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Probability – CBSE NCERT Study Resources

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10th

10th - Mathematics

Probability

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Overview

Probability is a branch of mathematics that deals with calculating the likelihood of a given event's occurrence, which is expressed as a number between 0 and 1. In this chapter, we explore the foundational concepts of probability, including experiments, outcomes, events, and theoretical vs. experimental probability, as per the CBSE Grade 10 syllabus.

Key Concepts

Basic Terms

  • Experiment: An action that results in one or more outcomes (e.g., rolling a die).
  • Outcome: A possible result of an experiment (e.g., getting a 4 on a die).
  • Sample Space (S): The set of all possible outcomes of an experiment.
  • Event: A subset of the sample space (e.g., getting an even number when rolling a die).

Types of Probability

Theoretical Probability: The ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely. Mathematically, P(E) = Number of favorable outcomes / Total number of outcomes.
Experimental Probability: The probability calculated based on actual experiments or observations. It is given by P(E) = Number of trials where the event occurred / Total number of trials.

Probability of Complementary Events

For an event E, the complementary event is denoted as E' (not E). The probability of the complementary event is given by:

P(E') = 1 - P(E)

Applications

Probability is widely used in real-life scenarios such as weather forecasting, sports strategies, and risk assessment in finance.

Examples

Example 1: Theoretical Probability

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

Solution: P(Red) = Number of red balls / Total balls = 3/8.

Example 2: Experimental Probability

In 50 tosses of a coin, heads appeared 28 times. Find the experimental probability of getting heads.

Solution: P(Heads) = 28/50 = 14/25.

Important Formulas

  • Probability of an event E: P(E) = n(E)/n(S)
  • Probability of complementary event: P(E') = 1 - P(E)

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:
A coin is tossed once. What is the probability of getting heads?
Answer:
1/2
Question 2:
A die is rolled. Find the probability of getting an even number.
Answer:
1/2
Question 3:
A bag has 3 red and 5 blue balls. What is the probability of drawing a red ball?
Answer:
3/8
Question 4:
What is the probability of an impossible event?
Answer:
0
Question 5:
What is the probability of a sure event?
Answer:
1
Question 6:
A card is drawn from a deck of 52. Find the probability of a king.
Answer:
1/13
Question 7:
Two coins are tossed. What is the probability of two tails?
Answer:
1/4
Question 8:
A die is rolled. Find the probability of a number > 4.
Answer:
1/3
Question 9:
A bag has 4 green and 6 yellow marbles. Find the probability of green.
Answer:
2/5
Question 10:
What is the probability of getting a prime number on a die?
Answer:
1/2
Question 11:
A spinner has 8 equal sectors. Find the probability of landing on 3.
Answer:
1/8
Question 12:
From a pack of cards, find the probability of a heart.
Answer:
1/4
Question 13:
Define probability in mathematical 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.
Mathematically, Probability (P) = Number of favorable outcomes / Total number of possible outcomes.

Question 14:
What is the probability of getting a head when tossing a fair coin?
Answer:

For a fair coin, there are two possible outcomes: head or tail.
Number of favorable outcomes (head) = 1.
Total outcomes = 2.
So, P(Head) = 1/2 = 0.5 or 50%.

Question 15:
If a die is rolled once, what is the probability of getting an even number?
Answer:

A die has six faces with numbers 1 to 6.
Even numbers: 2, 4, 6 (total 3).
Total outcomes = 6.
So, P(Even) = 3/6 = 1/2 = 0.5 or 50%.

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

Total balls = 5 (red) + 3 (blue) = 8.
Favorable outcomes (red) = 5.
So, P(Red) = 5/8 = 0.625 or 62.5%.

Question 17:
What is the probability of an impossible event?
Answer:

An impossible event has no favorable outcomes.
So, P(Impossible Event) = 0.

Question 18:
What is the probability of a certain event?
Answer:

A certain event will always occur.
So, the number of favorable outcomes equals the total outcomes.
Thus, P(Certain Event) = 1.

Question 19:
If two coins are tossed simultaneously, what is the probability of getting at least one head?
Answer:

Possible outcomes: HH, HT, TH, TT (total 4).
Favorable outcomes (at least one head): HH, HT, TH (total 3).
So, P(At least one head) = 3/4 = 0.75 or 75%.

Question 20:
A card is drawn from a well-shuffled deck of 52 cards. What is the probability of drawing a king?
Answer:

Total cards = 52.
Number of kings = 4.
So, P(King) = 4/52 = 1/13 ≈ 0.077 or 7.7%.

Question 21:
What is the sum of probabilities of all possible outcomes of an experiment?
Answer:

The sum of probabilities of all possible outcomes of an experiment is always 1.
This is because one of the outcomes must occur.

Question 22:
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: 2, 3, 5, 7 (total 4).
Total sectors = 8.
So, P(Prime) = 4/8 = 1/2 = 0.5 or 50%.

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:
A bag contains 5 red, 8 blue, and 7 green marbles. If one marble is drawn at random, what is the probability that it is not blue?
Answer:

Total marbles = 5 (red) + 8 (blue) + 7 (green) = 20.
Marbles that are not blue = 5 (red) + 7 (green) = 12.
Probability = Favorable outcomes / Total outcomes = 12/20 = 3/5.

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

Prime numbers on a die: 2, 3, 5 (total 3 outcomes).
Total outcomes = 6.
Probability = 3/6 = 1/2.

Question 3:
A card is drawn from a well-shuffled deck of 52 cards. What is the probability of drawing a king?
Answer:

Number of kings in a deck = 4.
Total cards = 52.
Probability = 4/52 = 1/13.

Question 4:
A box contains 10 discs numbered 1 to 10. One disc is drawn at random. Find the probability that the number on the disc is a multiple of 3.
Answer:

Multiples of 3 between 1-10: 3, 6, 9 (total 3 outcomes).
Total discs = 10.
Probability = 3/10.

Question 5:
Two coins are tossed simultaneously. What is the probability of getting at least one head?
Answer:

Possible outcomes: HH, HT, TH, TT (total 4).
Favorable outcomes for at least one head: HH, HT, TH (total 3).
Probability = 3/4.

Question 6:
A jar contains 24 marbles, some green and others blue. If the probability of drawing a green marble is 2/3, how many green marbles are there?
Answer:

Let green marbles = x.
Probability = x/24 = 2/3.
Solving: x = (2 × 24)/3 = 16.

Question 7:
A number is chosen from 1 to 50. Find the probability that it is a perfect square.
Answer:

Perfect squares between 1-50: 1, 4, 9, 16, 25, 36, 49 (total 7).
Total numbers = 50.
Probability = 7/50.

Question 8:
In a lottery, there are 10 prizes and 25 blanks. What is the probability of not winning a prize?
Answer:

Total tickets = 10 (prizes) + 25 (blanks) = 35.
Probability of not winning = Blanks/Total = 25/35 = 5/7.

Question 9:
A bag contains 12 balls, out of which x are black. If 6 more black balls are added, the probability of drawing a black ball doubles. Find x.
Answer:

Initial probability = x/12.
New probability = (x + 6)/18.
Given: (x + 6)/18 = 2 × (x/12).
Solving: x = 3.

Question 10:
A fair die is rolled. What is the probability of getting an even number or a number greater than 4?
Answer:

Even numbers: 2, 4, 6.
Numbers > 4: 5, 6.
Combined favorable outcomes: 2, 4, 5, 6 (total 4).
Probability = 4/6 = 2/3.

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:
In a lottery, there are 10 prizes and 25 blanks. What is the probability of not winning a prize?
Answer:

Total number of lottery tickets = Prizes + Blanks = 10 + 25 = 35.
Number of non-winning tickets (blanks) = 25.
Probability of not winning = Favorable outcomes / Total outcomes = 25/35 = 5/7.
Thus, the probability is 5/7.

Question 2:
A bag contains 5 red, 8 blue, and 7 green marbles. A marble is drawn at random. Find the probability that it is not a blue marble.
Answer:

Total number of marbles = 5 (red) + 8 (blue) + 7 (green) = 20.
Number of marbles that are not blue = 5 (red) + 7 (green) = 12.
Probability of drawing a non-blue marble = Favorable outcomes / Total outcomes = 12/20 = 3/5.
Thus, the probability is 3/5.

Question 3:
Two coins are tossed simultaneously. What is the probability of getting at least one head?
Answer:

Possible outcomes when two coins are tossed: HH, HT, TH, TT (Total = 4).
Outcomes with at least one head: HH, HT, TH (Favorable = 3).
Probability = Favorable outcomes / Total outcomes = 3/4.
Thus, the probability is 3/4.

Question 4:
A die is rolled once. Find the probability of getting a prime number.
Answer:

Total outcomes when a die is rolled: 1, 2, 3, 4, 5, 6 (Total = 6).
Prime numbers on a die: 2, 3, 5 (Favorable = 3).
Probability = Favorable outcomes / Total outcomes = 3/6 = 1/2.
Thus, the probability is 1/2.

Question 5:
A card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing a king or a queen.
Answer:

Total number of cards = 52.
Number of kings = 4.
Number of queens = 4.
Total favorable outcomes = 4 (kings) + 4 (queens) = 8.
Probability = Favorable outcomes / Total outcomes = 8/52 = 2/13.
Thus, the probability is 2/13.

Question 6:
A bag contains 5 red, 8 blue, and 7 green balls. One ball is drawn at random. Find the probability that the ball drawn is not red.
Answer:

Total number of balls = 5 (red) + 8 (blue) + 7 (green) = 20.
Number of balls that are not red = 8 (blue) + 7 (green) = 15.
Probability of drawing a ball that is not red = Favorable outcomes / Total outcomes = 15/20 = 3/4.

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

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

Question 8:
A card is drawn from a well-shuffled deck of 52 cards. Calculate the probability of drawing a king or a queen.
Answer:

Total number of cards = 52.
Number of kings = 4, and number of queens = 4.
Total favorable outcomes = 4 (kings) + 4 (queens) = 8.
Probability = Favorable outcomes / Total outcomes = 8/52 = 2/13.

Question 9:
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 number of 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.

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, 8 blue, and 7 green marbles. One marble is drawn at random. Find the probability that it is (a) red (b) not blue. Justify your answer with steps.
Answer:
Introduction

We studied probability in NCERT Chapter 15. Here, we find the likelihood of drawing marbles of specific colors.


Argument 1
  • Total marbles = 5 (red) + 8 (blue) + 7 (green) = 20.
  • P(red) = Favourable outcomes / Total outcomes = 5/20 = 1/4.

Argument 2
  • P(not blue) = 1 - P(blue) = 1 - (8/20) = 12/20 = 3/5.
  • Our textbook shows similar problems with complementary events.

Conclusion

Thus, probabilities are (a) 1/4 and (b) 3/5, derived using basic definitions.

Question 2:
In a survey of 200 students, 120 liked cricket and 80 liked football. If a student is chosen randomly, find the probability that they like only cricket. Assume no student dislikes both sports.
Answer:
Introduction

We apply probability to real-life surveys, as in NCERT Example 5.


Argument 1
  • Total students = 200.
  • Students liking only cricket = Total cricket lovers - those liking both = 120 - x.

Argument 2
  • Since no student dislikes both, x = 80 (football lovers).
  • Thus, only cricket lovers = 120 - 80 = 40.
  • P(only cricket) = 40/200 = 1/5.

Conclusion

The probability is 1/5, calculated using set theory and probability rules.

Question 3:
A bag contains 5 red, 8 blue, and 7 green marbles. A marble is drawn at random. Find the probability that it is (i) red (ii) not blue. Explain using step notation and basic probability formula.
Answer:
Introduction

We studied probability in Chapter 15. The probability of an event is calculated as P(E) = Number of favorable outcomes / Total outcomes.


Argument 1
  • Total marbles = 5 (red) + 8 (blue) + 7 (green) = 20.
  • (i) P(red) = 5/20 = 1/4.

Argument 2
  • (ii) P(not blue) = 1 - P(blue) = 1 - 8/20 = 12/20 = 3/5.
  • Our textbook shows similar problems in Example 5.

Conclusion

Thus, probabilities are (i) 1/4 and (ii) 3/5. This matches real-life scenarios like drawing colored balls.

Question 4:
Two coins are tossed simultaneously. Find the probability of getting (i) at least one head (ii) no tail. Derive the sample space and justify using NCERT methods.
Answer:
Introduction

We learned that tossing two coins gives outcomes like HH, HT, TH, TT. Probability is calculated using sample space.


Argument 1
  • Sample space = {HH, HT, TH, TT} → 4 outcomes.
  • (i) P(at least one head) = 3/4 (HH, HT, TH).

Argument 2
  • (ii) P(no tail) = P(HH) = 1/4.
  • Our textbook (Example 3) confirms this method.

Conclusion

Hence, probabilities are (i) 3/4 and (ii) 1/4. This applies to games like coin tosses in cricket.

Question 5:
A bag contains 5 red, 8 blue, and 7 green marbles. One marble is drawn at random. Find the probability that it is (a) red (b) not blue. Explain using step notation and basic probability rules.
Answer:
Introduction

We studied probability in Chapter 15 of NCERT. The probability of an event is calculated as favorable outcomes divided by total outcomes.


Argument 1
  • Total marbles = 5 (red) + 8 (blue) + 7 (green) = 20.
  • P(red) = 5/20 = 1/4.

Argument 2
  • P(not blue) = 1 - P(blue) = 1 - 8/20 = 12/20 = 3/5.
  • This follows the complement rule.

Conclusion

Thus, P(red) = 1/4 and P(not blue) = 3/5, as per NCERT examples.

Question 6:
A die is rolled twice. Find the probability that (a) the sum is 9 (b) both numbers are odd. Use tabular representation for possible outcomes and justify with NCERT methods.
Answer:
Introduction

Our textbook shows that a die has 6 outcomes. When rolled twice, total outcomes = 6 × 6 = 36.


Argument 1
  • Sum 9 occurs for (3,6), (4,5), (5,4), (6,3). So, P(sum=9) = 4/36 = 1/9.

Argument 2
  • Odd numbers: 1, 3, 5. Favorable pairs = (1,1), (1,3), ..., (5,5) = 9.
  • P(both odd) = 9/36 = 1/4.

Conclusion

Thus, P(sum=9) = 1/9 and P(both odd) = 1/4, verified using NCERT probability rules.

Question 7:
A bag contains 5 red, 8 blue, and 7 green balls. One ball is drawn at random. Find the probability that it is (i) red (ii) not blue. Explain using step notation and NCERT methods.
Answer:
Introduction

We studied probability in Chapter 15. Total balls = 5R + 8B + 7G = 20.


Argument 1
  • (i) P(Red) = Favourable outcomes/Total = 5/20 = 1/4
  • As per NCERT Example 5, P(E) = n(E)/n(S)

Argument 2
  • (ii) P(Not Blue) = 1 - P(Blue) = 1 - 8/20 = 12/20 = 3/5
  • Real-life: Like picking non-defective items.

Conclusion

Final probabilities are (i) 1/4 (ii) 3/5. Units are pure ratios.

Question 8:
Two coins are tossed simultaneously. List the sample space and find the probability of getting (i) exactly one head (ii) no tail. Validate using tabular representation.
Answer:
Introduction

Our textbook shows sample space S = {HH, HT, TH, TT} for two coins.


Argument 1
  • (i) P(Exactly 1 Head) = 2/4 = 1/2 (HT, TH)
  • Table: [Diagram: 2x2 grid with outcomes]

Argument 2
  • (ii) P(No Tail) = 1/4 (only HH)
  • NCERT Example 4 confirms this method.

Conclusion

Probabilities are (i) 0.5 (ii) 0.25. Results match theoretical expectations.

Question 9:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. If a ball is drawn at random, find the probability that it is (i) red (ii) not blue. Explain using probability formula and complementary events.
Answer:
Introduction

We studied probability as the likelihood of an event happening. Here, total balls = 5 + 3 + 2 = 10.


Argument 1
  • (i) P(red) = Favourable outcomes / Total outcomes = 5/10 = 1/2.

Argument 2
  • (ii) P(not blue) = 1 - P(blue) = 1 - 3/10 = 7/10 (using complementary events).

Conclusion

Our textbook shows similar problems. Thus, probabilities are (i) 0.5 (ii) 0.7.

Question 10:
In a class of 40 students, 25 like cricket and 15 like both cricket and football. If a student is chosen randomly, find the probability they like only cricket. Use Venn diagram logic and set notation.
Answer:
Introduction

We use set theory to solve such problems. Total students = 40.


Argument 1
  • Students liking only cricket = Total cricket lovers - Both = 25 - 15 = 10.

Argument 2
  • P(only cricket) = 10/40 = 1/4. [Diagram: Two overlapping circles for cricket and football]

Conclusion

Our NCERT examples confirm this method. The probability is 0.25.

Question 11:
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 (i) red (ii) not blue. Explain each step with reasoning.
Answer:

To solve this problem, we first determine the total number of balls in the bag and then apply the basic probability formula.


Total number of balls = Number of red balls + Number of blue balls + Number of green balls
= 5 + 3 + 2
= 10


(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 ball that is not blue:
Number of favorable outcomes (not blue) = Total balls - Blue balls
= 10 - 3
= 7
Probability = 7/10 or 0.7


Key takeaway: Always verify that the probability lies between 0 and 1, and ensure the total of all probabilities sums to 1 for exhaustive events.

Question 12:
Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is (i) 9 (ii) a perfect square. Justify your answer with proper calculations.
Answer:

When two dice are rolled, the total number of possible outcomes is 6 × 6 = 36 (since each die has 6 faces).


(i) Probability that the sum is 9:
Favorable outcomes where the sum is 9:
(3,6), (4,5), (5,4), (6,3) → Total 4 outcomes
Probability = Number of favorable outcomes / Total possible outcomes
= 4/36
= 1/9


(ii) Probability that the sum is a perfect square:
Perfect squares between 2 and 12 (minimum and maximum sums of two dice) are 4, 9.

  • For sum = 4: (1,3), (2,2), (3,1) → 3 outcomes
  • For sum = 9: 4 outcomes (as above)
Total favorable outcomes = 3 + 4 = 7
Probability = 7/36


Note: Listing all possible outcomes ensures accuracy. Cross-verify by checking the exhaustive cases.

Question 13:
A card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing (i) a king (ii) a red queen (iii) neither a heart nor a king. Provide a step-by-step explanation with reasoning.
Answer:

In a standard deck of 52 cards, there are 4 suits (Hearts, Diamonds, Clubs, Spades), each with 13 cards.


(i) Probability of drawing a king:
Number of kings in the deck = 4 (one from each suit)
Probability = 4/52
= 1/13


(ii) Probability of drawing a red queen:
Red suits are Hearts and Diamonds.
Number of red queens = 2 (Queen of Hearts and Queen of Diamonds)
Probability = 2/52
= 1/26


(iii) Probability of drawing neither a heart nor a king:

  • Total hearts = 13
  • Total kings = 4, but 1 king is already included in hearts (King of Hearts).
  • Number of unfavorable cards = Hearts + Kings - King of Hearts = 13 + 4 - 1 = 16
Number of favorable cards = Total cards - Unfavorable cards = 52 - 16 = 36
Probability = 36/52
= 9/13


Tip: For such problems, visualize the deck or use Venn diagrams to avoid overlapping errors.

Question 14:
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 (i) red (ii) not green. Explain each step clearly.
Answer:

To solve this problem, we first determine the total number of balls in the bag and then use the basic formula for probability:


Probability = Number of favorable outcomes / Total number of possible outcomes


Step 1: Calculate the total number of balls


Total balls = Red balls + Blue balls + Green balls = 5 + 3 + 2 = 10 balls.


Step 2: Find the probability of drawing a red ball


Number of red balls = 5


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


Step 3: Find the probability of drawing a ball that is not green


Number of non-green balls = Red balls + Blue balls = 5 + 3 = 8


Probability (Not Green) = 8/10 = 4/5 or 0.8.


Thus, the probabilities are:


  • (i) P(Red) = 1/2
  • (ii) P(Not Green) = 4/5
Question 15:
Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is (i) 9 (ii) a perfect square. Justify your answer with proper reasoning.
Answer:

When two dice are rolled, the total number of possible outcomes is 6 × 6 = 36 (since each die has 6 faces).


Step 1: Find the probability that the sum is 9


Favorable outcomes where the sum is 9:


  • (3, 6), (4, 5), (5, 4), (6, 3) → 4 outcomes

Probability (Sum = 9) = 4/36 = 1/9.


Step 2: Find the probability that the sum is a perfect square


Perfect squares between 2 and 12 (minimum and maximum sums of two dice) are 4 and 9.


Favorable outcomes for sum = 4:


  • (1, 3), (2, 2), (3, 1) → 3 outcomes

Favorable outcomes for sum = 9:


  • (3, 6), (4, 5), (5, 4), (6, 3) → 4 outcomes

Total favorable outcomes = 3 + 4 = 7


Probability (Sum is a perfect square) = 7/36.


Thus, the probabilities are:


  • (i) P(Sum = 9) = 1/9
  • (ii) P(Sum is a perfect square) = 7/36
Question 16:
A card is drawn from a well-shuffled deck of 52 playing cards. Find the probability that the card drawn is (i) a king (ii) a spade or a queen. Provide a step-by-step explanation.
Answer:

A standard deck has 52 cards divided into 4 suits: Hearts, Diamonds, Clubs, and Spades. Each suit has 13 cards.


Step 1: Find the probability of drawing a king


Number of kings in the deck = 4 (one from each suit)


Probability (King) = 4/52 = 1/13.


Step 2: Find the probability of drawing a spade or a queen


This is a case of non-mutually exclusive events because a card can be both a spade and a queen (Queen of Spades).


Number of spades = 13


Number of queens = 4


Number of cards that are both spade and queen = 1 (Queen of Spades)


Using the formula: P(A or B) = P(A) + P(B) - P(A and B)


Probability (Spade or Queen) = (13/52) + (4/52) - (1/52) = 16/52 = 4/13.


Thus, the probabilities are:


  • (i) P(King) = 1/13
  • (ii) P(Spade or Queen) = 4/13
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) If the red ball drawn is not replaced, find the probability of drawing a blue ball next.
(iii) Compare the probabilities in (i) and (ii) and explain the change, if any.
Answer:

(i) Probability of drawing a red ball initially:


Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10.
Probability (P) of drawing a red ball = Number of red balls / Total balls = 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 balls left = 10 - 1 = 9.
Number of blue balls remains unchanged = 3.
Probability (P) of drawing a blue ball now = 3/9 = 1/3 ≈ 0.333.

(iii) Comparison and explanation:


  • Initial probability of drawing a red ball (1/2) is higher than the subsequent probability of drawing a blue ball (1/3).
  • This change occurs because the total number of balls decreases after the first draw, altering the sample space.
  • The probability of drawing a blue ball increases relative to the remaining red balls but decreases in absolute terms due to the reduced total.

Key takeaway: Probabilities in sequential events without replacement depend on the updated sample space, demonstrating the dependent nature of such events.

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 the red ball drawn is not replaced, find the probability of drawing a blue ball next.
(br) Explain the concept of dependent events in this context.
Answer:

Solution:

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

(ii) After drawing one red ball (not replaced), total remaining balls = 10 - 1 = 9.
Number of blue balls remains unchanged = 3.
Probability of drawing a blue ball now = 3/9 = 1/3.

Explanation of dependent events:
In this scenario, the probability of drawing a blue ball depends on the earlier event of drawing a red ball (since the total number of balls reduced).
Such events, where the outcome of one affects the probability of the other, are called dependent events.

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) If the red ball drawn is not replaced, find the probability of drawing a blue ball next.
(iii) Compare the probabilities in (i) and (ii) and explain the change, if any.
Answer:

(i) Probability of drawing a red ball initially:


Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10.
Number of favorable outcomes (red balls) = 5.
Probability = Favorable outcomes / Total 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 balls left = 10 - 1 = 9.
Number of blue balls remains unchanged = 3.
Probability = 3/9 = 1/3 ≈ 0.333.

(iii) Comparison and explanation:


  • Initial probability of red ball (1/2) is higher than the subsequent probability of blue ball (1/3).
  • This change occurs because the total number of balls decreases after the first draw, altering the probability space.
  • The probability of drawing a blue ball increases slightly compared to its initial probability (3/10 = 0.3) because one red ball (a non-blue ball) is removed.

Value-added insight: This problem demonstrates the concept of dependent events in probability, where the outcome of the first event affects the second event's probability.

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) If the red ball drawn is not replaced, find the probability of drawing a blue ball next.
(br> Explain each step clearly and justify your answer with reasoning.
Answer:

To solve this problem, we will use the basic concept of probability, which is defined as the likelihood of an event occurring. The formula for probability is:


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


Part (i): Probability of drawing a red ball


1. Total number of balls in the bag = 5 (red) + 3 (blue) + 2 (green) = 10 balls.
2. Number of red balls = 5.
3. Probability of drawing a red ball (Pred) = Number of red balls / Total balls = 5/10 = 1/2 or 0.5.


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


1. After drawing one red ball, total remaining balls = 10 - 1 = 9 balls.
2. Number of blue balls remains unchanged at 3.
3. Probability of drawing a blue ball (Pblue) = Number of blue balls / Remaining total balls = 3/9 = 1/3.


Conclusion: The probability of drawing a red ball initially is 1/2, and the probability of drawing a blue ball next (without replacement) is 1/3.

Question 21:
Two dice are rolled simultaneously. Find the probability that:

(i) The sum of the numbers on the two dice is 8.
(ii) The numbers on both dice are even.

Show all possible outcomes and explain your reasoning step-by-step.
Answer:

When two dice are rolled, the total number of possible outcomes is 6 × 6 = 36 (since each die has 6 faces).


Part (i): Probability that the sum is 8


1. Favorable outcomes where the sum is 8:
(2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes.
2. Probability (Psum=8) = Favorable outcomes / Total outcomes = 5/36.


Part (ii): Probability that both numbers are even


1. Even numbers on a die: 2, 4, 6 → 3 choices per die.
2. Total favorable outcomes = 3 (for first die) × 3 (for second die) = 9 outcomes.
Possible pairs: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6).
3. Probability (Pboth even) = 9/36 = 1/4.


Conclusion: The probability of the sum being 8 is 5/36, and the probability of both numbers being even is 1/4.

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:
A bag contains 5 red, 3 blue, and 2 green marbles. A marble is drawn at random.
(i) Find the probability of drawing a red marble.
(ii) If the drawn marble is not green, what is the probability it is blue?
Answer:
Problem Interpretation

We must find probabilities for two scenarios using total marbles.


Mathematical Modeling
  • Total marbles = 5 + 3 + 2 = 10
  • (i) P(Red) = Red marbles/Total = 5/10 = 1/2
  • (ii) P(Blue|Not Green) = Blue/(Total - Green) = 3/8

Solution

(i) 1/2 (2m) (ii) 3/8 (2m). Our textbook shows similar problems.

Question 2:
In a class of 30 students, 18 like cricket, 12 like football, and 5 like both. A student is chosen randomly.
(i) Find P(liking only cricket).
(ii) Find P(not liking either sport).
Answer:
Problem Interpretation

We use set theory to find probabilities of preferences.


Mathematical Modeling
  • Only cricket = 18 - 5 = 13
  • Neither = 30 - (13 + 5 + 7) = 5

Solution

(i) P(Only cricket) = 13/30 (2m). (ii) P(Neither) = 5/30 = 1/6 (2m). We studied this in NCERT.

Question 3:
A die is rolled twice.
(i) Find P(sum is 9).
(ii) If the first roll shows 4, what is P(sum > 7)?
Answer:
Problem Interpretation

We calculate probabilities for two dice events.


Mathematical Modeling
  • (i) Favorable outcomes for sum=9: (3,6), (4,5), (5,4), (6,3) → 4/36 = 1/9
  • (ii) If first die=4, favorable sums: (4,4), (4,5), (4,6) → 3/6 = 1/2

Solution

(i) 1/9 (2m). (ii) 1/2 (2m). NCERT examples guided this solution.

Question 4:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. If a ball is drawn at random, find the probability that it is: (i) Red (ii) 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
  • (i) P(Red) = Favourable outcomes / Total outcomes = 5/10 = 1/2
  • (ii) P(Not green) = 1 - P(Green) = 1 - (2/10) = 4/5

Solution

Our textbook shows similar problems. The probabilities are (i) 1/2 and (ii) 4/5.

Question 5:
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 must find the probability of selecting a student who likes basketball.


Mathematical Modeling
  • Total students = 30
  • Basketball lovers = 30 - (12 + 8) = 10
  • P(Basketball) = 10/30 = 1/3

Solution

We studied such cases in NCERT. The probability is 1/3.

Question 6:
A bag contains 5 red, 3 blue, and 2 green marbles. A marble is drawn at random.
(i) What is the probability it is not blue?
(ii) If two marbles are drawn successively without replacement, find the probability both are red.
Answer:
Problem Interpretation

We studied probability in NCERT. Total marbles = 5 + 3 + 2 = 10.


Mathematical Modeling
  • (i) P(not blue) = 1 - P(blue) = 1 - 3/10 = 7/10
  • (ii) P(both red) = (5/10) × (4/9) = 20/90 = 2/9

Solution

Our textbook shows similar problems. (i) 7/10 (ii) 2/9.

Question 7:
In a class, 40% students like cricket, 30% like football, and 10% like both. A student is selected randomly.
(i) Find the probability they like only cricket.
(ii) What is the chance they like neither sport?
Answer:
Problem Interpretation

We use set theory from NCERT. Let total students = 100.


Mathematical Modeling
  • (i) P(only cricket) = 40% - 10% = 30% = 0.3
  • (ii) P(neither) = 100% - (30 + 20 + 10)% = 40% = 0.4

Solution

Our textbook shows Venn diagrams. (i) 0.3 (ii) 0.4.

Question 8:
A bag contains 5 red balls, 3 blue balls, and 2 green balls. If a ball is drawn at random, find the probability that it is (a) red (b) not blue. Show steps.
Answer:
Problem Interpretation

We need to find the probability of drawing a red ball and a non-blue 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 blue) = (Total - Blue) / Total = (10 - 3)/10 = 7/10

Solution

Our textbook shows similar examples. The probabilities are (a) 1/2 and (b) 7/10.

Question 9:
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

As per NCERT examples, the probabilities are (a) 2/5 and (b) 1/3.

Question 10:
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: (i) Cricket (ii) 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
  • (i) P(Cricket) = Cricket lovers / Total
  • (ii) P(Basketball) = Basketball lovers / Total

Solution

(i) P(Cricket) = 12/30 = 2/5. (ii) P(Basketball) = 10/30 = 1/3. We studied such problems in NCERT.

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 become 9.


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


Thus, the probabilities are 1/2 and 2/9 respectively.

Question 12:

In a class survey, 30 students were asked about their favorite subject. The results showed 12 preferred Mathematics, 8 preferred Science, and 10 preferred Social Studies. If a student is chosen at random:

  • Find the probability that the student prefers Mathematics.
  • If 5 more students join the survey and all prefer Mathematics, what is the updated probability?
Answer:

Total students initially = 30.


Probability of preferring Mathematics = Number of Mathematics lovers / Total students = 12/30 = 2/5.


After adding 5 more students who prefer Mathematics, total students become 35, and Mathematics lovers become 17.


Updated probability = 17/35.


Thus, the probabilities are 2/5 and 17/35 respectively.

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 scenario, answer the following:

  • What is the probability of drawing a red ball?
  • If a blue ball is drawn and not replaced, what is the new probability of drawing a green 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 or 0.5.

If a blue ball is drawn and not replaced, total remaining balls = 10 - 1 = 9 balls.

Number of green balls remains unchanged (2).

New probability of drawing a green ball = 2/9 ≈ 0.222 (or 22.2%).

Question 14:

In a class survey, 30 students were asked about their favorite sport. The results were: 12 chose cricket, 8 chose football, and 10 chose basketball. If one student is selected at random, find:

  • The probability that the student's favorite sport is cricket.
  • The probability that the student's favorite sport is not football.
Answer:

Total number of students surveyed = 30.

Probability of favorite sport being cricket = Number of cricket lovers / Total students = 12/30 = 2/5 or 0.4.

Number of students who did not choose football = Total students - Football lovers = 30 - 8 = 22.

Probability of favorite sport not being football = 22/30 = 11/15 ≈ 0.733 (or 73.3%).

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 one red ball is removed from the bag, find the new probability of drawing a red ball.
Answer:

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


(i) Probability of drawing a red ball:


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

(ii) After removing one red ball:


New number of red balls = 5 - 1 = 4
Total balls now = 10 - 1 = 9
Probability = 4/9.

Thus, the new probability of drawing a red ball is 4/9.

Question 16:
In a class of 40 students, 25 like cricket, 15 like football, and 10 like both. A student is selected at random.
(i) Find the probability that the student likes only cricket.
(ii) Find the probability that the student likes neither cricket nor football.
Answer:

Total students = 40.


(i) Probability of liking only cricket:


Students who like cricket (including both) = 25
Students who like both = 10
Students who like only cricket = 25 - 10 = 15
Probability = 15/40 = 3/8.

(ii) Probability of liking neither:


Students who like cricket or football or both = 25 (cricket) + 15 (football) - 10 (both) = 30
Students who like neither = Total students - Students who like either = 40 - 30 = 10
Probability = 10/40 = 1/4.

Thus, the probability that the student likes neither sport is 1/4.

Question 17:
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. A student is chosen at random.

Based on this data, answer the following:

(i) What is the probability that the student's favorite sport is Cricket? (ii) What is the probability that the student's favorite sport is not Football?
Answer:

(i) Probability that the student's favorite sport is Cricket:


Total number of students = 200
Number of students who prefer Cricket = 80
Probability = Number of favorable outcomes / Total number of outcomes
Probability = 80 / 200 = 2/5 or 0.4

(ii) Probability that the student's favorite sport is not Football:


Number of students who do not prefer Football = Total students - Football lovers
Number of students who do not prefer Football = 200 - 60 = 140
Probability = 140 / 200 = 7/10 or 0.7

Note: Probabilities are always between 0 and 1, inclusive. These results show that Cricket is preferred by 40% of students, while 70% do not prefer Football.

Question 18:
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:

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

(i) Probability that the ball drawn is red:


Total number of balls = 5 (red) + 3 (blue) + 2 (green) = 10
Number of red balls = 5
Probability = Number of red balls / Total number of balls
Probability = 5 / 10 = 1/2 or 0.5

(ii) New probability after removing one red ball:


Total number of balls now = 10 - 1 = 9
Number of red balls now = 5 - 1 = 4
Probability = 4 / 9 ≈ 0.444

Note: Removing a red ball decreases the probability of drawing a red ball from 0.5 to approximately 0.444, as the total number of favorable outcomes and total outcomes change.

Question 19:
A bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. Based on this, answer the following:

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


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


(ii) After removing 1 blue ball, total balls = 10 - 1 = 9.
Number of blue balls remaining = 3 - 1 = 2.
New probability of drawing a blue ball = 2/9.

Question 20:
In a class survey, 20 students like cricket, 15 like football, and 5 like both. If a student is chosen at random, find:

(i) The probability that the student likes only cricket.
(ii) The probability that the student likes neither cricket nor football.
Answer:

Let C represent students who like cricket, and F represent those who like football.


Total students = Students who like only cricket + only football + both + neither.
But since 'neither' is not given, assume total students = 20 (cricket) + 15 (football) - 5 (both) = 30 students.


(i) Students who like only cricket = Total cricket lovers - both = 20 - 5 = 15.
Probability = 15/30 = 1/2.


(ii) Students who like neither = Total students - (only cricket + only football + both) = 30 - (15 + 10 + 5) = 0.
Probability = 0/30 = 0 (since no student dislikes both).

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