Probability | NEB Class 11 Mathematics Notes

Chapter Overview

Probability is the mathematical measurement of uncertainty. It helps us predict the likelihood of an event occurring, from tossing coins and rolling dice to drawing cards and making real-world predictions. In the NEB Class 11 Mathematics exam, this chapter focuses heavily on counting outcomes, applying theorems, and recognizing when to use Addition vs. Multiplication rules.

What this chapter covers

Random Experiments & Sample SpaceClassical & Empirical ProbabilityComplementary EventsMutually Exclusive vs Independent EventsThe Addition TheoremThe Multiplication TheoremCoin, Dice, and Card Problems

Learning Objectives

1

Define probability and understand related terms like random experiment, trial, and event.

2

List out the complete Sample Space for experiments involving coins, dice, and cards using Tree Diagrams.

3

Calculate the probability of simple and compound events using the Classical formula.

4

Use Complementary Events to solve 'not' or 'none' type questions rapidly.

5

Apply the Addition Theorem for events joined by 'OR'.

6

Apply the Multiplication Theorem for events joined by 'AND'.

7

Confidently tackle NEB 4-mark and 5-mark word problems on probability.

The 'At Least One' Ultimate Shortcut

Key Insight

If an NEB question asks for the probability of 'at least one' event happening (e.g., at least one head in 3 coin tosses, or at least one student solving a math problem), DO NOT calculate the probability of 1, 2, and 3 separately and add them.

Use the Complement Shortcut:
$$P(\text{At least one}) = 1 - P(\text{None})$$
Find the probability that NO ONE does it, or it NEVER happens, and subtract from 1. This saves 5 minutes of calculation time in exams!

Example: Three students have 1/2, 1/3, and 1/4 chance to solve a problem. Chance NO ONE solves it = (1/2) × (2/3) × (3/4) = 1/4. So, chance AT LEAST ONE solves it = 1 - 1/4 = 3/4.

Key Definitions

Random Experiment

An operation whose exact outcome cannot be predicted in advance with certainty (e.g., tossing a coin).

Sample Space (S)

The set of all possible outcomes of a random experiment. For a die, S = {1, 2, 3, 4, 5, 6}. Denoted by n(S).

Event (E)

Any subset of the sample space. An outcome or set of outcomes we are interested in. E.g., getting an even number.

Mutually Exclusive Events

Events that CANNOT happen at the same time. If one happens, the other cannot. Example: Rolling a 2 and a 5 in a single throw.

Independent Events

Events where the occurrence of one does NOT affect the probability of the other. Example: Tossing a coin and rolling a die.

Complementary Event (Ē)

The event representing the 'non-occurrence' of event E. Mathematically, $P(\bar{E}) = 1 - P(E)$.

Mutually Exclusive vs. Independent Events

Students lose marks by confusing these two concepts. Memorize this difference:

Mutually Exclusive

Focuses on whether they can happen together.

They CANNOT happen at the same time.

Intersection is zero: $P(A \cap B) = 0$.

Used mainly with the Addition Theorem (OR).

Example: Getting a 'King' AND a 'Queen' on a single card draw is impossible.

VS

Independent Events

Focuses on whether one affects the other.

They CAN happen at the same time, but without influencing each other.

Intersection is product: $P(A \cap B) = P(A) \times P(B)$.

Used mainly with the Multiplication Theorem (AND).

Example: Tossing a coin (Heads) and rolling a die (Six).

Formula Sheet & Core Derivations

Classical (Mathematical) Probability

P(E) = \frac{n(E)}{n(S)} = \frac{m}{n}

Ratio of favorable outcomes (m) to total possible equally likely outcomes (n).

1If an experiment has $n$ total outcomes that are all equally likely to happen, and $m$ of those outcomes represent our specific event $E$.

2The 'share' of the total possibility belonging to $E$ is simply $m$ parts out of $n$.

3Therefore, $P(E) = \frac{m}{n}$.

Complementary Events Rule

P(E) + P(\bar{E}) = 1

An event either happens or it doesn't. Their probabilities must add up to 100% (or 1).

1Let $m$ be the favorable outcomes for E. Then the unfavorable outcomes (where E does not happen) are $n - m$.

2Probability of E not happening: $P(\bar{E}) = \frac{n - m}{n} = \frac{n}{n} - \frac{m}{n}$.

3$P(\bar{E}) = 1 - P(E)$. Rearranging gives $P(E) + P(\bar{E}) = 1$.

General Addition Theorem (OR)

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Used for 'OR' questions. If the events are Mutually Exclusive, $P(A \cap B) = 0$, so you just add.

1Let $n_1$ be outcomes in A, and $n_2$ be outcomes in B. Simply adding $n_1 + n_2$ double-counts the outcomes that belong to BOTH A and B ($n_3$).

2To get the correct union, subtract the overlap once: Total favorable = $n_1 + n_2 - n_3$.

3Divide everything by total outcomes $n$: $\frac{n_1}{n} + \frac{n_2}{n} - \frac{n_3}{n}$.

4This yields: $P(A) + P(B) - P(A \cap B)$.

Multiplication Theorem (AND)

P(A \cap B) = P(A) \times P(B)

Used for 'AND' questions when events A and B are Independent.

1For independent events, the total favorable combinations is the product of individual favorable outcomes ($m_1 \times m_2$).

2The total possible outcomes is the product of individual total outcomes ($n_1 \times n_2$).

3Probability = $\frac{m_1 \times m_2}{n_1 \times n_2} = \frac{m_1}{n_1} \times \frac{m_2}{n_2} = P(A) \times P(B)$.

Visualizing the Addition Theorem

Why do we subtract P(A ∩ B)? Because when you add the entire blue circle (A) and the entire red circle (B), you count the middle intersection TWICE. You must subtract it once to get the correct union.

The Venn diagram shows that P(A ∪ B) is the total shaded area. Subtracting the middle prevents double counting.

Visualizing Outcomes: The Cards and Coins Fact Sheet

Trick: You cannot solve probability without knowing the sample spaces of these classic systems.

Steps to Remember:

COINS (Heads/Tails): 1 Coin = 2 outcomes. 2 Coins = 4 outcomes. 3 Coins = 8 outcomes. The easiest way to find the sample space is to draw a Tree Diagram.
CARDS (52 Total): 4 Suits (Hearts, Diamonds, Clubs, Spades). Each suit has 13 cards. Red cards = 26, Black cards = 26. Face Cards: Jack, Queen, King (3 per suit = 12 total). Aces are NOT face cards.
DICE (6-sided): 1 Die = 6 outcomes. 2 Dice = $6 \times 6 = 36$ outcomes. Note: A sum of 7 has the highest probability on two dice (6/36 = 1/6).

Mnemonic: Cards: 52 total, 4 suits, 12 faces. Dice: 6^n. Coins: 2^n.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Question uses the word 'OR'	Use the Addition Theorem: P(A) + P(B) - P(A ∩ B).
Question uses 'AND' / 'BOTH' / 'SUCCESSIVELY'	Use the Multiplication Theorem: P(A) × P(B).
Question asks for 'AT LEAST ONE'	Use Complement shortcut: 1 - P(None).
Question says 'EITHER A OR B'	Use Addition Theorem.
Question says 'NEITHER A NOR B'	Find P(A ∪ B) and subtract from 1.

Common Mistakes (Lose Marks Here!)

Examiners specifically look for these errors in NEB answer sheets.

Forgetting the Overlap in OR questions: e.g., 'Probability of King or Heart'. Students write 4/52 + 13/52 = 17/52. WRONG! You must subtract the King of Hearts (1/52) because it was counted twice. Correct: 16/52.
With vs Without Replacement: If drawing two cards 'without replacement', the total cards (denominator) drops from 52 to 51 for the second draw. If 'with replacement', it stays 52.
Misunderstanding 'Face Cards': Many students include Aces as face cards. Face cards are strictly Jacks, Queens, and Kings (12 total in a deck).
Confusing Mutually Exclusive vs Independent: You ADD probabilities for mutually exclusive events. You MULTIPLY probabilities for independent events.

Solved Examples (Easy → Hard)

1

Look at the urn below containing colored balls. If one ball is drawn at random, what is the probability that it is NOT green?

Easy

Solution

First, count the total number of balls in the urn. There are 3 Red, 2 Blue, and 4 Green balls.
Total outcomes $n(S) = 3 + 2 + 4 = 9$.

Let G be the event of drawing a Green ball. $n(G) = 4$.
$$P(G) = \frac{4}{9}$$

We want the probability of NOT drawing a Green ball, which is $P(\bar{G})$.
Using the complement rule:
$$P(\bar{G}) = 1 - P(G) = 1 - \frac{4}{9} = \frac{5}{9}$$

Alternative Method: Favorable outcomes (Red + Blue) = 3 + 2 = 5. $P = 5/9$.

∴ 5/9

2

A card is drawn from a standard deck. What is the probability that it is a King OR a Heart?

Medium

Solution

Let $K$ be the event of drawing a King, and $H$ be the event of drawing a Heart.
$n(S) = 52$
$P(K) = \frac{4}{52}$
$P(H) = \frac{13}{52}$

Can they happen together? Yes, the King of Hearts. So, they are NOT mutually exclusive.
$P(K \cap H) = \frac{1}{52}$

By the Addition Theorem:
$$P(K \cup H) = P(K) + P(H) - P(K \cap H)$$
$$P(K \cup H) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$

∴ 4/13

3

Three students A, B, and C attempt a math problem. Their chances of solving it are 1/2, 1/3, and 1/4 respectively. Find the probability that the problem is solved.

Hard

Solution

The problem is considered 'solved' if at least one of them solves it. We use the Complement Trick.

Find probability of failing for each:
$P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$
$P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}$
$P(\bar{C}) = 1 - \frac{1}{4} = \frac{3}{4}$

Probability that NO ONE solves it (they all fail together, which is an AND operation):
$$P(\text{None}) = P(\bar{A}) \times P(\bar{B}) \times P(\bar{C})$$
$$P(\text{None}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4}$$

Probability that AT LEAST ONE solves it:
$$P(\text{Solved}) = 1 - P(\text{None}) = 1 - \frac{1}{4} = \frac{3}{4}$$

∴ 3/4

MCQ Practice (10 Questions)

Mixed Level — MCQ

1

Look at the Venn diagram below. What does the highlighted shaded region represent in probability?

Correct!

Incorrect. Correct: P(A ∩ B)

Solution

In Set Theory and Probability, the area where two sets overlap is called the intersection, denoted mathematically as $A \cap B$. It represents the probability that both events happen.

∴ P(A ∩ B)

2

If P(A) = 0.5 and P(B) = 0.3, and A and B are mutually exclusive, what is P(A ∪ B)?

Correct!

Incorrect. Correct: 0.8

Solution

Mutually exclusive means they cannot happen together ($P(A \cap B) = 0$). Therefore, the addition theorem simplifies to $P(A \cup B) = P(A) + P(B) = 0.5 + 0.3 = 0.8$.

∴ 0.8

3

The probability of a 'sure event' is:

Correct!

Incorrect. Correct: 1

Solution

By the axioms of probability, the probability of the entire sample space occurring is 1. Thus, a sure event has a probability of 1.

∴ 1

4

If a coin is tossed twice, what is the probability of getting at least one Tail?

Correct!

Incorrect. Correct: 3/4

Solution

Using the complement trick: $P(\text{at least one Tail}) = 1 - P(\text{No Tails})$. No tails means getting {HH}, which is 1/4. So, $1 - 1/4 = 3/4$.

∴ 3/4

5

Two events A and B are independent. P(A) = 1/3, P(B) = 1/4. Find P(A ∩ B).

Correct!

Incorrect. Correct: 1/12

Solution

The Multiplication Theorem states that for independent events, the probability of both happening is their product: $P(A \cap B) = P(A) \times P(B) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.

∴ 1/12

6

The probability of drawing a Red Queen from a deck of 52 cards is:

Correct!

Incorrect. Correct: 1/26

Solution

A standard deck has 4 Queens. Two are red (Queen of Hearts, Queen of Diamonds) and two are black. Probability = $\frac{2}{52} = \frac{1}{26}$.

∴ 1/26

7

P(A ∪ B) is equal to P(A) + P(B) ONLY if the events are:

Correct!

Incorrect. Correct: Mutually Exclusive

Solution

The general formula is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If $P(A \cap B) = 0$, the events are mutually exclusive, simplifying the formula to $P(A) + P(B)$.

∴ Mutually Exclusive

8

If the probability of winning a game is 4/9, what is the probability of losing?

Correct!

Incorrect. Correct: 5/9

Solution

Winning and losing are complementary events. $P(\text{Lose}) = 1 - P(\text{Win}) = 1 - \frac{4}{9} = \frac{5}{9}$.

∴ 5/9

9

Which of the following CANNOT be a probability?

Correct!

Incorrect. Correct: 1.2

Solution

A fundamental axiom of probability is that $0 \le P(E) \le 1$. Therefore, 1.2 is mathematically impossible as a probability.

∴ 1.2

10

If P(A ∩ B) = P(A) × P(B), then A and B are:

Correct!

Incorrect. Correct: Independent

Solution

The Multiplication Theorem states that for two independent events, the probability of their intersection is the product of their individual probabilities: $P(A \cap B) = P(A)P(B)$.

∴ Independent

Short Questions (2 Marks Each)

1

Look at the spinner below. If the spinner is spun once, what is the probability that it lands on an EVEN number?

[2 marks]

Solution

The spinner is divided into 4 equal sectors, so total possible outcomes $n(S) = 4$.
The even numbers on the spinner are 2 and 4. Favorable outcomes $n(E) = 2$.

$$P(\text{Even}) = \frac{n(E)}{n(S)} = \frac{2}{4} = \frac{1}{2}$$

∴ 1/2

2

Define a Sample Space and give an example of one for tossing three coins.

[2 marks]

Solution

Definition: A Sample Space (S) is the complete set of all possible outcomes of a random experiment.
Example (3 Coins): $n(S) = 2^3 = 8$ outcomes.
$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.

∴ S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

3

If P(A) = 0.6, P(B) = 0.5, and P(A ∩ B) = 0.2, calculate P(A ∪ B).

[2 marks]

Solution

Use the Addition Theorem:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$P(A \cup B) = 0.6 + 0.5 - 0.2$$
$$P(A \cup B) = 1.1 - 0.2 = 0.9$$

∴ 0.9

4

Explain why the probability of an impossible event is 0.

[2 marks]

Solution

By the classical definition, $P(E) = \frac{\text{favorable outcomes}}{\text{total outcomes}}$.
For an impossible event (like rolling a 7 on a standard 6-sided die), the number of favorable outcomes is exactly 0.
Therefore, $P(E) = \frac{0}{n} = 0$.

∴ Favorable outcomes = 0.

5

A card is drawn from a deck. What is the probability that it is a face card?

[2 marks]

Solution

Total possible outcomes $n(S) = 52$.
Face cards are Jacks, Queens, and Kings. There are 3 per suit, and 4 suits. So, $n(E) = 3 \times 4 = 12$.

$$P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13}$$

∴ 3/13

6

Are the events 'rolling an even number' and 'rolling a number greater than 4' mutually exclusive on a die? Why?

[2 marks]

Solution

Let $A$ = Even number = {2, 4, 6}.
Let $B$ = Number greater than 4 = {5, 6}.

The intersection $A \cap B = \{6\}$.
Since the intersection is NOT empty ($P(A \cap B) \neq 0$), they can happen at the same time (rolling a 6 satisfies both).
Therefore, they are NOT mutually exclusive.

∴ No, because intersection is {6}.

7

If the probability of Ram passing is 0.8 and Shyam passing is 0.7, find the probability that both pass.

[2 marks]

Solution

Assuming Ram and Shyam passing are independent events (one doesn't affect the other).
Use the Multiplication Theorem:
$$P(\text{Both pass}) = P(\text{Ram}) \times P(\text{Shyam})$$
$$P(\text{Both}) = 0.8 \times 0.7 = 0.56$$

∴ 0.56

8

Find the probability of getting a sum of 7 when two dice are thrown.

[2 marks]

Solution

Total outcomes $n(S) = 6 \times 6 = 36$.
Favorable outcomes for sum of 7:
E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
$n(E) = 6$

$$P(\text{Sum 7}) = \frac{6}{36} = \frac{1}{6}$$

∴ 1/6

9

A letter is chosen from the word 'NEPAL'. What is the probability it is a vowel?

[2 marks]

Solution

Total letters $n(S) = 5$.
Vowels in 'NEPAL' are E and A. So, $n(E) = 2$.

$$P(\text{Vowel}) = \frac{2}{5}$$

∴ 2/5

10

If P(Ē) = 0.45, find P(E).

[2 marks]

Solution

Using the Complementary Event formula:
$$P(E) + P(\bar{E}) = 1$$
$$P(E) = 1 - P(\bar{E})$$
$$P(E) = 1 - 0.45 = 0.55$$

∴ 0.55

Long Questions (4/5 Marks Each)

1

State and prove the Addition Theorem for any two events A and B.

[4 marks]

Solution

Statement: For any two events A and B, the probability of occurrence of at least one of them (A or B) is given by: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Proof:
Let $S$ be the sample space with $n$ total equally likely outcomes.
Let $m_1$ be outcomes favorable to A.
Let $m_2$ be outcomes favorable to B.
Let $m_3$ be outcomes favorable to BOTH A and B ($A \cap B$).

The number of outcomes favorable to either A OR B ($A \cup B$) is the outcomes in A, plus outcomes in B, minus the overlap (to avoid double counting):
$$n(A \cup B) = m_1 + m_2 - m_3$$

Divide both sides by total outcomes $n$:
$$\frac{n(A \cup B)}{n} = \frac{m_1}{n} + \frac{m_2}{n} - \frac{m_3}{n}$$

By the definition of classical probability ($P(E) = n(E)/n(S)$):
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) \quad \blacksquare$$

∴ Proved.

2

A card is drawn at random from a deck of 52. Find the probability that it is either a Club or an Ace.

[4 marks]

Solution

Let $C$ = drawing a Club, and $A$ = drawing an Ace.
Total outcomes $n(S) = 52$.

Probability of a Club: $P(C) = \frac{13}{52}$
Probability of an Ace: $P(A) = \frac{4}{52}$
Probability of both (Ace of Clubs): $P(C \cap A) = \frac{1}{52}$

Since the word is 'OR', apply the Addition Theorem:
$$P(C \cup A) = P(C) + P(A) - P(C \cap A)$$
$$P(C \cup A) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52}$$
$$P(C \cup A) = \frac{16}{52} = \frac{4}{13}$$

∴ 4/13

3

Two dice are thrown together. Find the probability that the sum is divisible by 3 or 4.

[5 marks]

Solution

Total outcomes $n(S) = 36$. Let $A$ = sum divisible by 3, $B$ = sum divisible by 4.
Possible sums from 2 dice range from 2 to 12.

Sums divisible by 3 are 3, 6, 9, 12.
Favorable pairs for A: (1,2), (2,1), (1,5), (2,4), (3,3), (4,2), (5,1), (3,6), (4,5), (5,4), (6,3), (6,6) $\rightarrow n(A) = 12$.
$$P(A) = \frac{12}{36}$$

Sums divisible by 4 are 4, 8, 12.
Favorable pairs for B: (1,3), (2,2), (3,1), (2,6), (3,5), (4,4), (5,3), (6,2), (6,6) $\rightarrow n(B) = 9$.
$$P(B) = \frac{9}{36}$$

Intersection ($A \cap B$): Sums divisible by BOTH 3 and 4. LCM of 3 and 4 is 12. So sum must be 12.
Favorable pair for sum 12: (6,6) $\rightarrow n(A \cap B) = 1$.
$$P(A \cap B) = \frac{1}{36}$$

By Addition Theorem:
$$P(A \cup B) = \frac{12}{36} + \frac{9}{36} - \frac{1}{36} = \frac{20}{36} = \frac{5}{9}$$

∴ 5/9

4

In a leap year, what is the probability of having 53 Tuesdays?

[4 marks]

Solution

A leap year has 366 days.
$$366 \text{ days} = 52 \text{ complete weeks} + 2 \text{ extra days}$$
52 complete weeks guarantee exactly 52 Tuesdays.
The remaining 2 extra days are consecutive. The sample space for these 2 days is:
$S$ = {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)}
$n(S) = 7$

To get 53 Tuesdays, the extra 2 days MUST include a Tuesday. The favorable outcomes are:
$E$ = {(Mon, Tue), (Tue, Wed)}
$n(E) = 2$

$$P(53\text{ Tuesdays}) = \frac{n(E)}{n(S)} = \frac{2}{7}$$

∴ 2/7

5

Prove that for two independent events A and B, $P(A \cup B) = 1 - P(\bar{A})P(\bar{B})$.

[4 marks]

Solution

We start with the General Addition Theorem:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Since A and B are independent, $P(A \cap B) = P(A)P(B)$. Substitute this:
$$P(A \cup B) = P(A) + P(B) - P(A)P(B)$$

Factor out $P(B)$ from the last two terms:
$$P(A \cup B) = P(A) + P(B)[1 - P(A)]$$

Recall the complement rule: $1 - P(A) = P(\bar{A})$. Substitute this in:
$$P(A \cup B) = P(A) + P(B)P(\bar{A})$$

Now, rewrite $P(A)$ as $1 - P(\bar{A})$:
$$P(A \cup B) = 1 - P(\bar{A}) + P(B)P(\bar{A})$$

Factor out $-P(\bar{A})$ from the last two terms:
$$P(A \cup B) = 1 - P(\bar{A})[1 - P(B)]$$

Substitute $1 - P(B) = P(\bar{B})$:
$$P(A \cup B) = 1 - P(\bar{A})P(\bar{B}) \quad \blacksquare$$

∴ Proved.

6

Derive the Multiplication Theorem for two independent events A and B.

[4 marks]

Solution

Statement: If A and B are independent events, the probability of both occurring simultaneously is $P(A \cap B) = P(A) \times P(B)$.

Proof:
Let $n_1$ be the total outcomes of experiment 1, and $m_1$ be the outcomes favorable to event A.
So, $P(A) = \frac{m_1}{n_1}$.

Let $n_2$ be the total outcomes of experiment 2, and $m_2$ be the outcomes favorable to event B.
So, $P(B) = \frac{m_2}{n_2}$.

By the fundamental principle of counting, the total possible outcomes of combining both experiments is $n = n_1 \times n_2$.
The total favorable outcomes for BOTH A and B happening is $m = m_1 \times m_2$.

By classical probability:
$$P(A \cap B) = \frac{m}{n} = \frac{m_1 \times m_2}{n_1 \times n_2}$$
$$P(A \cap B) = \left(\frac{m_1}{n_1}\right) \times \left(\frac{m_2}{n_2}\right)$$
$$P(A \cap B) = P(A) \times P(B) \quad \blacksquare$$

∴ Proved.

7

If P(A) = 2/5, P(B) = 1/3, and P(A ∩ B) = 1/5, are A and B independent? Justify your answer.

[4 marks]

Solution

For two events to be independent, they MUST satisfy the condition:
$$P(A \cap B) = P(A) \times P(B)$$

Let's calculate the product of their individual probabilities:
$$P(A) \times P(B) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$$

The given intersection probability is:
$$P(A \cap B) = \frac{1}{5} = \frac{3}{15}$$

Since $\frac{3}{15} \neq \frac{2}{15}$, $P(A \cap B) \neq P(A) \times P(B)$.

Conclusion: The events A and B are NOT independent.

∴ Not independent.

Past NEB Board Questions

1

Two dice are thrown. Find the probability that the sum of the numbers appearing is 7.

NEB 2079 [3 marks]

Solution

Total outcomes when two dice are thrown: $n(S) = 6 \times 6 = 36$.
Let $E$ be the event of getting a sum of 7.
Favorable outcomes $E$ = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}.
$n(E) = 6$.

$$P(E) = \frac{6}{36} = \frac{1}{6}$$.

∴ 1/6

2

A card is drawn from a well-shuffled deck of 52 cards. What is the probability of getting a King or a red card?

NEB 2078 [4 marks]

Solution

Let $K$ be the event of drawing a King, and $R$ be drawing a Red card.
$n(S) = 52$.
$P(K) = \frac{4}{52}$
$P(R) = \frac{26}{52}$

Overlap ($K \cap R$): There are 2 Red Kings (Hearts and Diamonds).
$P(K \cap R) = \frac{2}{52}$

By Addition Theorem:
$$P(K \cup R) = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}$$

∴ 7/13

3

Three students A, B and C solve a problem with probabilities 1/3, 1/4 and 1/5 respectively. What is the probability that the problem is solved?

NEB 2077 [5 marks]

Solution

Use the 'At least one' trick. The problem is solved if at least one student solves it.

Find failure probabilities:
$P(\bar{A}) = 1 - 1/3 = 2/3$
$P(\bar{B}) = 1 - 1/4 = 3/4$
$P(\bar{C}) = 1 - 1/5 = 4/5$

Probability NO ONE solves it:
$$P(\text{None}) = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{24}{60} = \frac{2}{5}$$

Probability AT LEAST ONE solves it:
$$P(\text{Solved}) = 1 - \frac{2}{5} = \frac{3}{5}$$

∴ 3/5

Chapter Test — Full Mixed Paper

Attempt this test in 45 minutes under exam conditions to prepare for the NEB finals.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

Probability cannot be:

Correct!

Incorrect. Correct: 1.5

Solution

Because $0 \le P(E) \le 1$.

∴ 1.5

2

If events are mutually exclusive, what is P(A ∩ B)?

Correct!

Incorrect. Correct: 0

Solution

By definition, mutually exclusive events cannot overlap.

∴ 0

2 Section B — Short Questions (2 Marks Each)

1

State the Addition Theorem for mutually exclusive events.

[2 marks]

Solution

If A and B are mutually exclusive events, then the probability of occurrence of either A or B is the sum of their individual probabilities: $P(A \cup B) = P(A) + P(B)$.

∴ P(A ∪ B) = P(A) + P(B)

2

Find the probability of drawing an Ace from a standard deck of cards.

[2 marks]

Solution

Total cards $n(S) = 52$. Number of Aces $n(E) = 4$.
$P(E) = \frac{4}{52} = \frac{1}{13}$.

∴ 1/13

3 Section C — Long Questions (4/5 Marks Each)

1

If P(A) = 0.6, P(B) = 0.5, and P(A ∩ B) = 0.2, find P(A ∪ B). Are A and B independent?

[5 marks]

Solution

1. Find $P(A \cup B)$ using Addition Theorem:
$P(A \cup B) = 0.6 + 0.5 - 0.2 = 1.1 - 0.2 = 0.9$.
2. Check for Independence:
Multiply $P(A)$ and $P(B)$: $0.6 \times 0.5 = 0.3$.
Check if it equals intersection: $0.3 \neq 0.2$.
Since $P(A \cap B) \neq P(A)P(B)$, they are NOT independent.

∴ Union = 0.9, Not Independent.

2

A coin is tossed two times. Draw the tree diagram and find the probability of getting exactly one head.

[4 marks]

Solution

From the tree diagram, Sample space $n(S) = 4$.
Favorable (exactly 1 head) = $\{HT, TH\}$.
$n(E) = 2$.
Probability = $2/4 = 1/2$.

∴ 1/2

One-Page Revision Cheat Sheet

1

Formula: $P(E) = \frac{\text{favorable outcomes}}{\text{total outcomes}}$.

2

Range: Probability is always between 0 (Impossible) and 1 (Sure).

3

Complement Rule: $P(E) + P(\bar{E}) = 1$. Use for 'NOT' questions.

4

Addition Theorem ('OR'): $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Subtract the overlap!

5

Mutually Exclusive: Events cannot happen together. $P(A \cap B) = 0$.

6

Multiplication Theorem ('AND'): $P(A \cap B) = P(A) \times P(B)$ (only if independent).

7

Independent: One event doesn't affect the other. $P(A \cap B) = P(A)P(B)$.

8

'At Least One' Trick: $P(\text{At least one}) = 1 - P(\text{None})$.

9

Cards Fact: 52 total. 4 suits (13 each). 12 Face cards (J, Q, K). Ace is NOT a face card.

10

Dice Fact: 2 dice have $6 \times 6 = 36$ outcomes. Sum of 7 is the most likely (1/6).