probability = 10/30 = 1/3. Our textbook shows similar examples with simple fractions.

Question 7:
In a class of 30 students, 12 prefer cricket, 8 prefer football, and the rest prefer badminton. A student is chosen at random. Find the probability that the student prefers badminton.
Answer:
Problem Interpretation

We need to find the probability of a student preferring badminton from a class of 30 students.

Mathematical Modeling
  • Total students = 30
  • Students preferring cricket = 12
  • Students preferring football = 8
  • Students preferring badminton = 30 - (12 + 8) = 10
Solution

Probability = Favourable outcomes / Total outcomes = 10/30 = 1/3.

Question 8:
A bag contains 5 red, 3 blue, and 2 green marbles. A marble is drawn at random. What is the probability that it is not blue?
Answer:
Problem Interpretation

We need to find the probability of drawing a marble that is not blue from the bag.

Mathematical Modeling
  • Total marbles = 5 (red) + 3 (blue) + 2 (green) = 10
  • Non-blue marbles = 5 (red) + 2 (green) = 7
Solution

Probability = Favourable outcomes / Total outcomes = 7/10.

Question 9:
A bag contains 5 red, 3 blue, and 2 green marbles. A marble is drawn at random.
(i) What is the probability of drawing a red marble?
(ii) If one blue marble is removed, what is the new probability of drawing a blue marble?
Answer:
Problem Interpretation

We have a bag with marbles of different colors. We need to find probabilities before and after removing a blue marble.


Mathematical Modeling
  • Total marbles initially: 5 + 3 + 2 = 10
  • After removal: Total marbles = 9

Solution
  • (i) P(Red) = 5/10 = 1/2
  • (ii) P(Blue after removal) = 2/9
Question 10:
In a class of 40 students, 25 like cricket, 15 like football, and 10 like both. A student is selected randomly.
(i) Find the probability that the student likes only cricket.
(ii) Find the probability that the student likes neither sport.
Answer:
Problem Interpretation

We studied probability using Venn diagrams in our textbook. Here, we need to find probabilities based on overlapping preferences.


Mathematical Modeling
  • Only cricket: 25 - 10 = 15
  • Neither: 40 - (15 + 10 + 5) = 10

Solution
  • (i) P(Only cricket) = 15/40 = 3/8
  • (ii) P(Neither) = 10/40 = 1/4
Question 11:

A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag. Based on this scenario, answer the following:

  • What is the probability of drawing a red ball?
  • If one blue ball is removed from the bag, what is the new probability of drawing a blue ball?
Answer:

Total number of balls initially = 5 (red) + 3 (blue) + 2 (green) = 10 balls.


Probability of drawing a red ball = Number of red balls / Total balls = 5/10 = 1/2.


After removing 1 blue ball, total balls = 10 - 1 = 9 balls.


Remaining blue balls = 3 - 1 = 2 blue balls.


New probability of drawing a blue ball = 2/9.


Note: Probability is always between 0 and 1, where 0 means impossible and 1 means certain.

Question 12:

In a class of 30 students, 12 students like cricket, 8 like football, and the rest like basketball. A student is selected at random. Answer the following:

  • What is the probability that the selected student likes basketball?
  • If 2 more students join the class and both like cricket, how does this affect the probability of selecting a cricket lover?
Answer:

Total students = 30.


Students who like basketball = Total students - (Cricket lovers + Football lovers) = 30 - (12 + 8) = 10 students.


Probability of selecting a basketball lover = 10/30 = 1/3.


After adding 2 more cricket lovers, total students = 30 + 2 = 32 students.


New number of cricket lovers = 12 + 2 = 14 students.


New probability of selecting a cricket lover = 14/32 = 7/16.


Note: The probability increased because the number of favorable outcomes (cricket lovers) increased while the total outcomes also increased.

Question 13:

A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag. Based on this, answer the following:

(i) What is the probability of drawing a red ball?

(ii) What is the probability of drawing a ball that is not blue?

Answer:

Total number of balls in the bag = 5 (red) + 3 (blue) + 2 (green) = 10 balls.

(i) Probability of drawing a red ball:


Number of red balls = 5
Total number of balls = 10
Probability = Number of red balls / Total number of balls = 5/10 = 1/2 or 0.5.

(ii) Probability of drawing a ball that is not blue:


Number of non-blue balls = 5 (red) + 2 (green) = 7 balls
Probability = Number of non-blue balls / Total number of balls = 7/10 or 0.7.
Question 14:

In a class survey, 20 students were asked about their favorite sport. The results are as follows: 8 chose cricket, 5 chose football, 4 chose basketball, and 3 chose tennis. If a student is selected at random, find:

(i) The probability that the student's favorite sport is cricket.

(ii) The probability that the student's favorite sport is not football.

Answer:

Total number of students surveyed = 20.

(i) Probability of favorite sport being cricket:


Number of students who chose cricket = 8
Probability = 8/20 = 2/5 or 0.4.

(ii) Probability of favorite sport not being football:


Number of students who did not choose football = 20 - 5 = 15
Probability = 15/20 = 3/4 or 0.75.
Question 15:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. One ball is drawn at random.
(i) Find the probability of drawing a red ball.
(ii) Find the probability of not drawing a blue ball.
Answer:

Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10 balls.


(i) Probability of drawing a red ball:


Number of red balls = 5
Probability = Favorable outcomes / Total outcomes = 5/10 = 1/2.

(ii) Probability of not drawing a blue ball:


Number of non-blue balls = 5 (red) + 2 (green) = 7
Probability = 7/10.

Key concept: Probability ranges from 0 (impossible) to 1 (certain). Here, the events are equally likely.

Question 16:
In a class survey, 18 students like pizza, 15 like burgers, and 7 like both. If a student is chosen at random, find:
(i) The probability that the student likes only pizza.
(ii) The probability that the student likes neither pizza nor burgers.
Answer:

Total students = Students who like only pizza + only burgers + both + neither.


Let’s break it down:


  • Students who like only pizza = 18 (total pizza lovers) - 7 (both) = 11.
  • Students who like only burgers = 15 - 7 = 8.
  • Students who like both = 7 (given).

Assume no student dislikes both (if not specified). Total = 11 + 8 + 7 = 26.


(i) Probability of liking only pizza:


Favorable outcomes = 11
Probability = 11/26.

(ii) Probability of liking neither:


If no data is given for 'neither', we assume all students like at least one. Thus, probability = 0.

Note: Always verify if the problem specifies 'neither' cases. Here, it’s implied all students are surveyed.

Question 17:
In a class of 30 students, 12 are girls. The teacher writes each student's name on a slip and puts them in a box. One slip is drawn at random.

(a) What is the probability that the drawn slip has a girl's name?
(b) If 5 more girls join the class, what will be the new probability of drawing a girl's name?
Answer:

(a) Probability of drawing a girl's name initially:


Total number of students = 30
Number of girls = 12
Probability = Number of favorable outcomes / Total outcomes
P(Girl) = 12/30 = 2/5 or 0.4

(b) New probability after 5 more girls join:


New number of girls = 12 + 5 = 17
New total students = 30 + 5 = 35
P(Girl) = 17/35 ≈ 0.486

Note: Probability is always between 0 and 1. The increase in girls raises the probability of drawing a girl's name.

Question 18:
A bag contains 5 red, 3 blue, and 2 green marbles. A marble is drawn, its color noted, and then replaced. The experiment is repeated 200 times.

(a) Predict how many times a red marble will be drawn.
(b) If 10 more red marbles are added, what will be the new probability of drawing a red marble?
Answer:

(a) Prediction for red marbles drawn:


Total marbles = 5 (red) + 3 (blue) + 2 (green) = 10
Probability of red marble = 5/10 = 0.5
Expected frequency in 200 trials = 0.5 × 200 = 100 times

(b) New probability after adding 10 red marbles:


New red marbles = 5 + 10 = 15
New total marbles = 10 + 10 = 20
P(Red) = 15/20 = 3/4 or 0.75

Note: Replacement ensures each draw is independent. Adding more red marbles increases the probability significantly.

Question 19:
In a school, a survey was conducted among 200 students to find their favorite sport. The results are as follows: Cricket - 80 students, Football - 60 students, Basketball - 40 students, and Others - 20 students. If a student is chosen at random, find the probability that the student's favorite sport is Football or Basketball.
Answer:

To find the probability that a randomly chosen student's favorite sport is Football or Basketball, follow these steps:


Step 1: Identify the total number of students surveyed.
Total students = 200

Step 2: Find the number of students who prefer Football and Basketball.
Football = 60 students
Basketball = 40 students
Total favorable outcomes = 60 + 40 = 100 students

Step 3: Calculate the probability using the formula:
Probability = (Number of favorable outcomes) / (Total number of outcomes)
Probability = 100 / 200 = 0.5

Thus, the probability that the student's favorite sport is Football or Basketball is 0.5 or 50%.

Question 20:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is not red. Show your steps clearly.
Answer:

To find the probability that the drawn ball is not red, follow these steps:


Step 1: Calculate the total number of balls in the bag.
Red balls = 5
Blue balls = 3
Green balls = 2
Total balls = 5 + 3 + 2 = 10

Step 2: Find the number of balls that are not red.
Non-red balls = Blue + Green = 3 + 2 = 5

Step 3: Compute the probability using the formula:
Probability = (Number of favorable outcomes) / (Total number of outcomes)
Probability = 5 / 10 = 0.5

Therefore, the probability that the ball drawn is not red is 0.5 or 50%.

Question 21:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random.

(i) Find the probability of drawing a red ball.
(ii) If one red ball is removed from the bag, what is the new probability of drawing a red ball?
Answer:

(i) Probability of drawing a red ball initially:


Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10.
Number of red balls = 5.
Probability = Number of favorable outcomes / Total outcomes = 5/10 = 1/2.

(ii) Probability after removing one red ball:


Total balls now = 10 - 1 = 9.
Red balls remaining = 5 - 1 = 4.
New probability = 4/9.

Key concept: Probability changes when the total or favorable outcomes are altered.

Question 22:
In a class survey, 20 students like cricket, 15 like football, and 5 like both. A student is selected at random.

(i) Find the probability that the student likes only cricket.
(ii) What is the probability that the student likes neither cricket nor football if the total class strength is 40?
Answer:

(i) Probability of liking only cricket:


Students who like only cricket = Total cricket lovers - Those who like both = 20 - 5 = 15.
Total students = 40.
Probability = 15/40 = 3/8.

(ii) Probability of liking neither:


Students who like either cricket or football or both = 20 + 15 - 5 = 30.
Students who like neither = Total - 30 = 40 - 30 = 10.
Probability = 10/40 = 1/4.

Key concept: Use the principle of inclusion-exclusion for overlapping events.

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