Chapter 14 — STATISTICS

Theory

When objects are drawn without replacement, each selection affects the next. In such cases, probability is computed using combinations because order does not matter.

Key concepts used:

• Total outcomes = Selecting r objects from n objects → \( ^nC_r \)
• Favourable outcomes depend on condition
• Complement Rule: \( P(A) = 1 - P(A') \)

Solution Roadmap

• Step 1: Calculate total number of marbles
• Step 2: Find total possible selections of 5 marbles
• Step 3: Solve part (i) using direct combination
• Step 4: Solve part (ii) using complement method

Visual Representation

Solution

Total number of marbles:

\[ 10 + 20 + 30 = 60 \]

Total number of ways of selecting 5 marbles from 60 marbles:

\[ ^{60}C_5 \]

(i) Probability that all marbles are blue

Number of blue marbles = 20

Number of ways to select 5 blue marbles from 20:

\[ ^{20}C_5 \]

Required probability:

\[ P(\text{all blue}) = \frac{^{20}C_5}{^{60}C_5} \]

(ii) Probability that at least one marble is green

Direct calculation is complex, so we use complement:

\[ P(\text{at least one green}) = 1 - P(\text{no green}) \]

If no green marble is selected, all marbles must come from red + blue marbles:

Number of non-green marbles:

\[ 10 + 20 = 30 \]

Number of ways to select 5 marbles from 30:

\[ ^{30}C_5 \]

Therefore,

\[ P(\text{no green}) = \frac{^{30}C_5}{^{60}C_5} \]

Final probability:

\[ P(\text{at least one green}) = 1 - \frac{^{30}C_5}{^{60}C_5} \]

Final Answer

\[ (i)\ \frac{^{20}C_5}{^{60}C_5} \]

\[ (ii)\ 1 - \frac{^{30}C_5}{^{60}C_5} \]

Significance for Exams

• This problem tests combination-based probability, a core topic in CBSE boards
• Use of complement method is frequently asked in JEE & NDA exams
• Understanding "without replacement" is crucial for advanced probability distributions
• Builds foundation for hypergeometric distribution (important in higher studies)

Theory

A standard deck consists of 52 cards divided into 4 suits:

• 13 Diamonds
• 13 Spades
• 13 Hearts
• 13 Clubs

Since cards are drawn without replacement and order does not matter, we use combinations.

Basic probability formula:

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Solution Roadmap

• Step 1: Find total ways to draw 4 cards from 52
• Step 2: Count ways to select 3 diamonds
• Step 3: Count ways to select 1 spade
• Step 4: Multiply favourable selections
• Step 5: Form probability ratio

Visual Representation

Solution

Total number of cards in the deck:

\[ 52 \]

Total number of ways of selecting 4 cards from 52:

\[ ^{52}C_4 \]

Number of diamonds = 13

Number of ways to select 3 diamonds from 13:

\[ ^{13}C_3 \]

Number of spades = 13

Number of ways to select 1 spade from 13:

\[ ^{13}C_1 \]

Since selections are independent (choosing diamonds and spade separately), total favourable ways:

\[ ^{13}C_3 \times ^{13}C_1 \]

Therefore, required probability:

\[ P(E) = \frac{^{13}C_3 \times ^{13}C_1}{^{52}C_4} \]

Final Answer

\[ \frac{^{13}C_3 \times ^{13}C_1}{^{52}C_4} \]

Significance for Exams

• Direct application of combinatorial probability — very common in CBSE board exams
• Tests understanding of selection without replacement
• Frequently appears in JEE as mixed-selection problems (multiple categories)
• Builds foundation for multivariate selection problems

Theory

In this problem, outcomes are not equally labeled but are still equally likely because each face of the die has equal chance of appearing.

Probability is defined as:

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Important concepts used:

• Counting frequency of each outcome
• Addition rule for mutually exclusive events
• Complement rule

Solution Roadmap

• Step 1: Write total number of faces
• Step 2: Count occurrences of 1, 2, 3
• Step 3: Solve each probability using definition
• Step 4: Apply addition rule and complement rule where required

Visual Representation

Solution

Total number of faces on the die:

\[ 6 \]

Distribution of numbers:

• Number 1 appears on 2 faces
• Number 2 appears on 3 faces
• Number 3 appears on 1 face

(i) Probability of getting 2

Number of favourable outcomes = 3

\[ P(2) = \frac{3}{6} \]

Simplifying:

\[ P(2) = \frac{1}{2} \]

(ii) Probability of getting 1 or 3

These events are mutually exclusive, so we add probabilities:

Number of favourable outcomes:

\[ 2 + 1 = 3 \]

\[ P(1 \text{ or } 3) = \frac{3}{6} \]

Simplifying:

\[ P(1 \text{ or } 3) = \frac{1}{2} \]

(iii) Probability of not getting 3

Using complement rule:

\[ P(\text{not } 3) = 1 - P(3) \]

\[ P(3) = \frac{1}{6} \]

\[ P(\text{not } 3) = 1 - \frac{1}{6} \]

\[ = \frac{5}{6} \]

Final Answer

\[ (i)\ \frac{1}{2} \]

\[ (ii)\ \frac{1}{2} \]

\[ (iii)\ \frac{5}{6} \]

Significance for Exams

• Tests understanding of non-uniform labeling with equal likelihood
• Very important for conceptual clarity in CBSE boards
• Forms base for discrete probability distributions (JEE level)
• Reinforces use of addition rule and complement rule

Theory

This is a classic case of selection without replacement.

Total tickets = 10,000
Prize-winning tickets = 10
Non-winning tickets = 10,000 − 10 = 9,990

When multiple tickets are purchased, selections become dependent, hence we must use combinations instead of simple multiplication of probabilities.

General model:

\[ P(\text{no prize in } r \text{ tickets}) = \frac{^{9990}C_r}{^{10000}C_r} \]

Solution Roadmap

• Step 1: Identify total, winning, and non-winning tickets
• Step 2: For each case, compute probability of zero prize tickets
• Step 3: Use direct probability for one ticket
• Step 4: Use combinations for multiple tickets

Visual Representation

Solution

Total tickets = 10,000

Prize tickets = 10

Non-prize tickets:

\[ 10000 - 10 = 9990 \]

(a) When one ticket is purchased

Probability of getting a prize:

\[ \frac{10}{10000} \]

Therefore, probability of not getting a prize:

\[ P(\text{no prize}) = 1 - \frac{10}{10000} \]

\[ = \frac{10000 - 10}{10000} \]

\[ = \frac{9990}{10000} \]

\[ = \frac{999}{1000} \]

(b) When two tickets are purchased

We need both tickets to be non-prize tickets.

Number of ways to select 2 non-prize tickets from 9990:

\[ ^{9990}C_2 \]

Total ways to select any 2 tickets from 10000:

\[ ^{10000}C_2 \]

Therefore:

\[ P(\text{no prize}) = \frac{^{9990}C_2}{^{10000}C_2} \]

(Alternate step-wise understanding)

First ticket not prize:

\[ \frac{9990}{10000} \]

Second ticket not prize:

\[ \frac{9989}{9999} \]

Combined probability:

\[ \frac{9990}{10000} \times \frac{9989}{9999} \]

This is equivalent to:

\[ \frac{^{9990}C_2}{^{10000}C_2} \]

(c) When 10 tickets are purchased

All 10 tickets must be non-prize tickets.

Number of ways to select 10 non-prize tickets:

\[ ^{9990}C_{10} \]

Total ways to select 10 tickets:

\[ ^{10000}C_{10} \]

Therefore:

\[ P(\text{no prize}) = \frac{^{9990}C_{10}}{^{10000}C_{10}} \]

Final Answer

\[ (a)\ \frac{999}{1000} \]

\[ (b)\ \frac{^{9990}C_2}{^{10000}C_2} \]

\[ (c)\ \frac{^{9990}C_{10}}{^{10000}C_{10}} \]

Significance for Exams

• Classic example of hypergeometric probability (very important for JEE)
• Tests understanding of dependent events
• Combination vs multiplication confusion is frequently tested
• Board exams often include 1-mark conceptual questions from this model

Theory

This is a problem of random grouping without replacement.

Total students = 100
Section A = 40 students
Section B = 60 students

Key idea: Placement of students is random, so we analyze probability using:

• Combinatorial approach
• Sequential probability approach
• Complement rule

Solution Roadmap

• Step 1: Fix one student (you)
• Step 2: Find probability of friend's placement
• Step 3: Compute same-section probability
• Step 4: Use complement for different-section probability

Visual Representation

Solution

Total number of students = 100

(a) Probability that both are in the same section

We consider two cases:

• Both in Section A
• Both in Section B

Case 1: Both in Section A

Probability that you are in Section A:

\[ \frac{40}{100} \]

Now, 39 seats remain in Section A out of 99 remaining students

Probability that your friend is also in Section A:

\[ \frac{39}{99} \]

Combined probability:

\[ \frac{40}{100} \times \frac{39}{99} \]

Case 2: Both in Section B

Probability that you are in Section B:

\[ \frac{60}{100} \]

Now, 59 seats remain in Section B out of 99 students

Probability that your friend is also in Section B:

\[ \frac{59}{99} \]

Combined probability:

\[ \frac{60}{100} \times \frac{59}{99} \]

Total probability (same section):

\[ P(\text{same}) = \frac{40}{100} \cdot \frac{39}{99} + \frac{60}{100} \cdot \frac{59}{99} \]

\[ = \frac{40 \times 39 + 60 \times 59}{100 \times 99} \]

\[ = \frac{1560 + 3540}{9900} \]

\[ = \frac{5100}{9900} \]

\[ = \frac{51}{99} \]

\[ = \frac{17}{33} \]

(b) Probability that both are in different sections

Using complement:

\[ P(\text{different}) = 1 - P(\text{same}) \]

\[ = 1 - \frac{17}{33} \]

\[ = \frac{33 - 17}{33} \]

\[ = \frac{16}{33} \]

Final Answer

\[ (a)\ \frac{17}{33} \]

\[ (b)\ \frac{16}{33} \]

Significance for Exams

• Tests understanding of dependent probability and conditional reasoning
• Classic CBSE board question type
• Important for JEE: reduces complex grouping into simple conditional steps
• Strengthens intuition of symmetry and partitioning

Theory

This is a classical problem of permutations with restrictions.

Each arrangement of letters into envelopes corresponds to a permutation.

Important concepts:

• Total arrangements = \( n! \)
• Derangements: arrangements where no object is in its correct position
• Complement Rule: \( P(A) = 1 - P(A') \)

Solution Roadmap

• Step 1: Find total permutations
• Step 2: List all arrangements explicitly
• Step 3: Identify derangements (no correct placement)
• Step 4: Use complement to find required probability

Visual Representation

Solution

Let the letters be L1, L2, L3 and envelopes be E1, E2, E3.

Total number of possible arrangements:

\[ 3! = 6 \]

Now, list all possible arrangements:

1. (L1 → E1, L2 → E2, L3 → E3) → All correct
2. (L1 → E1, L2 → E3, L3 → E2) → One correct
3. (L1 → E2, L2 → E1, L3 → E3) → One correct
4. (L1 → E2, L2 → E3, L3 → E1) → None correct
5. (L1 → E3, L2 → E1, L3 → E2) → None correct
6. (L1 → E3, L2 → E2, L3 → E1) → One correct

Number of arrangements with no letter in correct envelope:

\[ 2 \]

Therefore:

\[ P(\text{no correct}) = \frac{2}{6} \]

\[ = \frac{1}{3} \]

Required probability:

\[ P(\text{at least one correct}) = 1 - P(\text{none correct}) \]

\[ = 1 - \frac{1}{3} \]

\[ = \frac{2}{3} \]

Final Answer

\[ \frac{2}{3} \]

Significance for Exams

• Classic derangement problem — frequently asked conceptually in CBSE
• Forms base for advanced permutation topics in JEE
• Important to understand listing vs formula approach
• Strengthens use of complement probability

Theory

This problem is based on fundamental probability identities:

• Addition Law: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
• Complement Rule: \(P(A') = 1 - P(A)\)
• De Morgan’s Law: \(P(A' \cap B') = 1 - P(A \cup B)\)
• Partitioning: \(P(A \cap B') = P(A) - P(A \cap B)\)

Solution Roadmap

• Step 1: Compute union using addition law
• Step 2: Use complement for neither event
• Step 3: Subtract intersection from individual probabilities
• Step 4: Interpret results using Venn diagram logic

Visual Representation

Solution

Given:

\[ P(A) = 0.54,\quad P(B) = 0.69,\quad P(A \cap B) = 0.35 \]

(i) Find \(P(A \cup B)\)

Using addition law:

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

\[ = 0.54 + 0.69 - 0.35 \]

\[ = 1.23 - 0.35 \]

\[ = 0.88 \]

(ii) Find \(P(A' \cap B')\)

Using De Morgan’s Law:

\[ P(A' \cap B') = 1 - P(A \cup B) \]

\[ = 1 - 0.88 \]

\[ = 0.12 \]

(iii) Find \(P(A \cap B')\)

This represents part of A excluding intersection.

\[ P(A \cap B') = P(A) - P(A \cap B) \]

\[ = 0.54 - 0.35 \]

\[ = 0.19 \]

(iv) Find \(P(B \cap A')\)

This represents part of B excluding intersection.

\[ P(B \cap A') = P(B) - P(A \cap B) \]

\[ = 0.69 - 0.35 \]

\[ = 0.34 \]

Final Answer

\[ (i)\ 0.88 \]

\[ (ii)\ 0.12 \]

\[ (iii)\ 0.19 \]

\[ (iv)\ 0.34 \]

Significance for Exams

• Direct application of set-theoretic probability identities
• Very common CBSE 3–5 mark question pattern
• Critical for JEE: converting Venn diagram logic into equations
• Strengthens understanding of event partitioning

Theory

This problem is based on the Addition Rule of Probability:

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

Where:

• \(A\): Event that the person is male
• \(B\): Event that the person is over 35 years

Solution Roadmap

• Step 1: Count total persons
• Step 2: Identify males
• Step 3: Identify persons with age > 35
• Step 4: Identify overlap (male and age > 35)
• Step 5: Apply addition formula

Classification View

Solution

Total number of persons:

\[ 5 \]

Let \(M\): person is male
Let \(A\): person is over 35 years

Number of males = 3 (Harish, Rohan, Salim)

\[ P(M) = \frac{3}{5} \]

Persons with age greater than 35:

Sheetal (46), Salim (41)

Number of such persons = 2

\[ P(A) = \frac{2}{5} \]

Persons who are both male and over 35:

Only Salim

\[ P(M \cap A) = \frac{1}{5} \]

Applying addition rule:

\[ P(M \cup A) = P(M) + P(A) - P(M \cap A) \]

\[ = \frac{3}{5} + \frac{2}{5} - \frac{1}{5} \]

\[ = \frac{4}{5} \]

Final Answer

\[ \frac{4}{5} \]

Significance for Exams

• Direct application of addition theorem
• Tests ability to correctly identify intersection
• Very common CBSE 2–3 mark question
• Important foundation for set-based probability problems in JEE

Theory

• 4-digit number > 5000 ⇒ thousand’s place ∈ {5, 7}
• Divisible by 5 ⇒ unit digit ∈ {0, 5}
• Use multiplication principle for counting
• Handle restrictions (like excluding 5000) explicitly

Solution Roadmap

• Step 1: Fix thousand’s place (condition > 5000)
• Step 2: Count total outcomes
• Step 3: Count favourable outcomes using unit digit condition
• Step 4: Adjust for invalid cases
• Step 5: Compute probability

Visual Representation

Solution

Given digits: \(0,1,3,5,7\)

Thousand’s place must be \(5\) or \(7\)

(i) When repetition is allowed

Total outcomes

Thousand’s place: 2 choices
Remaining 3 places: each has 5 choices

\[ \text{Total} = 2 \times 5 \times 5 \times 5 = 250 \]

Favourable outcomes (unit digit = 0 or 5)

Case 1: Unit digit = 0

Thousand: 2 choices
Hundreds: 5 choices
Tens: 5 choices

\[ = 2 \times 5 \times 5 = 50 \]

Case 2: Unit digit = 5

Thousand: 2 choices
Hundreds: 5 choices
Tens: 5 choices

\[ = 2 \times 5 \times 5 = 50 \]

Total favourable before adjustment:

\[ 50 + 50 = 100 \]

Invalid case: \(5000\) (not > 5000)

\[ \text{Valid favourable} = 100 - 1 = 99 \]

Therefore:

\[ P = \frac{99}{250} \]

(This fraction is already in simplest form)

(ii) When repetition is not allowed

Total outcomes

Thousand: 2 choices
Remaining digits: choose from 4, then 3, then 2

\[ \text{Total} = 2 \times 4 \times 3 \times 2 = 48 \]

Favourable outcomes

Case 1: Unit digit = 0

Thousand: 2 choices (5 or 7)
Remaining 2 places: from remaining 3 digits

\[ = 2 \times 3 \times 2 = 12 \]

Case 2: Unit digit = 5

Thousand cannot be 5 (no repetition), so it must be 7 → 1 choice

Remaining 2 places: from remaining 3 digits

\[ = 1 \times 3 \times 2 = 6 \]

Total favourable:

\[ 12 + 6 = 18 \]

Therefore:

\[ P = \frac{18}{48} \]

\[ = \frac{3}{8} \]

Final Answer

\[ (i)\ \frac{99}{250} \]

\[ (ii)\ \frac{3}{8} \]

Significance for Exams

• Combines number formation + probability (very important in CBSE)
• Tests handling of constraints (≥5000, divisibility)
• Common JEE pattern: case-based counting
• Strengthens clarity between repetition vs no repetition

Theory

This is a permutation-based probability problem because:

• Order of digits matters (e.g., 1234 ≠ 4321)
• No repetition is allowed
• Total outcomes are arrangements of digits

Number of permutations of selecting 4 distinct digits from 10:

\[ {}^{10}P_4 = 10 \times 9 \times 8 \times 7 \]

Solution Roadmap

• Step 1: Count total possible sequences (permutations)
• Step 2: Identify favourable outcomes
• Step 3: Apply probability formula

Visual Representation

Solution

Each wheel has digits from 0 to 9 → total 10 digits

Since repetition is not allowed:

First digit can be chosen in 10 ways

Second digit can be chosen in 9 ways (one digit already used)

Third digit can be chosen in 8 ways

Fourth digit can be chosen in 7 ways

Therefore, total number of possible sequences:

\[ 10 \times 9 \times 8 \times 7 \]

\[ = 5040 \]

Only one sequence is correct (the actual lock code)

Therefore, number of favourable outcomes:

\[ 1 \]

Hence, required probability:

\[ P = \frac{1}{5040} \]

Final Answer

\[ \frac{1}{5040} \]

Significance for Exams

• Direct application of permutations in probability
• Tests understanding of order vs selection
• Very common in CBSE and JEE (lock/password type problems)
• Foundation for probability of arrangements in higher mathematics