Probability | Grade 11th - Mathematics Chapter Summary & NCERT CBSE Study Resources

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11th

11th - Mathematics

Probability

Overview of the Chapter: Probability

This chapter introduces the fundamental concepts of probability, which is a measure of the likelihood that an event will occur. Probability is widely used in various fields such as mathematics, statistics, science, and engineering. The chapter covers basic definitions, types of events, and different approaches to calculating probabilities.

Probability: Probability is a numerical measure of the likelihood of occurrence of an event. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

Key Topics Covered

Random Experiments and Sample Space
Events: Types of Events (Simple, Compound, Mutually Exclusive, Exhaustive, Independent)
Axiomatic Approach to Probability
Probability of Equally Likely Outcomes
Addition Theorem on Probability

Detailed Explanation

Random Experiments and Sample Space

A random experiment is an experiment where all possible outcomes are known, but the exact outcome is unpredictable. The set of all possible outcomes is called the sample space, denoted by S.

Sample Space (S): The collection of all possible outcomes of a random experiment.

Events

An event is a subset of the sample space. Events can be classified into different types:

Simple Event: An event with a single outcome.
Compound Event: An event with more than one outcome.
Mutually Exclusive Events: Events that cannot occur simultaneously.
Exhaustive Events: Events whose union covers the entire sample space.
Independent Events: Events where the occurrence of one does not affect the occurrence of the other.

Axiomatic Approach to Probability

The axiomatic approach defines probability based on three axioms:

For any event A, P(A) ≥ 0.
P(S) = 1.
If A and B are mutually exclusive events, then P(A ∪ B) = P(A) + P(B).

Probability of Equally Likely Outcomes

If all outcomes of a sample space are equally likely, the probability of an event A is given by:

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

Addition Theorem on Probability

For any two events A and B, the probability of their union is given by:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Conclusion

Probability is a fundamental concept in mathematics that helps quantify uncertainty. This chapter provides the foundational knowledge required to understand and apply probability in various real-world scenarios.

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 sample space in probability.

Answer:

Set of all possible outcomes of an experiment.

Question 2:

What is the probability of an impossible event?

Answer:

Numeric answer
0

Question 3:

If P(A) = 0.3, find P(A').

Answer:

Numeric answer
0.7

Question 4:

State the formula for conditional probability P(A|B).

Answer:

P(A|B) = P(A∩B) / P(B)

Question 5:

Two coins are tossed. Find P(exactly one head).

Answer:

Numeric answer
0.5

Question 6:

If A and B are independent, what is P(A∩B)?

Answer:

P(A) × P(B)

Question 7:

A die is rolled. Find P(prime number).

Answer:

Numeric answer
0.5

Question 8:

Define mutually exclusive events.

Answer:

Events that cannot occur simultaneously.

Question 9:

If P(A∪B) = 0.8 and P(A) = 0.5, find P(B) when A, B are independent.

Answer:

Numeric answer
0.6

Question 10:

A card is drawn from a deck. Find P(king or heart).

Answer:

Numeric answer
4/13

Question 11:

Prove P(A') = 1 - P(A) using axioms.

Answer:

A and A' are complementary events.

Question 12:

If P(A) = 0.4, P(B) = 0.5, P(A∩B) = 0.2, are A and B independent?

Answer:

Yes, since P(A∩B) = P(A)×P(B).

Question 13:

If P(A) = 0.3, what is P(A') where A' is the complement of event A?

Answer:

P(A') = 1 - P(A)
= 1 - 0.3
= 0.7.

Question 14:

A die is rolled once. What is the probability of getting a prime number?

Answer:

Prime numbers on a die: {2, 3, 5}
Total outcomes = 6
Probability = 3/6 = 0.5.

Question 15:

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

Answer:

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

Question 16:

If P(A ∪ B) = 0.8 and P(A) = 0.5, find P(B) if A and B are mutually exclusive.

Answer:

For mutually exclusive events, P(A ∪ B) = P(A) + P(B)
0.8 = 0.5 + P(B)
P(B) = 0.3.

Question 17:

What is the probability of a sure event?

Answer:

The probability of a sure event is 1 because it is certain to occur.

Question 18:

A card is drawn from a deck of 52 cards. What is the probability it is a red king?

Answer:

Total red kings = 2 (King of Hearts, King of Diamonds)
Probability = 2/52 = 1/26.

Question 19:

If P(A) = 0.4 and P(B) = 0.5, and A and B are independent, find P(A ∩ B).

Answer:

For independent events, P(A ∩ B) = P(A) × P(B)
= 0.4 × 0.5
= 0.2.

Question 20:

A bag contains 5 red and 7 blue balls. One ball is drawn at random. What is the probability it is blue?

Answer:

Total balls = 5 + 7 = 12
Probability of drawing a blue ball = 7/12.

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:

Define sample space in probability with an example.

Answer:

The sample space is the set of all possible outcomes of a random experiment.
For example, when tossing a coin, the sample space is {Heads, Tails}.

Question 2:

If P(A) = 0.3 and P(B) = 0.5, find P(A or B) if A and B are mutually exclusive events.

Answer:

For mutually exclusive events, P(A or B) = P(A) + P(B).
So, P(A or B) = 0.3 + 0.5 = 0.8.

Question 3:

Define independent events in probability.

Answer:

Two events are independent if the occurrence of one does not affect the probability of the other.
Mathematically, P(A ∩ B) = P(A) × P(B).

Question 4:

A bag contains 4 red and 6 blue balls. One ball is drawn at random. Find the probability of drawing a red ball.

Answer:

Total balls = 4 (red) + 6 (blue) = 10.
Probability of red ball = 4/10 = 2/5.

Question 5:

If P(A) = 0.4, find P(A'), where A' is the complement of event A.

Answer:

P(A') = 1 - P(A) = 1 - 0.4 = 0.6.

Question 6:

Explain the difference between mutually exclusive and exhaustive events.

Answer:

Mutually exclusive events cannot occur simultaneously (P(A ∩ B) = 0).
Exhaustive events cover all possible outcomes (P(A) + P(B) + ... = 1).

Question 7:

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

Answer:

Number of kings = 4, queens = 4.
Total favorable = 8.
Probability = 8/52 = 2/13.

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:

Differentiate between mutually exclusive and independent events in probability.

Answer:

Mutually Exclusive Events: Two events are mutually exclusive if they cannot occur simultaneously.
Example: Getting a Head and a Tail in a single coin toss.

Independent Events: Two events are independent if the occurrence of one does not affect the probability of the other.
Example: Tossing a coin and rolling a die are independent events.

Key difference: Mutually exclusive events cannot happen together, while independent events can occur simultaneously without influencing each other.

Question 2:

Explain the concept of conditional probability with a real-life example.

Answer:

Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), meaning the probability of event A given event B.

Example: The probability of a student passing an exam (Event A) given that they studied regularly (Event B).

Formula: P(A|B) = P(A ∩ B) / P(B), provided P(B) ≠ 0.

Conditional probability helps in making informed decisions based on prior information.

Question 3:

A bag contains 5 red and 3 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are red.

Answer:

Step 1: Total balls = 5 (red) + 3 (blue) = 8.

Step 2: Probability of first ball being red = 5/8.

Step 3: After drawing one red ball, remaining red balls = 4, total balls = 7.

Step 4: Probability of second ball being red = 4/7.

Step 5: Combined probability = (5/8) × (4/7) = 20/56 = 5/14.

Thus, the probability that both balls are red is 5/14.

Question 4:

State and prove the addition theorem of probability for two events.

Answer:

Addition Theorem: For any two events A and B, the probability of occurrence of at least one of them is given by:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Proof:
1. A ∪ B can be written as the union of two disjoint sets: A and (B − A ∩ B).
2. So, P(A ∪ B) = P(A) + P(B − A ∩ B).
3. But P(B − A ∩ B) = P(B) − P(A ∩ B).
4. Therefore, P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

This theorem is fundamental for calculating probabilities of combined events.

Question 5:

Define sample space in probability with an example of rolling a die.

Answer:

The sample space is the set of all possible outcomes of a random experiment. For example, when rolling a fair six-sided die, the sample space S is:
S = {1, 2, 3, 4, 5, 6}.

Each outcome is equally likely, and the sample space helps in calculating probabilities of events, such as getting an even number.

Question 6:

Differentiate between mutually exclusive and independent events with examples.

Answer:

Mutually exclusive events cannot occur simultaneously, while independent events do not affect each other's probability.

Mutually exclusive example: Getting a head or tail in a single coin toss (both cannot happen at once).
Independent example: Rolling a die and tossing a coin (the outcome of one does not influence the other).

Question 7:

A bag contains 5 red and 3 blue balls. If one ball is drawn at random, find the probability of getting a blue ball.

Answer:

Total balls = 5 (red) + 3 (blue) = 8.
Number of favorable outcomes (blue balls) = 3.
Probability = Favorable outcomes / Total outcomes = 3/8.

Thus, the probability of drawing a blue ball is 3/8.

Question 8:

Explain the law of total probability with a simple scenario.

Answer:

The law of total probability states that if B₁, B₂, ..., B_n are mutually exclusive and exhaustive events, then for any event A:
P(A) = P(A|B₁)P(B₁) + P(A|B₂)P(B₂) + ... + P(A|B_n)P(B_n).

Example: A factory has two machines producing bulbs. Machine 1 makes 60% of bulbs with a 2% defect rate, and Machine 2 makes 40% with a 3% defect rate. The total probability of a defective bulb is:
P(Defective) = (0.02 × 0.6) + (0.03 × 0.4) = 0.024.

Question 9:

If P(A) = 0.4 and P(B) = 0.5, and A and B are independent events, find P(A ∩ B).

Answer:

For independent events, P(A ∩ B) = P(A) × P(B).
Given: P(A) = 0.4 and P(B) = 0.5.
Thus, P(A ∩ B) = 0.4 × 0.5 = 0.2.

Therefore, the probability of both A and B occurring is 0.2.

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, 4 blue, and 3 green balls. Two balls are drawn at random without replacement. Calculate the probability that both balls are of the same color. Show the derivation step-by-step.

Answer:

Theoretical Framework

We studied that probability of an event is the ratio of favorable outcomes to total possible outcomes. Here, we use combinations since order doesn't matter.

Evidence Analysis

Total balls = 5 (R) + 4 (B) + 3 (G) = 12
Total ways to draw 2 balls: C(12,2) = 66
Same color cases:
- Both Red: C(5,2) = 10
- Both Blue: C(4,2) = 6
- Both Green: C(3,2) = 3
Favorable outcomes = 10 + 6 + 3 = 19

Critical Evaluation

Probability = 19/66 ≈ 0.2879. Our textbook shows similar problems validating the combination approach for unordered selection.

Question 2:

Prove that for any two independent events A and B, P(A ∪ B) = P(A) + P(B) - P(A)P(B). Derive this using the definition of independence.

Answer:

Theoretical Framework

We know P(A ∪ B) = P(A) + P(B) - P(A ∩ B) for any events. For independent events, P(A ∩ B) = P(A)P(B).

Evidence Analysis

Given A and B are independent ⇒ P(A ∩ B) = P(A) × P(B)
Substitute into union formula: P(A ∪ B) = P(A) + P(B) - P(A)P(B)
Example: If P(A)=0.4, P(B)=0.5, then P(A ∪ B) = 0.4 + 0.5 - (0.4×0.5) = 0.7

Critical Evaluation

The derivation matches our textbook's theorem on independent events. This formula is crucial for calculating combined probabilities in real-world scenarios like quality testing.

Question 3:

A bag contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random without replacement. Calculate the probability that both balls are of the same color. Justify your answer using conditional probability and combinatorial principles.

Answer:

Theoretical Framework

We studied that probability of an event is the ratio of favorable outcomes to total outcomes. Here, we use combinations to calculate total ways of drawing 2 balls from 12.

Evidence Analysis

Total outcomes: C(12,2) = 66.
Favorable outcomes: C(5,2) + C(4,2) + C(3,2) = 10 + 6 + 3 = 19.
Probability = 19/66 ≈ 0.2879.

Critical Evaluation

Conditional probability ensures accuracy since the second draw depends on the first. Our textbook shows this aligns with the multiplication rule for dependent events.

Question 4:

Prove that for any two independent events A and B, P(A ∪ B) = P(A) + P(B) - P(A)P(B). Derive this using the axioms of probability and explain its significance in real-world scenarios.

Answer:

Theoretical Framework

We studied the addition rule for probability: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). For independent events, P(A ∩ B) = P(A)P(B).

Evidence Analysis

Substitute P(A ∩ B) into the addition rule.
Derivation: P(A ∪ B) = P(A) + P(B) - P(A)P(B).

Critical Evaluation

This formula is crucial in real-world modeling, like predicting overlapping risks in finance. Our textbook shows it simplifies calculations for independent systems.

Question 5:

A bag contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random without replacement. Find the probability that both balls are of the same color. Justify your answer using the addition and multiplication rules of probability.

Answer:

Theoretical Framework

We studied that probability of an event is calculated using the formula P(E) = Number of favorable outcomes / Total outcomes. Here, we apply the multiplication rule for dependent events since balls are drawn without replacement.

Evidence Analysis

Total balls = 5 (red) + 4 (blue) + 3 (green) = 12.
P(Both red) = (5/12) × (4/11) = 20/132.
P(Both blue) = (4/12) × (3/11) = 12/132.
P(Both green) = (3/12) × (2/11) = 6/132.

Critical Evaluation

Using the addition rule, total probability = 20/132 + 12/132 + 6/132 = 38/132 = 19/66. Our textbook confirms this method for combined events.

Question 6:

Prove that the probability of the complementary event of E is given by P(E') = 1 - P(E). Derive this using the axiomatic approach to probability and illustrate with an example of a fair die roll.

Answer:

Theoretical Framework

We learned the axiomatic approach defines probability as P(E) + P(E') = 1, where E' is the complementary event. This is derived from the sample space S = E ∪ E' and P(S) = 1.

Evidence Analysis

For a fair die, let E = {1, 3, 5} (odd numbers).
P(E) = 3/6 = 0.5.
E' = {2, 4, 6} (even numbers).
P(E') = 3/6 = 0.5 = 1 - 0.5.

Critical Evaluation

The derivation aligns with the axiom P(S) = 1. Our textbook uses this to validate the complement rule, which simplifies complex probability calculations.

Question 7:

Define conditional probability and derive its formula using the multiplication theorem. Illustrate with an example from NCERT.

Answer:

Theoretical Framework

Conditional probability measures the likelihood of event A given event B has occurred, denoted as P(A|B). Our textbook shows it relies on the intersection of A and B relative to B's probability.

Evidence Analysis

From multiplication theorem: P(A∩B) = P(A|B) × P(B)
Rearranged: P(A|B) = P(A∩B)/P(B) (derivation shown)

Critical Evaluation

Example: A bag has 3 red/2 blue balls. Probability of drawing red after one blue is removed becomes P(Red|Blue removed) = 3/4, demonstrating dependence.

Question 8:

Prove the law of total probability for mutually exclusive events A₁, A₂, A₃ partitioning sample space S. Validate with a real-world scenario.

Answer:

Theoretical Framework

The law states P(B) = ΣP(B|Aᵢ)P(Aᵢ) for events partitioning S. We studied this as foundational for Bayes' Theorem.

Evidence Analysis

Proof: Since Aᵢ are mutually exclusive and exhaustive, B = ∪(B∩Aᵢ). Thus, P(B) = ΣP(B∩Aᵢ) = ΣP(B|Aᵢ)P(Aᵢ)

Future Implications

Real-world: Weather prediction uses partitions like rainy/sunny days to calculate overall probability of flight delays (P(Delay) = P(Delay|Rain)P(Rain) + ...).

Question 9:

Compare independent vs dependent events using Venn diagrams. Solve: If P(A)=0.4, P(B)=0.5, and P(A∪B)=0.7, are A and B independent?

Answer:

Theoretical Framework

Independent events satisfy P(A∩B)=P(A)P(B). Dependent events violate this due to conditional influence.

Evidence Analysis

Given P(A∪B)=P(A)+P(B)-P(A∩B), we find P(A∩B)=0.2
But P(A)P(B)=0.4×0.5=0.2. Since P(A∩B)=P(A)P(B), events are independent

Critical Evaluation

[Diagram: Overlapping circles A/B with 0.2 intersection] Textbook confirms this matches the independence criterion despite initial union probability.

Question 10:

Using the axiomatic approach, prove that the probability of an impossible event is zero. Discuss its significance in real-world applications.

Answer:

Theoretical Framework

We studied that probability is defined using three axioms: non-negativity, normalization, and additivity. An impossible event has no outcomes in the sample space.

Evidence Analysis

Let S be the sample space and ∅ the impossible event.
From axiom 2: P(S) = 1.
S = S ∪ ∅ (disjoint union).
By axiom 3: P(S) = P(S) + P(∅).
Thus, 1 = 1 + P(∅) ⇒ P(∅) = 0.

Critical Evaluation

This proof aligns with our textbook's treatment of probability. It's foundational for modeling scenarios like zero-chance outcomes in quality control.

Question 11:

A bag contains 5 red, 3 blue and 2 green balls. Using conditional probability, find the chance of drawing two red balls consecutively without replacement. Validate with a tree diagram.

Answer:

Theoretical Framework

We use conditional probability P(A∩B) = P(A) × P(B|A). Total balls = 10.

Evidence Analysis

First draw: P(Red₁) = 5/10 = 0.5
Second draw: P(Red₂|Red₁) = 4/9 ≈ 0.444
Compound probability: 0.5 × 0.444 ≈ 0.222

[Diagram: Two-level tree with Red/Blue/Green branches]
Critical Evaluation

Our calculation matches the tree diagram's path probabilities. This models real scenarios like defective item sampling.

Question 12:

Define probability and explain its significance in real-life scenarios with suitable examples. Also, differentiate between theoretical probability and experimental probability.

Answer:

Probability is a branch of mathematics that measures the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). It helps quantify uncertainty and is widely used in fields like statistics, finance, and science.

Significance in real-life:

Weather forecasting: Predicting rain with a 70% chance.
Games: Calculating the odds of rolling a six in dice.
Finance: Assessing risks in stock market investments.

Theoretical probability is based on reasoning and mathematical principles (e.g., probability of drawing a red card from a deck is 26/52 = 0.5).

Experimental probability is derived from actual experiments or observations (e.g., flipping a coin 100 times and getting heads 55 times gives an experimental probability of 0.55).

The two differ in their approach: theoretical is ideal, while experimental is empirical.

Question 13:

A bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. Calculate the probability that the ball is:
(i) Red
(ii) Not green
(iii) Either red or blue

Answer:

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

(i) Probability of drawing a red ball:
P(Red) = Number of red balls / Total balls
P(Red) = 5/10 = 0.5 or 50%.

(ii) Probability of not drawing a green ball:
P(Not Green) = 1 - P(Green)
P(Green) = 2/10 = 0.2
P(Not Green) = 1 - 0.2 = 0.8 or 80%.

(iii) Probability of drawing either red or blue:
P(Red or Blue) = P(Red) + P(Blue)
P(Red) = 5/10
P(Blue) = 3/10
P(Red or Blue) = 5/10 + 3/10 = 8/10 = 0.8 or 80%.

Question 14:

Explain the addition rule of probability for mutually exclusive and non-mutually exclusive events with examples. How is it applied in practical situations?

Answer:

The addition rule of probability calculates the probability of either of two events occurring.

1. Mutually Exclusive Events: Events that cannot occur simultaneously (e.g., rolling a die cannot result in both 2 and 5).
Formula: P(A or B) = P(A) + P(B).
Example: Probability of rolling a 2 or 5 on a die:
P(2) = 1/6, P(5) = 1/6
P(2 or 5) = 1/6 + 1/6 = 2/6 = 1/3.

2. Non-Mutually Exclusive Events: Events that can occur together (e.g., drawing a king or a heart from a deck).
Formula: P(A or B) = P(A) + P(B) - P(A and B).
Example: Probability of drawing a king or a heart:
P(King) = 4/52, P(Heart) = 13/52, P(King and Heart) = 1/52
P(King or Heart) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

Practical Application: Used in risk assessment, such as calculating the probability of machine failure due to either electrical or mechanical issues.

Question 15:

Define conditional probability with an example. Explain its significance in real-life scenarios using a mathematical approach.

Answer:

Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), which reads as 'the probability of event A given event B'.

Example: Consider a deck of 52 playing cards. The probability of drawing a king (P(K)) is 4/52 = 1/13. However, if we know that the card drawn is a heart (H), the conditional probability of it being a king (P(K|H)) is 1/13, since there is only one king of hearts in the 13 hearts.

Significance in real-life: Conditional probability is widely used in fields like medicine, finance, and weather forecasting. For instance, in medical testing, it helps determine the probability of having a disease given a positive test result. Mathematically, if P(D) is the probability of having the disease and P(T|D) is the probability of testing positive given the disease, then P(D|T) can be calculated using Bayes' Theorem.

Formula: P(A|B) = P(A ∩ B) / P(B), provided P(B) ≠ 0.

Question 16:

A bag contains 5 red, 3 blue, and 2 green balls. Two balls are drawn at random without replacement. Find the probability that both balls are of the same color. Show all steps clearly.

Answer:

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

Step 1: Calculate total ways to draw 2 balls.
Total outcomes = C(10, 2) = 10! / (2! * 8!) = 45.

Step 2: Calculate favorable outcomes for same color.

Both red: C(5, 2) = 10.
Both blue: C(3, 2) = 3.
Both green: C(2, 2) = 1.

Total favorable outcomes = 10 (red) + 3 (blue) + 1 (green) = 14.

Step 3: Compute probability.
Probability = Favorable outcomes / Total outcomes = 14 / 45.

Final Answer: The probability that both balls are of the same color is 14/45.

Question 17:

Define conditional probability and explain its significance with a real-life example. Also, derive the formula for conditional probability using the basic definition of probability.

Answer:

Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), which means the probability of event A occurring given that event B has already occurred.

Significance: Conditional probability is widely used in real-life scenarios like weather forecasting, medical testing, and risk assessment. For example, the probability of a person testing positive for a disease (A) given that they have symptoms (B) is a conditional probability.

Derivation of the formula:
1. The basic definition of probability is P(A) = Number of favorable outcomes / Total number of outcomes.
2. For conditional probability, the sample space reduces to the outcomes where event B has occurred.
3. Thus, P(A|B) = Number of outcomes where both A and B occur / Number of outcomes where B occurs.
4. Dividing numerator and denominator by the total number of outcomes, we get:
P(A|B) = P(A ∩ B) / P(B).

Question 18:

A bag contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random without replacement. Find the probability that both balls are of the same color. Show all steps clearly.

Answer:

Step 1: Total number of balls
Total balls = 5 (red) + 4 (blue) + 3 (green) = 12 balls.

Step 2: Calculate probabilities for each color
1. Probability both are red:
P(Red) = (5/12) × (4/11) = 20/132.
2. Probability both are blue:
P(Blue) = (4/12) × (3/11) = 12/132.
3. Probability both are green:
P(Green) = (3/12) × (2/11) = 6/132.

Step 3: Add the probabilities
Since these are mutually exclusive events, the total probability is:
P(Same color) = P(Red) + P(Blue) + P(Green)
= 20/132 + 12/132 + 6/132
= 38/132
= 19/66 (simplified).

Question 19:

Define conditional probability with an example. Explain its significance in real-life situations using a detailed scenario.

Answer:

Example: Consider a deck of 52 playing cards. The probability of drawing a king (Event A) is 4/52. However, if we know that the card drawn is a heart (Event B), the conditional probability P(A|B) becomes 1/13, as there is only one king in the 13 hearts.

Real-life significance: Conditional probability is widely used in medical testing. For instance, suppose a disease affects 1% of a population (Event A), and a test for the disease is 99% accurate (Event B). The conditional probability P(A|B) helps determine the likelihood that a person actually has the disease given a positive test result, considering false positives. This is crucial for making informed medical decisions.

Question 20:

Explain the multiplication theorem of probability with a mathematical derivation. Provide a practical application where this theorem is useful.

Answer:

The multiplication theorem states that the probability of the joint occurrence of two events A and B is given by:
P(A ∩ B) = P(A) × P(B|A), provided P(A) ≠ 0.

Derivation:
1. From the definition of conditional probability, P(B|A) = P(A ∩ B) / P(A).
2. Rearranging the terms, we get P(A ∩ B) = P(A) × P(B|A).

Practical application: This theorem is essential in quality control. For example, in a factory, suppose the probability of a machine being defective (Event A) is 0.05, and the probability of a defective product given a defective machine (Event B) is 0.8. The multiplication theorem helps calculate the joint probability P(A ∩ B) = 0.05 × 0.8 = 0.04, which is the likelihood of both the machine being defective and producing a defective product. This aids in risk assessment and decision-making.

Question 21:

A bag contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random without replacement. Find the probability that both balls are of the same color. Explain each step clearly.

Answer:

To find the probability that both balls drawn are of the same color, we need to consider the three possible cases: both are red, both are blue, or both are green.

Step 1: Total number of balls
Total balls = 5 (red) + 4 (blue) + 3 (green) = 12 balls.

Step 2: Total number of ways to draw 2 balls
Number of ways to choose 2 balls out of 12 = C(12, 2) = (12 × 11) / (2 × 1) = 66.

Step 3: Calculate favorable outcomes for each color

Both red: C(5, 2) = (5 × 4) / (2 × 1) = 10 ways.
Both blue: C(4, 2) = (4 × 3) / (2 × 1) = 6 ways.
Both green: C(3, 2) = (3 × 2) / (2 × 1) = 3 ways.

Step 4: Total favorable outcomes
Total favorable = 10 (red) + 6 (blue) + 3 (green) = 19.

Step 5: Calculate probability
Probability = Favorable outcomes / Total outcomes = 19 / 66.

Thus, the probability that both balls drawn are of the same color is 19/66.

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. Two marbles are drawn without replacement. Find the probability that:
(i) Both are red.
(ii) One is blue and the other is green.

Answer:

Problem Interpretation
We need to find probabilities for two scenarios involving dependent events (no replacement).
Mathematical Modeling

Total marbles = 5 + 3 + 2 = 10
Total ways to draw 2 marbles: C(10,2) = 45

Solution
(i) P(Both red) = C(5,2)/C(10,2) = 10/45 = 2/9
(ii) P(One blue and one green) = (C(3,1) × C(2,1))/45 = 6/45 = 2/15

Question 2:

In a survey, 60% of students own a bicycle, 30% own a scooter, and 10% own both. If a student is chosen randomly, find:
(i) P(owns only bicycle)
(ii) P(owns neither).

Answer:

Problem Interpretation
We apply set theory to probability using given percentages.
Mathematical Modeling
Let:

P(B) = 0.6 (bicycle)
P(S) = 0.3 (scooter)
P(B∩S) = 0.1

Solution
(i) P(only bicycle) = P(B) - P(B∩S) = 0.6 - 0.1 = 0.5
(ii) P(neither) = 1 - P(B∪S) = 1 - (0.6 + 0.3 - 0.1) = 0.2

Question 3:

Prove: If A and B are independent events, then P(A' ∩ B') = P(A') × P(B'). Use this to find P(A' ∩ B') when P(A) = 0.4, P(B) = 0.7.

Answer:

Problem Interpretation
We verify the independence property for complements.
Mathematical Modeling
Given independence:

P(A∩B) = P(A)P(B)
P(A') = 1 - P(A)

Solution
Proof: P(A'∩B') = P((A∪B)') = 1 - P(A∪B) = 1 - [P(A)+P(B)-P(A∩B)] = (1-P(A))(1-P(B)) = P(A')P(B').
For P(A)=0.4, P(B)=0.7:
P(A'∩B') = (0.6)(0.3) = 0.18

Question 4:

A bag contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random without replacement.
Problem Interpretation: Find the probability that both balls are of the same color.
Mathematical Modeling: Use combinations to calculate favorable and total outcomes.

Answer:

Problem Interpretation: We need to find the probability of drawing two balls of the same color (red, blue, or green) without replacement.
Mathematical Modeling: Total balls = 12. Favorable cases:

Both red: C(5,2) = 10
Both blue: C(4,2) = 6
Both green: C(3,2) = 3

Total favorable = 19. Total outcomes = C(12,2) = 66.
Solution: Probability = 19/66 ≈ 0.2879.

Question 5:

In a survey, 60% of students prefer online classes, 30% prefer hybrid, and 10% prefer offline. If 3 students are selected randomly, find the probability that at least two prefer the same mode.
Problem Interpretation: Calculate P(at least two identical preferences).
Mathematical Modeling: Use complementary probability (all three preferences different).

Answer:

Problem Interpretation: We calculate P(at least two identical) = 1 - P(all three different).
Mathematical Modeling: P(online) = 0.6, P(hybrid) = 0.3, P(offline) = 0.1. P(all different) = 3! × (0.6 × 0.3 × 0.1) = 0.108.
Solution: P(at least two identical) = 1 - 0.108 = 0.892.

Question 6:

A bag contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random. Find the probability that: (i) Both are red, (ii) One is blue and the other is green.

Answer:

Problem Interpretation

We need to find probabilities for two scenarios when drawing two balls from a bag with given counts.

Mathematical Modeling

Total balls = 5 + 4 + 3 = 12
Total ways to draw 2 balls = C(12, 2) = 66

Solution

(i) P(Both red) = C(5, 2)/66 = 10/66 = 5/33
(ii) P(One blue and one green) = (C(4, 1) × C(3, 1))/66 = 12/66 = 2/11

Question 7:

In a survey, 60% of students prefer online classes, 30% prefer offline, and 10% prefer hybrid. If 3 students are chosen randomly, find the probability that at least two prefer the same mode.

Answer:

Problem Interpretation

We calculate the probability of at least two students sharing a preference among three randomly chosen students.

Mathematical Modeling

P(Online) = 0.6, P(Offline) = 0.3, P(Hybrid) = 0.1
Total cases = 3^3 = 27

Solution

P(All different) = 3! × (0.6 × 0.3 × 0.1) = 0.108
P(At least two same) = 1 - 0.108 = 0.892

Question 8:

A survey in a school found that 40% of students play football, 30% play cricket, and 20% play both. Using probability rules, find the probability that a randomly selected student plays neither sport. Justify your steps.

Answer:

Problem Interpretation

We need to find the probability of a student playing neither football nor cricket using given percentages.

Mathematical Modeling

Let P(F) = 0.40 (football)
P(C) = 0.30 (cricket)
P(F ∩ C) = 0.20 (both)

Solution

Using the addition rule: P(F ∪ C) = P(F) + P(C) - P(F ∩ C) = 0.40 + 0.30 - 0.20 = 0.50. Probability of neither = 1 - P(F ∪ C) = 1 - 0.50 = 0.50.

Question 9:

Two dice are rolled. Derive the probability that the sum is 7 given that the first die shows a prime number. Show all steps.

Answer:

Problem Interpretation

We must find P(Sum=7 | First die is prime). Our textbook shows conditional probability as P(A|B) = P(A ∩ B)/P(B).

Mathematical Modeling

Prime outcomes for first die: {2, 3, 5} → P(B) = 3/6 = 0.5
Favorable pairs: (2,5), (3,4), (5,2) → P(A ∩ B) = 3/36

Solution

P(A|B) = (3/36)/(0.5) = 1/6. Thus, the probability is 1/6.

Question 10:

A survey of 200 students shows that 120 prefer online classes, while 80 prefer offline classes. If 2 students are selected at random, find the probability that:
(i) Both prefer online classes.
(ii) One prefers online and the other prefers offline.

Answer:

Problem Interpretation

We need to find probabilities for two scenarios involving student preferences.

Mathematical Modeling

Total students = 200
Online preference = 120
Offline preference = 80

Solution

(i) P(Both online) = (120/200) × (119/199) ≈ 0.357. (ii) P(One online and one offline) = [(120/200) × (80/199)] + [(80/200) × (120/199)] ≈ 0.482.

Question 11:

A bag contains 5 red, 4 blue, and 3 green balls. Three balls are drawn without replacement. Derive the probability distribution of X, where X is the number of red balls drawn.

Answer:

Problem Interpretation

We must derive the probability distribution for the number of red balls drawn.

Mathematical Modeling

Total balls = 12
Red balls = 5
Non-red balls = 7

Solution

P(X=0) = C(7,3)/C(12,3) ≈ 0.159. P(X=1) = [C(5,1) × C(7,2)]/C(12,3) ≈ 0.477. P(X=2) = [C(5,2) × C(7,1)]/C(12,3) ≈ 0.318. P(X=3) = C(5,3)/C(12,3) ≈ 0.045.

Question 12:

A bag contains 5 red balls, 3 blue balls, and 2 green balls. Two balls are drawn at random without replacement.
Problem Interpretation: Find the probability that both balls are of the same color.
Mathematical Modeling: Use combinations to calculate favorable and total outcomes.

Answer:

Problem Interpretation: We need to find the probability of drawing two balls of the same color (red, blue, or green).
Mathematical Modeling: Total balls = 10. Favorable cases:

Both red: C(5,2) = 10
Both blue: C(3,2) = 3
Both green: C(2,2) = 1

Solution: Total outcomes = C(10,2) = 45. Probability = (10 + 3 + 1)/45 = 14/45 ≈ 0.311.

Question 13:

In a survey, 60% of students prefer online classes, 30% prefer offline classes, and 10% have no preference. If 3 students are selected randomly, find the probability that exactly 2 prefer online classes.
Problem Interpretation: Apply binomial probability for independent events.
Mathematical Modeling: Use P(online) = 0.6, P(not online) = 0.4.

Answer:

Problem Interpretation: We calculate the probability of exactly 2 out of 3 students preferring online classes.
Mathematical Modeling: This is a binomial distribution problem with n=3, k=2, p=0.6.
Solution: P(X=2) = C(3,2) × (0.6)² × (0.4)¹ = 3 × 0.36 × 0.4 = 0.432. Our textbook shows similar examples for binomial probability.

Question 14:

A survey in a school found that 40% of students play football, 30% play cricket, and 20% play both. Using probability rules, determine the likelihood that a randomly selected student plays neither sport.

Answer:

Problem Interpretation

We need to find the probability of a student playing neither football nor cricket, given individual and joint probabilities.

Mathematical Modeling

Let P(F) = 0.4 (football)
P(C) = 0.3 (cricket)
P(F ∩ C) = 0.2 (both)

Solution

Using the addition rule, P(F ∪ C) = P(F) + P(C) - P(F ∩ C) = 0.4 + 0.3 - 0.2 = 0.5. Thus, P(neither) = 1 - P(F ∪ C) = 0.5.

Question 15:

A bag contains 5 red, 3 blue, and 2 green marbles. Two marbles are drawn without replacement. Derive the probability that both are red using conditional probability.

Answer:

Problem Interpretation

We must calculate the probability of drawing two red marbles consecutively without replacement.

Mathematical Modeling

Total marbles = 5 + 3 + 2 = 10
First draw: P(Red₁) = 5/10
Second draw: P(Red₂|Red₁) = 4/9

Solution

By multiplication rule, P(Red₁ ∩ Red₂) = P(Red₁) × P(Red₂|Red₁) = (5/10) × (4/9) = 2/9.

Question 16:

A bag contains 5 red balls, 3 blue balls, and 2 green balls. Two balls are drawn at random without replacement. Based on this, answer the following:

(i) Find the probability that both balls are red.

(ii) If one ball is red, what is the probability that the second ball is blue?

Answer:

Solution:

(i) Probability both balls are red:

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

Number of ways to draw 2 red balls = ⁵C₂ = 10.

Total ways to draw any 2 balls = ¹⁰C₂ = 45.

Probability = ¹⁰⁄₄₅ = ²⁄₉.

(ii) Conditional probability (second ball is blue given first is red):

After drawing 1 red ball, remaining balls = 9 (4 red, 3 blue, 2 green).

Probability second ball is blue = ³⁄₉ = ¹⁄₃.

Note: Since the first ball is already red (given), we calculate probability based on the reduced sample space.

Question 17:

In a class of 30 students, 18 play cricket, 12 play basketball, and 8 play both. A student is selected at random. Answer the following:

(i) Find the probability that the student plays neither sport.

(ii) If the student plays cricket, what is the probability they also play basketball?

Answer:

Solution:

(i) Probability student plays neither sport:

Total students = 30.

Students playing at least one sport = Cricket + Basketball − Both = 18 + 12 − 8 = 22.

Students playing neither = 30 − 22 = 8.

Probability = ⁸⁄₃₀ = ⁴⁄₁₅.

(ii) Conditional probability (plays basketball given they play cricket):

Students playing cricket = 18.

Students playing both = 8.

Probability = ⁸⁄₁₈ = ⁴⁄₉.

Note: This is an application of conditional probability where the sample space is reduced to cricket players only.

Question 18:

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

What is the probability that the ball drawn is not red?
If two balls are drawn successively without replacement, what is the probability that both are blue?

Answer:

Part 1: Probability that the ball drawn is not red:

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

Part 2: Probability both drawn balls are blue (without replacement):

First draw: Probability = 3/10.
Second draw (after removing one blue ball): Probability = 2/9.
Combined probability = (3/10) × (2/9) = 6/90 = 1/15.

Note: Always simplify fractions and consider dependency in successive events.

Question 19:

In a class of 40 students, 25 play cricket, 15 play basketball, and 10 play both. A student is selected randomly. Answer:

What is the probability that the student plays only cricket?
What is the probability that the student plays neither sport?

Answer:

Part 1: Probability of playing only cricket:

Students playing only cricket = Total cricket players - Students playing both = 25 - 10 = 15.
Probability = 15/40 = 3/8.

Part 2: Probability of playing neither sport:

Total students playing at least one sport = Cricket + Basketball - Both = 25 + 15 - 10 = 30.
Students playing neither = Total students - Students playing at least one sport = 40 - 30 = 10.
Probability = 10/40 = 1/4.

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

Question 20:

A bag contains 5 red balls, 3 blue balls, and 2 green balls. Two balls are drawn at random without replacement. Based on this, answer the following:

Find the probability that both balls are red.
Find the probability that one ball is blue and the other is green.

Answer:

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

Probability that both balls are red:
Number of ways to choose 2 red balls from 5 = ⁵C₂ = 10.
Total ways to choose any 2 balls = ¹⁰C₂ = 45.
Probability = ¹⁰⁄₄₅ = ²⁄₉.

Probability that one is blue and the other is green:
Number of ways to choose 1 blue and 1 green = 3 (blue) × 2 (green) = 6.
Probability = ⁶⁄₄₅ = ²⁄₁₅.

Question 21:

In a class of 30 students, 12 are girls and 18 are boys. A committee of 4 students is to be formed. Answer the following:

Find the probability that the committee has exactly 2 girls and 2 boys.
Find the probability that the committee has at least one girl.

Answer:

Total number of students = 30 (12 girls + 18 boys).

Probability of exactly 2 girls and 2 boys:
Number of ways to choose 2 girls from 12 = ¹²C₂ = 66.
Number of ways to choose 2 boys from 18 = ¹⁸C₂ = 153.
Total ways to form the committee = ³⁰C₄ = 27405.
Probability = (66 × 153) / 27405 = ¹⁰⁰⁹⁸⁄₂₇₄₀₅ ≈ 0.3684.

Probability of at least one girl:
Easier to calculate using the complement: P(at least 1 girl) = 1 − P(no girls).
Number of ways to choose 4 boys from 18 = ¹⁸C₄ = 3060.
Probability of no girls = ³⁰⁶⁰⁄₂₇₄₀₅ ≈ 0.1117.
Thus, P(at least 1 girl) = 1 − 0.1117 = 0.8883.

Question 22:

A bag contains 5 red balls, 3 blue balls, and 2 green balls. Two balls are drawn at random without replacement. Based on this, answer the following:

(i) Find the probability that both balls drawn are red.

(ii) If one ball is red, what is the probability that the second ball is blue?

Answer:

Solution:

(i) Probability both balls are red:

Total balls = 5 (red) + 3 (blue) + 2 (green) = 10.
Number of ways to draw 2 red balls = ⁵C₂ = 10.
Total ways to draw any 2 balls = ¹⁰C₂ = 45.
Probability = ¹⁰⁄₄₅ = ²⁄₉.

(ii) Conditional probability (second ball is blue given first is red):

After drawing 1 red ball, remaining balls = 4 red, 3 blue, 2 green = 9.
Probability second ball is blue = ³⁄₉ = ¹⁄₃.

Note: Conditional probability accounts for the change in sample space after the first event.

Question 23:

In a class of 30 students, 12 play cricket, 8 play basketball, and 5 play both. A student is selected at random. Answer the following:

(i) Find the probability that the student plays either cricket or basketball.

(ii) If the student plays basketball, what is the probability they also play cricket?

Answer:

Solution:

(i) Probability of playing cricket or basketball:

Using the principle of inclusion-exclusion:
P(Cricket ∪ Basketball) = P(Cricket) + P(Basketball) − P(Both).
P(Cricket) = ¹²⁄₃₀ = 0.4.
P(Basketball) = ⁸⁄₃₀ ≈ 0.2667.
P(Both) = ⁵⁄₃₀ ≈ 0.1667.
P(Cricket ∪ Basketball) = 0.4 + 0.2667 − 0.1667 = 0.5.

(ii) Conditional probability (plays cricket given they play basketball):

P(Cricket | Basketball) = P(Both) / P(Basketball).
P(Both) = ⁵⁄₃₀.
P(Basketball) = ⁸⁄₃₀.
P(Cricket | Basketball) = (⁵⁄₃₀) / (⁸⁄₃₀) = ⁵⁄₈.

Note: Conditional probability narrows the sample space to only basketball players.

Question 24:

A bag contains 5 red, 4 blue, and 3 green balls. Two balls are drawn at random without replacement. Using the concept of conditional probability, find the probability that both balls are red given that at least one ball is red.

Answer:

To solve this problem, we use the definition of conditional probability: P(A|B) = P(A ∩ B) / P(B), where A is the event that both balls are red, and B is the event that at least one ball is red.

Step 1: Total number of balls
Total balls = 5 (red) + 4 (blue) + 3 (green) = 12.

Step 2: Find P(A ∩ B)
Since A is a subset of B (if both are red, at least one is red), P(A ∩ B) = P(A).
P(A) = Probability both balls are red = (5/12) × (4/11) = 20/132 = 5/33.

Step 3: Find P(B)
P(B) = Probability at least one ball is red = 1 - P(no red balls).
P(no red balls) = (7/12) × (6/11) = 42/132 = 7/22.
Thus, P(B) = 1 - (7/22) = 15/22.

Step 4: Compute P(A|B)
P(A|B) = (5/33) / (15/22) = (5/33) × (22/15) = 110/495 = 2/9.

Final Answer: The probability that both balls are red given that at least one is red is 2/9.

Question 25:

In a class of 30 students, 18 play cricket, 12 play basketball, and 8 play both. A student is selected at random. Using the concept of probability of independent events, determine whether the events 'playing cricket' and 'playing basketball' are independent.

Answer:

To check if two events are independent, we verify if P(A ∩ B) = P(A) × P(B), where:
A = Event that a student plays cricket.
B = Event that a student plays basketball.

Step 1: Calculate individual probabilities
P(A) = Number of cricket players / Total students = 18/30 = 3/5.
P(B) = Number of basketball players / Total students = 12/30 = 2/5.

Step 2: Calculate P(A ∩ B)
P(A ∩ B) = Number playing both / Total students = 8/30 = 4/15.

Step 3: Check independence condition
P(A) × P(B) = (3/5) × (2/5) = 6/25 ≈ 0.24.
P(A ∩ B) = 4/15 ≈ 0.267.

Since 0.267 ≠ 0.24, the events are not independent.

Conclusion: Playing cricket and playing basketball are dependent events in this class.

Probability – CBSE NCERT Study Resources

Overview of the Chapter: Probability

Key Topics Covered

Detailed Explanation

Random Experiments and Sample Space

Events

Axiomatic Approach to Probability

Probability of Equally Likely Outcomes

Addition Theorem on Probability

Conclusion

All Question Types with Solutions – CBSE Exam Pattern

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

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

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

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

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