Class 11 Mathematics Chapter 14 Probability

Key Concept: Mutually Exclusive Events

c) Pair $A$ and $B$

[Solution Description] For events to be mutually exclusive, their intersection must be empty. Calculate the intersections:

$$A \cap B = \{1, 3, 5\} \cap \{2, 4, 6\} = \emptyset$$

$$C \cap D = \{1, 2, 3\} \cap \{3, 4, 5\} = \{3\}$$

Only the pair $A$ and $B$ has an empty intersection, so they are mutually exclusive.

Key Concept: Probability in Games, Conditional Probability

a) 0.4167

[Solution Description] Apply the formula for the union of two events:

$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$

Substitute the provided values:

$$P(A \cup B) = 0.2778 + 0.1667 – 0.0278$$

Calculate:

$$P(A \cup B) = 0.4167$$

So, the probability is 0.4167.

Key Concept: Complex Union and Intersection, Conditional Probability

c) 0.9

[Solution Description] Let $P(A)$ be the probability of reading newspaper A, and $P(B)$ be the probability of reading newspaper B. We have $$P(A) = 0.7$$, $$P(B) = 0.5$$, and $$P(A \cap B) = 0.3$$. We need to find $$P(A \cup B)$$ which is given by:

$$P(A \cup B) = P(A) + P(B) – P(A \cap B).$$

Substituting the given values:

$$P(A \cup B) = 0.7 + 0.5 – 0.3 = 0.9.$$

Therefore, the probability that a randomly selected person reads at least one of the newspapers is 0.9.

Key Concept: Complex Conditional Probability, Non-Mutually Exclusive Events

c) 0.4737

[Solution Description] Let $D$ be the event “defective” and $A$ be the event “produced by Line A”. We need to find $P(A | D)$ using Bayes’ theorem:

$$P(A | D) = \frac{P(D | A) \cdot P(A)}{P(D)}$$

Where:

$P(D | A) = 0.03$, $P(A) = 0.60$, $P(D | B) = 0.05$, and $P(B) = 0.40$.

Calculate $P(D)$:

$$P(D) = P(D | A) \cdot P(A) + P(D | B) \cdot P(B)$$

$$= 0.03 \times 0.60 + 0.05 \times 0.40 = 0.018 + 0.020 = 0.038$$

Now, calculate $P(A | D)$:

$$P(A | D) = \frac{0.03 \times 0.60}{0.038} = \frac{0.018}{0.038} \approx 0.4737$$

Therefore, the probability is approximately 0.4737.

Key Concept: Conditional Probability Basics

b) 0.5

[Solution Description] To find $P(B|A)$, we use the formula for conditional probability:

$$P(B|A) = \frac{P(A \cap B)}{P(A)}$$

Substitute the given values:

$$P(B|A) = \frac{0.3}{0.6} = 0.5$$

So, $$P(B|A) = 0.5$$.

Key Concept: Complex Event Analysis, Probability Calculation

b) $\frac{11}{26}$

[Solution Description] There are 13 hearts in a deck of 52 cards. The total number of face cards (Jack, Queen, King) across all suits is 12. However, among these, 3 are also hearts (Jack, Queen, King of hearts). Hence, the union of hearts and face cards consists of $13 + 12 – 3 = 22$ cards, as we must subtract the overlapping face cards that are also hearts.

The probability is then calculated by dividing the number of favorable outcomes by the total number of possible outcomes:

$$P(\text{Heart or Face Card}) = \frac{22}{52} = \frac{11}{26}$$

Key Concept: Complex Probability Assignment, Advanced Mutually Exclusive

b) 0.9

[Solution Description] The given events $A$ and $B$ are mutually exclusive, hence $$P(A \cup B) = P(A) + P(B)$$. Firstly, verify the validity of the probabilities:

$$P(S) = P(\omega_1) + P(\omega_2) + P(\omega_3) + P(\omega_4) = 0.2 + 0.3 + 0.25 + 0.15 = 0.9$$

This does not sum to 1, meaning the probability assignment is incorrect according to axiomatic rules. However, since we need to find $P(A \cup B)$:

First calculate $P(A)$:

$$P(A) = P(\omega_1) + P(\omega_2) = 0.2 + 0.3 = 0.5$$

Then calculate $P(B)$:

$$P(B) = P(\omega_3) + P(\omega_4) = 0.25 + 0.15 = 0.4$$

Since $A$ and $B$ are mutually exclusive:

$$P(A \cup B) = P(A) + P(B) = 0.5 + 0.4 = 0.9$$

The probability assignments are invalid, but $P(A \cup B)$ calculation yields 0.9.

Key Concept: Exhaustive and Exclusive Analysis

a) Yes

[Solution Description] Events are mutually exclusive if $$P \cap Q = P \cap R = P \cap S = Q \cap R = Q \cap S = R \cap S = \phi$$; all intersections must be empty:

– $P \cap Q = \phi$

– $P \cap R = \phi$

– $P \cap S = \phi$

– $Q \cap R = \phi$

– $Q \cap S = \phi$

– $R \cap S = \phi$

All intersections are indeed empty, hence they are mutually exclusive.

To check if they are exhaustive, we find their union:

$$P \cup Q \cup R \cup S = \{2, 4, 6, 1, 3, 5, 7, 8, 9, 10, 11\}$$

Assume the sample space $$T = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$$. Since $$P \cup Q \cup R \cup S = T$$, they are exhaustive.

Therefore, the events are both mutually exclusive and exhaustive.

Key Concept: Complex Event Analysis, Probability with Conditions

c) No; $\frac{13}{204}$

[Solution Description] Two events are mutually exclusive if they cannot happen at the same time, meaning $E \cap F = \phi$. However, in this scenario, it is possible to draw a heart first and then a diamond. Therefore, these events are not mutually exclusive.

To find $P(E \cap F)$:

– Number of ways to choose a heart as the first card = 13.

– Remaining cards = 51.

– Number of ways to choose a diamond as the second card = 13.

– Therefore, $$P(E \cap F) = \frac{13}{52} \times \frac{13}{51}$$.

Calculating:

$$P(E \cap F) = \frac{13 \times 13}{52 \times 51} = \frac{169}{2652} = \frac{13}{204}.$$

Thus, $$P(E \cap F) = \frac{13}{204}$$.

Key Concept: Axiomatic Proofs, Complex Probability Problems

b) Independent

[Solution Description] For $X$ and $Y$ to be independent, the following condition must hold:

$$P(X \cap Y) = P(X) \cdot P(Y)$$

First, calculate each probability:

Since there are 4 equally likely outcomes,

$$P(X) = \frac{2}{4} = \frac{1}{2}, \quad P(Y) = \frac{2}{4} = \frac{1}{2}$$

The intersection of $X$ and $Y$ is:

$$X \cap Y = \{b\}$$

So,

$$P(X \cap Y) = \frac{1}{4}$$

Check independence:

$$P(X) \cdot P(Y) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

Since $$P(X \cap Y) = P(X) \cdot P(Y)$$, $X$ and $Y$ are independent.

Key Concept: Real-World Application, Conditional Probability

b) $\frac{24}{91}$

[Solution Description] The total number of ways to draw 3 balls from 15 is given by $${}_{15}C_3 = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$$.

Now, let’s find the number of ways to draw one ball of each color:

– Choose 1 red out of 5: ${}_{5}C_1 = 5$

– Choose 1 blue out of 4: ${}_{4}C_1 = 4$

– Choose 1 green out of 6: ${}_{6}C_1 = 6$

So the number of favorable outcomes = $$5 \times 4 \times 6 = 120$$.

Hence, the probability that all balls are of different colors is $$\frac{120}{455} = \frac{24}{91}$$.

Key Concept: Probability with Axiomatic Approach, Complementary Events

c) 0.6

[Solution Description] The sum of probabilities for all elements should be 1 as per the axioms:

$$P(S) = P(\omega_1) + P(\omega_2) + P(\omega_3) + P(\omega_4) + P(\omega_5) = 0.1 + 0.2 + 0.3 + 0.1 + 0.3 = 1$$

Since the total is 1, the assignment satisfies the axioms. Now finding $$P(E)$$:

$$P(E) = P(\omega_1) + P(\omega_3) = 0.1 + 0.3 = 0.4$$

The probability of the complementary event $E’$ is given by:

$$P(E’) = 1 – P(E) = 1 – 0.4 = 0.6$$

Hence, the probability of the complementary event $E’$ is 0.6.

Key Concept: Event Classification

b) Compound event

[Solution Description] When two dice are rolled, the possible sums range from 2 (when both dice show 1) to 12 (when both dice show 6). The event ‘rolling a sum of 3’ corresponds to the following outcome pairs: (1,2) and (2,1).

Since this event consists of more than one individual outcome, it is classified as a compound event.

Key Concept: Probability of Union of Events

b) 0.6

[Solution Description] To find $P(A \cup B)$, we use the formula for the probability of the union of two events: $$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$. Substituting the given values:

$$P(A \cup B) = 0.3 + 0.4 – 0.1 = 0.6$$

So, $$P(A \cup B) = 0.6$$.

Key Concept: Complex Probability Assignment, Probability of Combined Events

a) $\frac{7}{13}$

[Solution Description] Since $A$ and $B$ are independent, we have $$P(A \cap B) = P(A) \times P(B)$$. Using the formula for combined events:

$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$

Substitute the given values into the equation:

$$0.7 = 0.35 + x – (0.35 \times x)$$

Simplifying the expression:

$$0.7 = 0.35 + x – 0.35x$$

Solving for $x$:

$$0.7 = 0.35(1-x) + x$$

$$0.7 = 0.35 + 0.65x$$

$$0.35 = 0.65x$$

$$x = \frac{0.35}{0.65} = \frac{7}{13}$$

Key Concept: Event Difference

c) $\{D_1, D_2, …, D_{13}\}$

[Solution Description] The event A represents all red cards in the deck, which are hearts and diamonds. The set for A is $$\{H_1, H_2, …, H_{13}, D_1, D_2, …, D_{13}\}$$. The event B represents only hearts, so its set is $$\{H_1, H_2, …, H_{13}\}$$. The difference between these events, $A – B$, gives us diamonds as these are red but not hearts. Hence, $$A – B = \{D_1, D_2, …, D_{13}\}$$.

Key Concept: Complex Event Analysis, Exhaustive and Mutually Exclusive

a) Only A and C are mutually exclusive

[Solution Description] The condition for mutually exclusive events is that they cannot occur simultaneously. Let’s analyze each pair:

– A: Event A occurs if both balls drawn are red. Probability of first red = $$\frac{5}{12}$$, and second red = $$\frac{4}{11}$$. Total probability $$P(A) = \frac{5}{12} \times \frac{4}{11}$$.

– B: Event B occurs if both balls are the same color. Consider cases:

– Both red: $$P(\text{both red}) = P(A)$$,

– Both blue: $$\frac{4}{12} \times \frac{3}{11}$$,

– Both green: $$\frac{3}{12} \times \frac{2}{11}$$.

Total $$P(B) = P(A) + \frac{4}{12} \times \frac{3}{11} + \frac{3}{12} \times \frac{2}{11}$$.

– C: Event C occurs when the second ball is green. This can happen in different sequences of the first draw.

Checking pairs:

– A and B: Not mutually exclusive since A is a subset of B.

– A and C: Mutually exclusive because drawing two reds precludes the second being green.

– B and C: Not mutually exclusive because one green could be drawn, followed by another of the same or different color.

Thus, only events A and C are mutually exclusive.

Key Concept: Exhaustive Events, Complex Event Calculation

c) 0.20

[Solution Description] The events being mutually exclusive means $$E_1 \cap E_2 = E_2 \cap E_3 = E_1 \cap E_3 = \phi$$. Since they are also exhaustive, we have $$E_1 \cup E_2 \cup E_3 = S$$. Thus, using $$P(E_1 \cup E_2 \cup E_3) = 1$$, we find $$P(E_3) = 1 – P(E_1) – P(E_2)$$.

$$P(E_3) = 1 – 0.25 – 0.55$$

$$= 0.20$$

To calculate $$E_3 \cap (E_1 \cup E_2)’$$:

$(E_1 \cup E_2)’ = E_3$ because it’s complementary to $E_1 \cup E_2$. Therefore, $$E_3 \cap (E_1 \cup E_2)’ = E_3 \cap E_3 = E_3$$.

Hence, $$P(E_3 \cap (E_1 \cup E_2)’) = P(E_3) = 0.20$$.

Key Concept: Mutually Exclusive Events, Analyzing Consistency

b) No, they are not mutually exclusive.

[Solution Description] Three events $A$, $B$, and $C$ are mutually exclusive if no two events can occur at the same time, which means:

$$P(A \cup B \cup C) = P(A) + P(B) + P(C).$$

Calculate:

$$P(A) + P(B) + P(C) = 0.4 + 0.3 + 0.2 = 0.9.$$

Since $$P(A \cup B \cup C) = 0.85$$ differs from $0.9$, the events are not mutually exclusive; there must be some overlap between them.

Correct answer option: b)

Key Concept: Basic Definition

c) 0

[Solution Description] The probability of an impossible event, which cannot occur, is defined as 0 according to probability theory.

Key Concept: Event Combination

b) $A: \{1, 2, 3, 4\}, B: \{5\}, C: \{6\}$

[Solution Description] For events to be exhaustive, their union must cover all outcomes in the sample space. For a six-sided die, the sample space is $\{1, 2, 3, 4, 5, 6\}$. Let’s examine each option:

– Option A: The union is $$A \cup B = \{1, 2, 3, 4, 5\}$$, which does not include 6.

– Option B: The union is $$A \cup B \cup C = \{1, 2, 3, 4, 5, 6\}$$, covering the entire sample space.

– Option C: The union is $$A \cup B = \{1, 2, 4, 5, 6\}$$, which does not include 3.

– Option D: The union is $$A \cup B = \{2, 3, 4, 5, 6\}$$, which does not include 1.

Thus, only Option B is exhaustive as it covers every possible outcome from 1 to 6.

Key Concept: Probability with Non-Standard Assignments, Complex Event Probability

c) 0.65

[Solution Description] The even numbers on a die are 2, 4, and 6. Sum their individual probabilities:

$$P(\text{even}) = P(2) + P(4) + P(6)$$

Calculation:

$$= 0.2 + 0.25 + 0.2 = 0.65$$

Therefore, the probability of rolling an even number is 0.65.

Key Concept: Basic Formula

c) $P(A \cup B) = P(A) + P(B) – P(A \cap B)$

[Solution Description] The formula for the probability of the union of two events $A$ and $B$ is given by:

$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$

This formula accounts for the overlap between the probabilities of $A$ and $B$.

Key Concept: Advanced Historical Analysis, Axiomatic Proofs

c) A. N. Kolmogorov; $0.8$

[Solution Description] The modern axiomatic approach to probability was formulated by A. N. Kolmogorov. According to his axioms, $$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$. Substituting the given values:

$$P(A \cup B) = 0.4 + 0.5 – 0.1 = 0.8$$

Key Concept: Complex Probability Calculation, Probability with Conditions

c) 0.50

[Solution Description] To find the probability of event $A$, sum the probabilities of the outcomes in $A$:

$$P(A) = P(\omega_2) + P(\omega_4) + P(\omega_5) = 0.25 + 0.10 + 0.15 = 0.50$$

The probability of the complementary event is:

$$P(\text{not } A) = 1 – P(A) = 1 – 0.50 = 0.50$$

Key Concept: Simple Event Probability

b) $\frac{1}{2}$

[Solution Description] An even number on a die can be 2, 4, or 6. So there are 3 favorable outcomes.

Total possible outcomes when a die is rolled = 6

Number of favorable outcomes for rolling an even number = 3

Using the formula for probability:

$$P(\text{Even Number}) = \frac{\text{Number of Even Numbers}}{\text{Total Outcomes}} = \frac{3}{6} = \frac{1}{2}$$

Key Concept: Union of Events

b) $\{2, 4, 5, 6\}$

[Solution Description] Event A consists of outcomes: $$\{4, 5, 6\}$$.

Event B consists of outcomes: $$\{2, 4, 6\}$$.

The union $$A \cup B$$ includes all unique outcomes from both events: $$\{2, 4, 5, 6\}$$.

Therefore, the union of A and B is $$\{2, 4, 5, 6\}$$.

Key Concept: Complex Probability Calculations, Probability with Combinations

a) $\frac{7}{22}$

[Solution Description] To solve this problem, we use combinations to calculate the different ways of choosing the balls.

– Total ways to choose any 3 balls out of 12:

$$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$

– Ways to choose 2 red balls out of 5:

$$\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$$

– Ways to choose 1 blue ball out of 7:

$$\binom{7}{1} = 7$$

– Therefore, the number of favorable outcomes is:

$$10 \times 7 = 70$$

– The probability is then given by:

$$P(\text{exactly 2 red}) = \frac{70}{220} = \frac{7}{22}$$

Hence, the probability is $$\frac{7}{22}$$.

Key Concept: Probability of Union

b) $\frac{2}{3}$

[Solution Description] To find the probability of drawing either a blue or yellow marble, we use the probability of union of two events.

Total number of marbles = 4 + 3 + 2 = 9.

Probability of drawing a blue marble:

$$P(\text{blue}) = \frac{4}{9}$$

Probability of drawing a yellow marble:

$$P(\text{yellow}) = \frac{2}{9}$$

Since these two events are mutually exclusive, the combined probability is simply the sum of the probabilities:

$$P(\text{blue or yellow}) = P(\text{blue}) + P(\text{yellow}) = \frac{4}{9} + \frac{2}{9} = \frac{6}{9} = \frac{2}{3}$$

Key Concept: Probability with Conditions, Non-Straightforward Path

a) $\frac{1}{6}$

[Solution Description] For the sum to be 4, the favorable outcomes are (1,3), (2,2), and (3,1). For the sum to be 10, the favorable outcomes are (4,6), (5,5), and (6,4).

Total possible outcomes when two dice are rolled is $$6 \times 6 = 36$$.

Total favorable outcomes for the sum being either 4 or 10 = 3 (for 4) + 3 (for 10) = 6.

Thus, the probability is

$$P(\text{{sum is 4 or 10}}) = \frac{6}{36} = \frac{1}{6}$$

Key Concept: Complementary Events

c) 0.7

[Solution Description] By the rule of complementary events, $$P(\text{not } A) = 1 – P(A)$$. Given $P(A) = 0.3$, we calculate:

$$P(\text{not } A) = 1 – 0.3 = 0.7$$

Key Concept: Probability of Complementary Events, Probability with Conditions

b) 0.1323

[Solution Description] This is a binomial probability problem where the number of trials $n = 5$, the number of successes $k = 3$, the probability of success (rain) $p = 0.3$, and the probability of failure (no rain) $q = 0.7$.

Using the binomial probability formula:

$$P(X = k) = \binom{n}{k} p^k q^{n-k}$$

Substitute the values:

$$P(X = 3) = \binom{5}{3} (0.3)^3 (0.7)^{2}$$

Calculate the combinations:

$$\binom{5}{3} = 10$$

Now calculate the probability:

$$P(X = 3) = 10 \times (0.027) \times (0.49)$$

$$= 10 \times 0.01323 = 0.1323$$

Key Concept: Complex Event Probability, Probability with Real-World Application

c) 0.42

[Solution Description] Since the events are independent, the probability of both events occurring is the product of their probabilities:

$$P(A \cap B) = P(A) \cdot P(B)$$

where $P(A) = 0.7$ is the probability of clear skies in the morning, and $P(B) = 0.6$ is the probability of no precipitation in the afternoon. Thus,

$$P(A \cap B) = 0.7 \times 0.6 = 0.42$$

Therefore, the probability of having clear skies in the morning and no precipitation in the afternoon is 0.42.

Key Concept: Simple Event Identification

b) $\{1\}$

[Solution Description] A simple event consists of exactly one outcome from the sample space. Here, each number in the list $\{1, 2, 3, 4, 5, 6\}$ represents a single outcome. Therefore, any individual number like $\{1\}$ or $\{2\}$ etc., is a simple event.

Key Concept: Multi-step Calculation

c) 0.12

[Solution Description] Let $P(C)$ be the probability that a person owns a car, and $P(B|C)$ be the probability that a person owns a bike given that they own a car. We are asked to find $P(C \cap B)$.

Given:

$$P(C) = 0.6$$

$$P(B|C) = 0.2$$

Using the formula for conditional probability:

$$P(C \cap B) = P(C) \cdot P(B|C)$$

Substituting the values:

$$P(C \cap B) = 0.6 \times 0.2 = 0.12$$

Therefore, the probability that a randomly selected person owns both a car and a bike is 0.12.

Key Concept: Event Probability Calculation

b) $\frac{1}{3}$

[Solution Description] The sample space for rolling a die is $$S = \{1, 2, 3, 4, 5, 6\}$$. The event of getting a number greater than 4 is $$A = \{5, 6\}$$. The probability of an event is calculated as $$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{6} = \frac{1}{3}$$.

Key Concept: Complementary Events

c) $\frac{3}{4}$

[Solution Description] The probability of the complement of an event A, denoted by $A^c$, is given by $$P(A^c) = 1 – P(A)$$. Given that $$P(A) = \frac{1}{4}$$, we calculate:

$$P(A^c) = 1 – \frac{1}{4} = \frac{3}{4}$$

Thus, the probability of not A is $$\frac{3}{4}$$.

Key Concept: Compound Event

c) $\{(1, 2), (3, 4)\}$

[Solution Description] A compound event consists of more than one outcome. For two dice, the sample space $S$ includes outcomes like $(1, 1)$, $(2, 3)$, etc. The set $$\{(1, 2), (3, 4)\}$$ involves multiple outcomes, making it a compound event.