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CBSE - Class 11 Mathematics Probability Worksheet
What is the probability of getting at least one head when 3 coins are tossed together?
1/8
b.1/4
c.1/2
d.7/8
Worksheet Answers
Solution:
When three coins are tossed simultaneously, each coin has $2$ possible outcomes: Head ($H$) or Tail ($T$). Since the coins are independent, the total number of outcomes in the sample space $S$ is given by $2^n$, where $n$ is the number of coins.
$n(S) = 2^3 = 8$
The exhaustive list of possible outcomes is defined as follows:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
The event $E$ is defined as "getting no tail." In terms of the sample space, this signifies that all three coins must result in Heads ($H$).
Favorable outcomes: $E = \{HHH\}$
Number of favorable outcomes, $n(E) = 1$
Per the classical definition of probability, the probability of an event $E$ occurring is the ratio of the number of favorable outcomes to the total number of equally likely outcomes in the sample space [Laplace's Definition of Probability]:
$P(E) = \frac{n(E)}{n(S)}$
$P(E) = \frac{1}{8}$
| Outcome | Tail Count | Condition Met (Zero Tails) |
|---|---|---|
| HHH | 0 | Yes |
| HHT, HTH, THH | 1 | No |
| HTT, THT, TTH | 2 | No |
| TTT | 3 | No |
Final Solution: The probability of getting no tail when three coins are tossed is $\frac{1}{8}$ or $0.125$.
Solution:
Let the total number of students in the class be $n(S)$.
We are required to find the number of students who are in $B$ but not in $A$. This set is denoted as $(B - A)$ or $(B \cap A^c)$. [By Set Theory: $n(B - A) = n(B) - n(A \cap B)$]
Substituting the given values:
$n(B - A) = 32 - 24$ $n(B - A) = 8$The probability $P(E)$ of an event occurring is defined by the ratio of favorable outcomes to the total sample space. [By Classical Definition of Probability: $P(E) = \frac{n(E)}{n(S)}$]
Let $E$ be the event that a student opted for NSS but not NCC:
$P(E) = \frac{n(B - A)}{n(S)}$ $P(E) = \frac{8}{60}$Reducing the fraction to its simplest form:
$P(E) = \frac{8 \div 4}{60 \div 4} = \frac{2}{15}$Final Solution: The probability that the student has opted for NSS but not NCC is $\frac{2}{15}$.
Solution:
To determine if the given values for $P(A)$, $P(B)$, and $P(A \cap B)$ are consistently defined, we must verify them against the fundamental axioms of probability theory, specifically the Axiom of Monotonicity and the properties of subsets within a sample space $S$.
A set of probabilities is consistent if and only if it satisfies the following conditions:
Given values:
We test the condition $P(A \cap B) \leq P(A)$:
$0.6 \leq 0.5$
This inequality is false. By the definition of the intersection of two sets, the probability of the intersection cannot exceed the probability of any individual event that constitutes that intersection [Per the Monotonicity Property of Measures]. If $A \cap B \subseteq A$, then the measure $P(A \cap B)$ must be less than or equal to the measure $P(A)$.
Since the condition $P(A \cap B) \leq P(A)$ is violated ($0.6 > 0.5$), the event $(A \cap B)$ contains elements that are not contained within event $A$. This is a logical contradiction in the context of set theory and Kolmogorov's axioms of probability.
Final Solution: The probabilities $P(A) = 0.5, P(B) = 0.7, P(A \cap B) = 0.6$ are not consistently defined because they violate the axiom $P(A \cap B) \leq P(A)$.
Solution:
In a standard deck of cards, the total number of distinct outcomes ($n(S)$) is defined by the total number of cards in the pack.
$n(S) = 52$ [Definition of a standard deck probability space].
Let $E$ be the event of drawing an 'Ace'. A standard deck contains four suits (Hearts, Diamonds, Clubs, and Spades), and each suit contains exactly one Ace.
$n(E) = 4$ [By the structure of standard 52-card decks].
The probability $P(E)$ of an event occurring is defined by the ratio of favorable outcomes to the total sample space:
$P(E) = \frac{n(E)}{n(S)}$ [Laplace's Definition of Classical Probability].
Substituting the values:
$P(E) = \frac{4}{52}$
To reduce the fraction, find the greatest common divisor (GCD) of 4 and 52, which is 4:
$P(E) = \frac{4 \div 4}{52 \div 4} = \frac{1}{13}$
Final Solution: The probability of selecting an Ace from a deck of 52 cards is $\frac{1}{13}$.
Solution:
When two six-sided dice are tossed, the sample space $S$ consists of all ordered pairs $(x, y)$, where $x$ represents the result of the first die and $y$ represents the result of the second die. The total number of outcomes is $6 \times 6 = 36$.
$S = \{ (1,1), (1,2), \dots, (6,6) \}$
Event A: Getting an even number on the first die.
$A = \{ (x, y) \in S : x \in \{2, 4, 6\} \}$
$A = \{ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \}$
Event B: Getting an odd number on the first die.
$B = \{ (x, y) \in S : x \in \{1, 3, 5\} \}$
$B = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \}$
In set theory, "A or B" corresponds to the union of sets $A$ and $B$, denoted as $A \cup B$. This set contains all elements that are in $A$, in $B$, or in both [Per the Definition of Union in Probability Theory].
$A \cup B = \{ \omega : \omega \in A \text{ or } \omega \in B \}$
Since the first die must be either even $\{2, 4, 6\}$ or odd $\{1, 3, 5\}$, and every integer from 1 to 6 falls into one of these two categories, the union covers the entire sample space $S$.
| Die 1 | Die 2 (Possible outcomes) |
|---|---|
| 1 | 1, 2, 3, 4, 5, 6 |
| 2 | 1, 2, 3, 4, 5, 6 |
| 3 | 1, 2, 3, 4, 5, 6 |
| 4 | 1, 2, 3, 4, 5, 6 |
| 5 | 1, 2, 3, 4, 5, 6 |
| 6 | 1, 2, 3, 4, 5, 6 |
Because $A \cap B = \emptyset$ (the first die cannot be both even and odd simultaneously) and $A \cup B = S$, the event "A or B" represents the Certain Event, as the first die will necessarily yield either an even or an odd number.
Final Solution: The event "A or B" is equal to the sample space S, containing all 36 possible outcomes of throwing two dice.
Solution:
The problem provides the following probabilities for events $E$ and $F$ within a sample space $S$:
We are required to determine $P(E \cup F)$, which represents the probability of event "$E$ or $F$".
The probability of the union of two events is governed by the Addition Theorem of Probability, which accounts for the inclusion-exclusion principle to prevent double-counting of the intersection [Per the Axioms of Probability].
The formula is defined as:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
Substituting the given fractional values into the established equation:
$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$
To perform the arithmetic operation, identify the Least Common Multiple (LCM) of the denominators $4, 2,$ and $8$. The LCM is $8$.
Convert each term to an equivalent fraction with a denominator of $8$:
Perform the summation and subtraction:
$P(E \cup F) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8}$
$P(E \cup F) = \frac{2 + 4 - 1}{8}$
$P(E \cup F) = \frac{5}{8}$
Final Solution: $P(E \text{ or } F) = \frac{5}{8}$
Solution:
Based on the provided probability space, we define the following parameters:
The event "not $E$ and not $F$" is represented symbolically as $P(E^c \cap F^c)$, where $E^c$ and $F^c$ denote the complements of events $E$ and $F$, respectively.
Per De Morgan's Law of Sets, the intersection of the complements of two sets is equal to the complement of their union:
$E^c \cap F^c = (E \cup F)^c$
[By De Morgan’s Law: $(A^c \cap B^c) = (A \cup B)^c$]
To find the probability of the complement of the union, we must first determine $P(E \cup F)$ using the Addition Theorem of Probability:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
Substituting the given values:
$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$
Finding a common denominator (8):
$P(E \cup F) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}$
[Per the Addition Theorem for non-mutually exclusive events]
Given the Complement Rule of Probability:
$P(E^c \cap F^c) = P((E \cup F)^c) = 1 - P(E \cup F)$
$P(E^c \cap F^c) = 1 - \frac{5}{8}$
$P(E^c \cap F^c) = \frac{8 - 5}{8} = \frac{3}{8}$
[Per the Complement Rule: $P(A^c) = 1 - P(A)$]
Final Solution: $P(\text{not } E \text{ and not } F) = \frac{3}{8}$
Solution:
The experiment consists of tossing a fair, six-faced die. The sample space $S$ is defined as the set of all possible outcomes. Since a standard die has faces numbered 1 through 6, the sample space is:
$S = \{1, 2, 3, 4, 5, 6\}$
[Per the Classical Definition of Probability, the total number of outcomes $n(S) = 6$.]
Event $A$ is defined as the occurrence of a "number less than 7." We must identify all elements $x \in S$ such that $x < 7$.
By set notation, event $A$ is a subset of the sample space $S$ [By the definition of an event as a subset of the sample space].
$A = \{1, 2, 3, 4, 5, 6\}$
Since every outcome in the sample space satisfies the condition of being less than 7, event $A$ is equivalent to the sample space itself.
[In probability theory, such an event is classified as a Sure Event or a Certain Event because $A = S$.]
Final Solution: The event A is {1, 2, 3, 4, 5, 6}.
Solution:
When a standard fair six-sided die is thrown, the set of all possible outcomes, known as the sample space ($S$), consists of all integers from 1 to 6. [Per the Axioms of Probability]
$S = \{1, 2, 3, 4, 5, 6\}$
The total number of elementary events is given by $n(S) = 6$.
Let $E$ be the event that the number obtained is less than 6. We identify all elements in $S$ that satisfy the condition $x < 6$.
The favorable outcomes are: $\{1, 2, 3, 4, 5\}$.
The number of favorable outcomes is $n(E) = 5$.
The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes in the sample space. [By the Classical Definition of Probability]
$P(E) = \frac{n(E)}{n(S)}$
$P(E) = \frac{5}{6}$
| Metric | Value |
|---|---|
| Total Outcomes ($n(S)$) | 6 |
| Favorable Outcomes ($n(E)$) | 5 |
| Probability ($P(E)$) | $0.833$ (approx) |
Final Solution: The probability of getting a number less than 6 is $\frac{5}{6}$.
Solution:
When three unbiased coins are tossed simultaneously, each coin has $2$ possible outcomes: Heads ($H$) or Tails ($T$). The total number of outcomes is determined by the Fundamental Counting Principle: $2 \times 2 \times 2 = 8$.
The sample space $S$ is defined as:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Two events $A$ and $B$ are defined as mutually exclusive if they cannot occur simultaneously. Formally, $A \cap B = \emptyset$, where $\emptyset$ is the null (empty) set. [By the Axiomatic Definition of Probability].
To satisfy the condition of mutual exclusivity, we must identify two subsets of $S$ that share no common elements.
To verify that $A$ and $B$ are mutually exclusive, we calculate their intersection:
$A \cap B = \{HHH\} \cap \{TTT\} = \emptyset$
[By the Principle of Disjoint Sets, since no outcome exists in both sets, the probability $P(A \cap B) = 0$.]
Since the set of outcomes where three heads appear and the set of outcomes where three tails appear have no intersection, they are mutually exclusive.
Final Solution: Event $A = \{HHH\}$ (getting three heads) and Event $B = \{TTT\}$ (getting three tails) are mutually exclusive.
Solution:
When a standard six-faced die is rolled, the set of all possible outcomes, denoted as the sample space $S$, is defined as:
$S = \{1, 2, 3, 4, 5, 6\}$
[By the definition of a discrete sample space for a fair die]
Event $E$ is defined as the event where the die shows a 4. Thus:
$E = \{4\}$
Event $F$ is defined as the event where the die shows an even number. The even numbers in the sample space $S$ are 2, 4, and 6. Thus:
$F = \{2, 4, 6\}$
To determine if two events are mutually exclusive, we must calculate the intersection of the two sets. Two events are mutually exclusive if and only if $E \cap F = \emptyset$, meaning the events share no common outcomes.
$E \cap F = \{4\} \cap \{2, 4, 6\}$
$E \cap F = \{4\}$
[Per the set-theoretic definition of intersection]
Since $E \cap F = \{4\}$, and $\{4\} \neq \emptyset$ (the intersection is not the empty set), there is an outcome common to both events.
Because the die showing "4" satisfies both the condition of being a "4" and the condition of being an "even number," the events are not mutually exclusive.
Final Solution: No, E and F are not mutually exclusive because E $\cap$ F = {4} $\neq$ $\emptyset$.
Solution:
When tossing three unbiased coins simultaneously, each coin has two possible outcomes: Heads ($H$) or Tails ($T$). The total number of elementary outcomes in the sample space $S$ is calculated as $2^n$, where $n=3$.
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
The total number of outcomes $|S| = 8$.
We define three events based on the number of heads appearing in the toss:
To confirm mutual exclusivity, we check the intersection of all pairs:
[By the definition of intersection of disjoint sets, the events are mutually exclusive].
To confirm the events are exhaustive, we take the union of $A, B, \text{ and } C$:
$A \cup B \cup C = \{TTT\} \cup \{HTT, THT, TTH\} \cup \{HHH, HHT, HTH, THH\}$
$A \cup B \cup C = \{TTT, HTT, THT, TTH, HHH, HHT, HTH, THH\} = S$
[Since the union of the events spans all possible outcomes in $S$, the collection is exhaustive].
Final Solution: The three events $A = \{TTT\}$, $B = \{HTT, THT, TTH\}$, and $C = \{HHH, HHT, HTH, THH\}$ are mutually exclusive and exhaustive.
Solution:
When a standard fair six-sided die is tossed, the possible outcomes are represented by the set of integers from 1 to 6. Per the definition of a sample space in classical probability:
$S = \{1, 2, 3, 4, 5, 6\}$
The total number of possible outcomes is denoted by $n(S)$:
$n(S) = 6$
Let $E$ be the event that "a number more than 6 appears." Mathematically, we define $E$ as the subset of $S$ such that every element $x \in E$ satisfies $x > 6$.
Reviewing the set $S = \{1, 2, 3, 4, 5, 6\}$, we observe that there is no integer in the sample space that satisfies the condition $x > 6$.
$E = \emptyset$ (The Null Set or Impossible Event)
Therefore, the number of favorable outcomes is:
$n(E) = 0$
Per the axiomatic definition of probability, the probability of an event $E$ is the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space [Laplace's Definition of Probability]:
$P(E) = \frac{n(E)}{n(S)}$
Substituting the known values:
$P(E) = \frac{0}{6}$
$P(E) = 0$
Since the condition "a number more than 6" cannot be satisfied by any element within the sample space $S$, the event is classified as an impossible event. By the axioms of probability, the probability of an impossible event is always 0.
Final Solution: The probability of the event is $0$.
Solution:
When one coin is tossed, the outcomes are $\{H, T\}$. Since three coins are tossed, the sample space $S$ consists of all possible triplets of heads ($H$) and tails ($T$). The total number of outcomes is given by $2^n$, where $n=3$.
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
The total number of elements in the sample space, $n(S)$, is $8$. [Per the Fundamental Counting Principle].
The event $E$ represents "at least 2 heads." This implies the outcome must contain either exactly 2 heads or exactly 3 heads. We identify these outcomes from the set $S$:
Therefore, $E = \{HHT, HTH, THH, HHH\}$.
The number of favorable outcomes, $n(E)$, is $4$.
The probability of an event is defined by the ratio of favorable outcomes to the total number of outcomes in the sample space: $P(E) = \frac{n(E)}{n(S)}$. [Per the Classical Definition of Probability].
$P(E) = \frac{4}{8}$
Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor (GCD), which is $4$:
$P(E) = \frac{4 \div 4}{8 \div 4} = \frac{1}{2}$
| Event Criteria | Sample Points | Count |
|---|---|---|
| At least 2 Heads | HHH, HHT, HTH, THH | 4 |
| Total Sample Space | HHH, HHT, HTH, THH, HTT, THT, TTH, TTT | 8 |
Final Solution: The probability of getting at least 2 heads is $\frac{1}{2}$ or $0.5$.
Solution:
To determine the probability of the union of two events $A$ and $B$, denoted as $P(A \cup B)$ or $P(A \text{ or } B)$, we utilize the General Addition Rule of Probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
[Per the Axioms of Probability defined by Kolmogorov]
The problem states that events $A$ and $B$ are mutually exclusive. By definition, two events are mutually exclusive if they cannot occur simultaneously. Mathematically, this implies that the intersection of the two events is an empty set:
$A \cap B = \emptyset$
Consequently, the probability of the intersection is zero:
$P(A \cap B) = 0$
[By the definition of mutually exclusive events in set theory]
Substituting the given values and the constraint derived in Step 2 into the General Addition Rule:
$P(A) = \frac{3}{5}$
$P(B) = \frac{1}{5}$
$P(A \cup B) = \frac{3}{5} + \frac{1}{5} - 0$
$P(A \cup B) = \frac{3 + 1}{5}$
$P(A \cup B) = \frac{4}{5}$
Final Solution: The probability of A or B occurring is $\frac{4}{5}$.
Solution:
When three unbiased coins are tossed simultaneously, each coin has two possible outcomes: Head ($H$) or Tail ($T$). For three coins, the total number of elementary outcomes is calculated by $2^n$ where $n=3$.
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Total number of outcomes, $n(S) = 2^3 = 8$ [Per the Multiplication Principle of Counting].
The event $E$ is defined as "at most two tails." This signifies that the number of tails ($T$) in any outcome must be less than or equal to 2. Mathematically, $T \in \{0, 1, 2\}$.
The set of outcomes for event $E$ is $\{HHH, HHT, HTH, THH, HTT, THT, TTH\}$.
Total number of favorable outcomes, $n(E) = 7$.
The probability $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of outcomes in the sample space [Per the Classical Definition of Probability]:
$P(E) = \frac{n(E)}{n(S)}$
$P(E) = \frac{7}{8}$
The event "at most two tails" is the complement of the event "three tails" (getting 3 tails). Let $E'$ be the event of getting 3 tails: $E' = \{TTT\}$.
$n(E') = 1$
$P(E') = \frac{1}{8}$
Using the Complementary Probability Rule: $P(E) = 1 - P(E')$ [By the Axioms of Probability].
$P(E) = 1 - \frac{1}{8} = \frac{7}{8}$
Final Solution: The probability of getting at most two tails is $\frac{7}{8}$ or $0.875$.
Solution:
To determine the probability of the union of two events $A$ and $B$, we identify the following parameters provided by the problem set:
To find the probability of event $A$ or event $B$ occurring, we utilize the Addition Theorem of Probability [Per the Axioms of Probability]:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
This formula accounts for the inclusion-exclusion principle, ensuring that the intersection (the overlapping region of the two sets) is not double-counted.
Substitute the given variables into the derived formula:
$P(A \cup B) = 0.42 + 0.48 - 0.16$
Perform the summation and subtraction operations with high precision:
The calculation $0.42 + 0.48 - 0.16$ yields a result within the valid probability range $[0, 1]$. As $0.74 \leq 1$, the solution is mathematically consistent with the Kolmogorov axioms of probability.
Final Solution: $P(A \cup B) = 0.74$
Solution:
When three coins are tossed, each coin has two possible outcomes: Head ($H$) or Tail ($T$). Since there are three coins, the total number of elementary outcomes is given by $2^n$ where $n=3$.
The sample space $S$ is defined as:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Total number of outcomes, $n(S) = 8$.
Two events $A$ and $B$ are defined as not mutually exclusive if they have at least one common outcome, i.e., $A \cap B \neq \emptyset$ [Per the definition of Mutually Exclusive Events in Probability Theory].
Let us define the following two events:
Set A (At least two heads):
$A = \{HHH, HHT, HTH, THH\}$
Set B (At least two tails):
$B = \{HTT, THT, TTH, TTT\}$
To verify if these events are not mutually exclusive, we evaluate the intersection $A \cap B$.
By observing the sets $A$ and $B$:
$A \cap B = \emptyset$ (These specific sets are mutually exclusive).
Revision: To satisfy the requirement of "not mutually exclusive," we must redefine the events to ensure a non-empty intersection.
Let us define new events:
Set C: $\{HHT, HTH, THH\}$
Set D: $\{HHH, HHT, HTH, HTT\}$
Checking the intersection of $C$ and $D$:
$C \cap D = \{HHT, HTH\}$
Since $C \cap D \neq \emptyset$, the events share common outcomes ($HHT$ and $HTH$).
Final Solution: Two events that are not mutually exclusive are: Event C (getting exactly two heads) and Event D (getting a head on the first coin), because $C \cap D = \{HHT, HTH\}$, which is not an empty set.
Solution:
Let $C$ be the random variable representing the outcome of the coin toss, where $C \in \{1, 6\}$. Let $D$ be the random variable representing the outcome of the standard fair die toss, where $D \in \{1, 2, 3, 4, 5, 6\}$.
The sample space $S$ consists of all possible ordered pairs $(c, d)$ where $c \in \{1, 6\}$ and $d \in \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes $n(S)$ is calculated by the Fundamental Counting Principle [Per the Product Rule of Combinatorics]:
$n(S) = n(C) \times n(D) = 2 \times 6 = 12$
The set $S$ is as follows:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let $E$ be the event that the sum of the numbers on the coin and the die is equal to $12$. We denote the sum as $S_{sum} = c + d$. We seek the subset of $S$ such that $c + d = 12$.
Thus, the event $E = \{(6, 6)\}$. The number of favorable outcomes is $n(E) = 1$.
The probability $P(E)$ of an event in a sample space with equally likely outcomes is defined by the ratio of favorable outcomes to the total number of outcomes [Per the Classical Definition of Probability]:
$P(E) = \frac{n(E)}{n(S)}$
$P(E) = \frac{1}{12}$
Final Solution: The probability that the sum of the numbers is 12 is $\frac{1}{12}$.
