Chapter 14 — STATISTICS

Solution Roadmap

• Step 1: Check whether any probability is negative.
• Step 2: Compute the total sum of probabilities.
• Step 3: Verify whether the sum equals exactly \(1\).
• Step 4: Conclude validity or invalidity.

Concept Visualization

Solution

We check each assignment one by one using the probability axioms.

(a)

\[ \begin{aligned} \text{Sum} &= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 \\ &= (0.1 + 0.01) + (0.05 + 0.03) + (0.01 + 0.2) + 0.6 \\ &= 0.11 + 0.08 + 0.21 + 0.6 \\ &= (0.11 + 0.08) + (0.21 + 0.6) \\ &= 0.19 + 0.81 \\ &= 1.00 \end{aligned} \]

All probabilities are non-negative and sum equals 1 ⇒ Valid.

(b)

\[ \begin{aligned} \text{Sum} &= \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} \\ &= 7 \times \frac{1}{7} \\ &= 1 \end{aligned} \]

All probabilities are positive and sum equals 1 ⇒ Valid.

(c)

\[ \begin{aligned} \text{Sum} &= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 \\ &= (0.1 + 0.2) + (0.3 + 0.4) + (0.5 + 0.6) + 0.7 \\ &= 0.3 + 0.7 + 1.1 + 0.7 \\ &= (0.3 + 0.7) + (1.1 + 0.7) \\ &= 1.0 + 1.8 \\ &= 2.8 \end{aligned} \]

Since total probability is greater than 1 ⇒ Invalid.

(d)

Here, probabilities \(-0.1\) and \(-0.2\) are negative.

Negative probabilities are not allowed ⇒ Invalid.

(e)

One probability is missing and also \(\frac{15}{14} > 1\), which violates the rule \(P(\omega_i) \leq 1\).

Hence, invalid.

Final Answer: (c), (d), (e)

Significance for Exams

• This is a foundational concept for all probability problems in CBSE boards.
• Directly tested in MCQs and assertion-reason questions.
• Highly important for JEE Main, NDA, CUET where validity checks are frequently asked.
• Builds base for random variables and distributions in higher classes.

Theory

When an experiment consists of repeated independent trials (like tossing a coin twice), the sample space is constructed using all possible ordered outcomes.

• Each outcome is equally likely.
• Probability of an event is given by \(P(A)=\frac{\text{Number of favourable outcomes}}{\text{Total outcomes}}\).
• For “at least one” type questions, the complement method is often more efficient.

Solution Roadmap

• Step 1: Construct the sample space.
• Step 2: Identify outcomes satisfying “at least one tail”.
• Step 3: Apply classical probability formula.
• Step 4: Verify using complement method.

Tree Diagram Representation

Solution

When a coin is tossed twice, the sample space is:

\[ S = \{HH, HT, TH, TT\} \]

Total number of outcomes:

\[ n(S) = 4 \]

Let \(A\) be the event “at least one tail occurs”.

Outcomes having at least one tail are:

\[ A = \{HT, TH, TT\} \]

Number of favourable outcomes:

\[ n(A) = 3 \]

Applying probability formula:

\[ \begin{aligned} P(A) &= \frac{n(A)}{n(S)} \\ &= \frac{3}{4} \end{aligned} \]

Alternative Method (Complement)

The complement of event \(A\) is “no tail occurs”.

\[ A' = \{HH\} \] \[ n(A') = 1 \] \[ \begin{aligned} P(A') &= \frac{1}{4} \end{aligned} \]

Using complement rule:

\[ \begin{aligned} P(A) &= 1 - P(A') \\ &= 1 - \frac{1}{4} \\ &= \frac{3}{4} \end{aligned} \]

Final Answer: \(\frac{3}{4}\)

Significance for Exams

• This is a core probability model used in CBSE board exams.
• “At least one” problems are very common in MCQs.
• Complement method is a high-speed trick in JEE Main and NDA.
• Forms the base for binomial probability and independent events.

Theory

When a fair die is thrown, it produces six equally likely outcomes. Therefore, the classical probability formula applies:

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

Important classifications:

• Certain Event: Probability = 1
• Impossible Event: Probability = 0
• Finite Uniform Sample Space: All outcomes equally likely

Solution Roadmap

• Step 1: Write the sample space.
• Step 2: Identify favourable outcomes for each case.
• Step 3: Count outcomes carefully.
• Step 4: Apply probability formula.

Sample Space Visualization

Solution

Sample space:

\[ S = \{1,2,3,4,5,6\} \] \[ n(S) = 6 \]

(i) Prime number will appear

Prime numbers in the sample space are:

\[ A = \{2,3,5\} \]

Number of favourable outcomes:

\[ n(A) = 3 \] \[ \begin{aligned} P(A) &= \frac{n(A)}{n(S)} \\ &= \frac{3}{6} \\ &= \frac{1}{2} \end{aligned} \]

(ii) Number greater than or equal to 3

Favourable outcomes are:

\[ B = \{3,4,5,6\} \] \[ n(B) = 4 \] \[ \begin{aligned} P(B) &= \frac{4}{6} \\ &= \frac{2}{3} \end{aligned} \]

(iii) Number less than or equal to 1

Only one number satisfies this condition:

\[ C = \{1\} \] \[ n(C) = 1 \] \[ \begin{aligned} P(C) &= \frac{1}{6} \end{aligned} \]

(iv) Number more than 6

No outcome in the sample space satisfies this condition.

\[ D = \varnothing \] \[ n(D) = 0 \] \[ \begin{aligned} P(D) &= \frac{0}{6} \\ &= 0 \end{aligned} \]

(v) Number less than 6

Favourable outcomes are:

\[ E = \{1,2,3,4,5\} \] \[ n(E) = 5 \] \[ \begin{aligned} P(E) &= \frac{5}{6} \end{aligned} \]

Final Answers: \(\frac{1}{2},\ \frac{2}{3},\ \frac{1}{6},\ 0,\ \frac{5}{6}\)

Significance for Exams

• Direct application of classical probability frequently asked in CBSE boards.
• Helps build clarity on event types (impossible, certain).
• Forms base for probability inequalities and ranges in JEE.
• Quick elimination technique useful in MCQ-based exams.

Theory

A standard deck of playing cards contains 52 cards divided into 4 suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards.

• Total cards = 52
• Each outcome is equally likely
• Probability formula: \(P(A)=\frac{n(A)}{n(S)}\)

Solution Roadmap

• Step 1: Identify total outcomes (sample space).
• Step 2: Count favourable outcomes carefully.
• Step 3: Apply probability formula.
• Step 4: Simplify fractions.

Deck Structure Visualization

Solution

(a) Number of sample points

Total number of cards in a standard deck:

\[ n(S) = 52 \]

(b) Probability of ace of spades

There is exactly one ace of spades in the deck.

\[ n(A) = 1 \] \[ \begin{aligned} P(\text{ace of spades}) &= \frac{n(A)}{n(S)} \\ &= \frac{1}{52} \end{aligned} \]

(c)(i) Probability of an ace

There are 4 aces (one in each suit):

\[ n(A) = 4 \] \[ \begin{aligned} P(\text{ace}) &= \frac{4}{52} \\ &= \frac{2}{26} \\ &= \frac{1}{13} \end{aligned} \]

(c)(ii) Probability of a black card

Black cards consist of clubs and spades.

Number of clubs = 13

Number of spades = 13

\[ \begin{aligned} n(\text{black cards}) &= 13 + 13 \\ &= 26 \end{aligned} \] \[ \begin{aligned} P(\text{black card}) &= \frac{26}{52} \\ &= \frac{13}{26} \\ &= \frac{1}{2} \end{aligned} \]

Final Answers: (a) \(52\), (b) \(\frac{1}{52}\), (c)(i) \(\frac{1}{13}\), (c)(ii) \(\frac{1}{2}\)

Significance for Exams

• This is a standard model problem in CBSE board exams.
• Very common in JEE Main and NDA MCQs.
• Tests understanding of sample space and counting.
• Forms the base for conditional probability and combinations.

Theory

When two independent experiments are performed together, the sample space is formed using ordered pairs.

• Coin outcomes: \(\{1,6\}\)
• Die outcomes: \(\{1,2,3,4,5,6\}\)
• Total outcomes: \(2 \times 6 = 12\)
• Each outcome is equally likely

Sum-based problems require careful identification of combinations producing a given total.

Solution Roadmap

• Step 1: Construct the sample space using ordered pairs.
• Step 2: Identify outcomes producing required sum.
• Step 3: Count favourable outcomes.
• Step 4: Apply probability formula.

Sample Space Visualization

Correct Sample Space Visualization (Coin × Die)

Solution

Sample space:

\[ \begin{aligned} S=\{&(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ &(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned} \] \[ n(S) = 12 \]

(i) Sum = 3

We find all ordered pairs whose sum is 3.

Checking systematically:

\[ \begin{aligned} (1,1) &\rightarrow 1+1=2 \\ (1,2) &\rightarrow 1+2=3 \quad \checkmark \\ (1,3) &\rightarrow 1+3=4 \\ (6,1) &\rightarrow 6+1=7 \\ \text{All others} &> 3 \end{aligned} \]

Only one favourable outcome:

\[ A = \{(1,2)\}, \quad n(A)=1 \] \[ \begin{aligned} P(A) &= \frac{1}{12} \end{aligned} \]

(ii) Sum = 12

Checking all outcomes:

\[ \begin{aligned} (6,6) &\rightarrow 6+6=12 \quad \checkmark \end{aligned} \]

Only one favourable outcome:

\[ B = \{(6,6)\}, \quad n(B)=1 \] \[ \begin{aligned} P(B) &= \frac{1}{12} \end{aligned} \]

Final Answer: (i) \(\frac{1}{12}\), (ii) \(\frac{1}{12}\)

Significance for Exams

• Important example of combined experiments.
• Builds base for ordered pair logic used in JEE.
• Helps avoid common mistake of treating coin as {H,T} instead of {1,6}.
• Frequently asked in MCQs and case-based questions.

Theory

When selecting one element randomly from a finite group, each member has an equal probability of being chosen.

Probability is given by:

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

This type of problem is a direct application of classical probability and counting of favourable cases.

Solution Roadmap

• Step 1: Determine total number of council members.
• Step 2: Identify favourable outcomes (women).
• Step 3: Apply probability formula.
• Step 4: Simplify the fraction.

Population Visualization

Solution

Number of men = 4

Number of women = 6

Total number of council members:

\[ \begin{aligned} n(S) &= 4 + 6 \\ &= 10 \end{aligned} \]

Let \(W\) be the event that the selected member is a woman.

Number of favourable outcomes:

\[ n(W) = 6 \]

Applying probability formula:

\[ \begin{aligned} P(W) &= \frac{n(W)}{n(S)} \\ &= \frac{6}{10} \end{aligned} \]

Simplifying the fraction:

\[ \begin{aligned} \frac{6}{10} &= \frac{3}{5} \end{aligned} \]

Final Answer: \(\frac{3}{5}\)

Significance for Exams

• Basic application of probability with real-life data.
• Common in CBSE 1-mark and 2-mark questions.
• Frequently appears in CUET, NDA, SSC exams.
• Builds base for conditional probability (selection problems).

Theory

When a coin is tossed multiple times, outcomes depend only on the number of heads. This leads to a binomial distribution framework.

• Total outcomes for 4 tosses: \(2^4 = 16\)
• Number of ways to get \(h\) heads: \(^{4}C_h\)
• Each outcome has probability \(\frac{1}{16}\)

Net amount depends on number of heads and tails.

Solution Roadmap

• Step 1: Let number of heads = \(h\), tails = \(4-h\).
• Step 2: Form the money expression.
• Step 3: Evaluate for all \(h = 0,1,2,3,4\).
• Step 4: Count outcomes using combinations.
• Step 5: Compute probabilities.

Distribution Visualization

Solution

Total number of outcomes:

\[ n(S) = 2^4 = 16 \]

Let number of heads = \(h\), then number of tails = \(4 - h\).

Money gained:

• Each head gives +1
• Each tail gives -1.5

Net amount:

\[ \begin{aligned} \text{Amount} &= h(1) - (4-h)(1.5) \\ &= h - 1.5(4-h) \\ &= h - 6 + 1.5h \\ &= 2.5h - 6 \end{aligned} \]

Now evaluate for all possible values of \(h\):

Case 1: \(h = 0\)

\[ \text{Amount} = 2.5(0) - 6 = -6 \] \[ n = {}^4C_0 = 1 \] \[ P = \frac{1}{16} \]

Case 2: \(h = 1\)

\[ \text{Amount} = 2.5(1) - 6 = -3.5 \] \[ n = {}^4C_1 = 4 \] \[ P = \frac{4}{16} = \frac{1}{4} \]

Case 3: \(h = 2\)

\[ \text{Amount} = 2.5(2) - 6 = -1 \] \[ n = {}^4C_2 = 6 \] \[ P = \frac{6}{16} = \frac{3}{8} \]

Case 4: \(h = 3\)

\[ \text{Amount} = 2.5(3) - 6 = 1.5 \] \[ n = {}^4C_3 = 4 \] \[ P = \frac{4}{16} = \frac{1}{4} \]

Case 5: \(h = 4\)

\[ \text{Amount} = 2.5(4) - 6 = 4 \] \[ n = {}^4C_4 = 1 \] \[ P = \frac{1}{16} \]

Final Results:

• Amount = -6, Probability = \(\frac{1}{16}\)
• Amount = -3.5, Probability = \(\frac{1}{4}\)
• Amount = -1, Probability = \(\frac{3}{8}\)
• Amount = 1.5, Probability = \(\frac{1}{4}\)
• Amount = 4, Probability = \(\frac{1}{16}\)

Significance for Exams

• Direct introduction to binomial distribution concept.
• Very important for JEE Main and Advanced probability.
• Teaches transformation of outcomes into random variable.
• Frequently asked in case-study and value-based questions.

Theory

When multiple coins are tossed, the total number of outcomes is given by:

\[ 2^n \]

For 3 coins:

\[ n(S) = 2^3 = 8 \]

Events can be grouped based on number of heads or tails using combinations.

Solution Roadmap

• Step 1: Write complete sample space.
• Step 2: Group outcomes based on number of heads/tails.
• Step 3: Count favourable outcomes.
• Step 4: Apply probability formula.

Tree Diagram Representation

Solution

\[ S = \{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\} \] \[ n(S) = 8 \]

(i) 3 heads

\[ A = \{HHH\}, \quad n(A)=1 \] \[ P(3H) = \frac{1}{8} \]

(ii) 2 heads

\[ B = \{HHT,HTH,THH\}, \quad n(B)=3 \] \[ P(2H) = \frac{3}{8} \]

(iii) At least 2 heads

Includes 2 heads or 3 heads:

\[ P = \frac{3+1}{8} = \frac{4}{8} = \frac{1}{2} \]

(iv) At most 2 heads

Includes 0H, 1H, 2H:

\[ P = \frac{1+3+3}{8} = \frac{7}{8} \]

(v) No head

\[ \{TTT\} \Rightarrow P = \frac{1}{8} \]

(vi) 3 tails

\[ \{TTT\} \Rightarrow P = \frac{1}{8} \]

(vii) Exactly 2 tails

\[ \{HTT,THT,TTH\}, \quad P = \frac{3}{8} \]

(viii) No tail

\[ \{HHH\} \Rightarrow P = \frac{1}{8} \]

(ix) At most 2 tails

\[ P = \frac{1+3+3}{8} = \frac{7}{8} \]

Final Answers: \(\frac{1}{8}, \frac{3}{8}, \frac{1}{2}, \frac{7}{8}, \frac{1}{8}, \frac{1}{8}, \frac{3}{8}, \frac{1}{8}, \frac{7}{8}\)

Significance for Exams

• Classic binomial distribution base problem.
• Very frequent in CBSE + JEE MCQs.
• Builds clarity for “at least” vs “at most” traps.
• Foundation for random variables and probability distributions.

Theory

For any event \(A\), the complement of the event (denoted by \(A'\)) represents “the event does not occur”.

The fundamental rule of probability states:

\[ P(A) + P(A') = 1 \]

Therefore:

\[ P(A') = 1 - P(A) \]

Solution Roadmap

• Step 1: Identify given probability \(P(A)\).
• Step 2: Apply complement rule.
• Step 3: Perform fraction subtraction carefully.

Complement Visualization

Solution

Given:

\[ P(A) = \frac{2}{11} \]

Using complement rule:

\[ P(A') = 1 - P(A) \]

Express 1 as a fraction with denominator 11:

\[ 1 = \frac{11}{11} \]

Now subtract:

\[ \begin{aligned} P(A') &= \frac{11}{11} - \frac{2}{11} \ &= \frac{11 - 2}{11} \ &= \frac{9}{11} \end{aligned} \]

Final Answer: \(\frac{9}{11}\)

Significance for Exams

• This is one of the most fundamental probability identities.
• Frequently used in “at least” type problems in JEE.
• Helps simplify complex probability problems using complement.
• Common in MCQs and assertion-reason questions.

Theory

When selecting a letter randomly from a word, each position is treated as an equally likely outcome.

• Total outcomes = total number of letters
• Repeated letters are counted separately
• Probability = favourable outcomes ÷ total outcomes

Vowels are: \(A, E, I, O, U\)

Solution Roadmap

• Step 1: Count total letters.
• Step 2: Count vowels carefully.
• Step 3: Find consonants using total − vowels.
• Step 4: Apply probability formula.

Letter Distribution Visualization

Solution

Word: ASSASSINATION

Total number of letters:

\[ n(S) = 13 \]

Counting vowels

Letters in the word:

\[ A, S, S, A, S, S, A, T, I, O, N, A, I \]

Vowels are:

\[ A, A, A, A, I, I, O \]

Counting step-by-step:

• Number of A = 4
• Number of I = 2
• Number of O = 1

\[ \begin{aligned} \text{Total vowels} &= 4 + 2 + 1 \\ &= 7 \end{aligned} \]

(i) Probability of vowel

\[ \begin{aligned} P(V) &= \frac{7}{13} \end{aligned} \]

(ii) Probability of consonant

Number of consonants:

\[ \begin{aligned} \text{Consonants} &= 13 - 7 \\ &= 6 \end{aligned} \] \[ \begin{aligned} P(C) &= \frac{6}{13} \end{aligned} \]

Final Answers: (i) \(\frac{7}{13}\), (ii) \(\frac{6}{13}\)

Significance for Exams

• Tests careful counting with repeated elements.
• Common mistake: ignoring repetition (important for MCQs).
• Foundation for probability with words and arrangements.
• Frequently asked in CBSE + CUET exams.

Theory

When selecting items where order does not matter, we use combinations.

Number of ways to choose \(r\) objects from \(n\) objects:

\[ ^nC_r = \frac{n!}{r!(n-r)!} \]

In lottery problems:

• Total outcomes = all possible selections
• Favourable outcome = exact match

Solution Roadmap

• Step 1: Identify total selections using combinations.
• Step 2: Identify favourable outcome.
• Step 3: Apply probability formula.
• Step 4: Expand factorial carefully.

Selection Visualization

Solution

Total numbers available: \(1\) to \(20\)

Player selects 6 numbers.

Total possible selections:

\[ n(S) = {}^{20}C_6 \]

Using formula:

\[ {}^{20}C_6 = \frac{20!}{6! \cdot 14!} \]

Favourable outcome

Only one case: exact match with lottery numbers.

\[ n(E) = 1 \]

Applying probability formula:

\[ P(E) = \frac{1}{^{20}C_6} \]

Expanding factorial

\[ \begin{aligned} {}^{20}C_6 &= \frac{20!}{6! \cdot 14!} \\ &= \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14!}{6! \cdot 14!} \end{aligned} \]

Cancel \(14!\):

\[ = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6!} \]

Now expand \(6!\):

\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \]

So:

\[ {}^{20}C_6 = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{720} \]

Simplify step-by-step:

\[ \begin{aligned} &= \frac{(20 \times 19 \times 18 \times 17 \times 16 \times 15)}{720} \\ &= 38760 \end{aligned} \]

Therefore:

\[ \begin{aligned} P(E) &= \frac{1}{38760} \end{aligned} \]

Final Answer: \(\frac{1}{38760}\)

Significance for Exams

• Direct application of combinations in probability.
• Very important for JEE Main & Advanced.
• Classic example of extremely low probability event.
• Foundation for lottery, selection, and random sampling problems.

Theory

For any two events \(A\) and \(B\), the following conditions must always hold:

• \(0 \le P(A), P(B), P(A \cap B), P(A \cup B) \le 1\)
• \(P(A \cap B) \le \min(P(A), P(B))\)
• \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
• \(P(A \cup B) \le 1\)

Solution Roadmap

• Step 1: Check basic bounds (0 to 1).
• Step 2: Verify intersection constraint.
• Step 3: Use addition law.
• Step 4: Conclude consistency.

Venn Concept Visualization

Solution

(i) Given:

\[ P(A)=0.5,\quad P(B)=0.7,\quad P(A \cap B)=0.6 \]

Check intersection condition:

\[ \min(P(A), P(B)) = \min(0.5, 0.7) = 0.5 \]

But:

\[ P(A \cap B) = 0.6 > 0.5 \]

This violates the rule:

\[ P(A \cap B) \le \min(P(A), P(B)) \]

Hence, inconsistent.

(ii) Given:

\[ P(A)=0.5,\quad P(B)=0.4,\quad P(A \cup B)=0.8 \]

Using addition law:

\[ \begin{aligned} P(A \cap B) &= P(A)+P(B)-P(A \cup B) \\ &= 0.5 + 0.4 - 0.8 \\ &= 0.1 \end{aligned} \]

Check conditions:

• \(0 \le 0.1 \le 1\) ✓
• \(0.1 \le \min(0.5, 0.4) = 0.4\) ✓

All conditions satisfied ⇒ consistent.

Final Answer: (i) Not consistent, (ii) Consistent

Significance for Exams

• Tests deep understanding of probability axioms.
• Very important for JEE Main MCQs.
• Common trap: ignoring \(P(A \cap B) \le P(A)\).
• Forms base for conditional probability and Bayes theorem.

Theory

For any two events \(A\) and \(B\), the fundamental relation is:

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

This formula is used to compute any missing value when the other three are known.

Solution Roadmap

• Step 1: Identify known values.
• Step 2: Substitute into formula.
• Step 3: Simplify carefully.
• Step 4: Solve for missing term.

Formula Visualization

Solution

(i)

\[ \begin{aligned} P(A)=\frac{1}{3}, \\ P(B)=\frac{1}{5}, \\ P(A \cap B)=\frac{1}{15} \end{aligned} \]

Using formula:

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

Convert into common denominator (15):

\[ \begin{aligned} \frac{1}{3} &= \frac{5}{15} \\ \frac{1}{5} &= \frac{3}{15} \end{aligned} \]

Substitute:

\[ \begin{aligned} P(A \cup B) &= \frac{5}{15} + \frac{3}{15} - \frac{1}{15} \\ &= \frac{5+3-1}{15} \\ &= \frac{7}{15} \end{aligned} \]

(ii)

\[ \begin{aligned} P(A)=0.35,\\ P(A \cap B)=0.25,\\ P(A \cup B)=0.6 \end{aligned} \]

Formula:

\[ 0.6 = 0.35 + P(B) - 0.25 \]

Simplify RHS:

\[ 0.35 - 0.25 = 0.10 \] \[ 0.6 = 0.10 + P(B) \] \[ \begin{aligned} P(B) &= 0.6 - 0.10 \\ &= 0.50 \end{aligned} \]

(iii)

\[ \begin{aligned} P(A)=0.5,\\ P(B)=0.35,\\ P(A \cup B)=0.7 \end{aligned} \]

Using formula:

\[ P(A \cap B) = P(A)+P(B)-P(A \cup B) \] \[ \begin{aligned} P(A \cap B) &= 0.5 + 0.35 - 0.7 \\ &= 0.85 - 0.7 \\ &= 0.15 \end{aligned} \]

Final Answers: (i) \(\frac{7}{15}\), (ii) \(0.50\), (iii) \(0.15\)

Significance for Exams

• Core application of addition law of probability.
• Frequently asked in CBSE short-answer questions.
• Very important for JEE MCQ elimination.
• Foundation for conditional probability and Bayes theorem.

Theory

Two events are said to be mutually exclusive if they cannot occur simultaneously.

Therefore:

\[ P(A \cap B) = 0 \]

General addition rule:

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

For mutually exclusive events:

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

Solution Roadmap

• Step 1: Identify given probabilities.
• Step 2: Use mutual exclusivity condition.
• Step 3: Apply addition rule.
• Step 4: Simplify.

Mutually Exclusive Visualization

Solution

Given:

\[ P(A) = \frac{3}{5}, \quad P(B) = \frac{1}{5} \]

Since events are mutually exclusive:

\[ P(A \cap B) = 0 \]

Using addition rule:

\[ \begin{aligned} P(A \cup B) &= P(A) + P(B) - P(A \cap B) \\ &= \frac{3}{5} + \frac{1}{5} - 0 \end{aligned} \]

Add fractions:

\[ \begin{aligned} \frac{3}{5} + \frac{1}{5} &= \frac{3+1}{5} \\ &= \frac{4}{5} \end{aligned} \]

Final Answer: \(\frac{4}{5}\)

Significance for Exams

• Key concept: mutually exclusive vs non-mutually exclusive.
• Very common in CBSE and JEE MCQs.
• Fast-solving trick: directly add probabilities when no overlap.
• Foundation for advanced probability laws.

Theory

For any two events \(E\) and \(F\), the addition law is:

\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]

Also, using complement:

\[ P(E' \cap F') = 1 - P(E \cup F) \]

This is based on De Morgan’s Law:

\[ (E \cup F)' = E' \cap F' \]

Solution Roadmap

• Step 1: Use addition law to find \(P(E \cup F)\).
• Step 2: Convert fractions carefully.
• Step 3: Use complement to find \(P(E' \cap F')\).

Concept Visualization

Solution

Given:

\[ \begin{aligned} P(E)=\frac{1}{4}, \\ P(F)=\frac{1}{2}, \\ P(E \cap F)=\frac{1}{8} \end{aligned} \]

(i) Find \(P(E \cup F)\)

Using addition law:

\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]

Convert all terms to denominator 8:

\[ \begin{aligned} \frac{1}{4} = \frac{2}{8}, \\ \frac{1}{2} = \frac{4}{8} \end{aligned} \]

Substitute:

\[ \begin{aligned} P(E \cup F) &= \frac{2}{8} + \frac{4}{8} - \frac{1}{8} \\ &= \frac{2+4-1}{8} \\ &= \frac{5}{8} \end{aligned} \]

(ii) Find \(P(E' \cap F')\)

Using complement rule:

\[ P(E' \cap F') = 1 - P(E \cup F) \]

Convert 1 into fraction:

\[ 1 = \frac{8}{8} \]

Substitute:

\[ \begin{aligned} P(E' \cap F') &= \frac{8}{8} - \frac{5}{8} \\ &= \frac{8-5}{8} \\ &= \frac{3}{8} \end{aligned} \]

Final Answers: (i) \(\frac{5}{8}\), (ii) \(\frac{3}{8}\)

Significance for Exams

• Tests combined use of addition law + complement.
• Very common in CBSE 2–3 mark questions.
• Important for JEE probability + set theory linkage.
• Builds base for conditional probability & Bayes theorem.

Theory

De Morgan’s Law states:

\[ (E \cap F)' = E' \cup F' \]

Also, complement rule:

\[ P(A') = 1 - P(A) \]

For mutually exclusive events:

\[ P(E \cap F) = 0 \]

Solution Roadmap

• Step 1: Apply De Morgan’s law.
• Step 2: Convert given probability into intersection form.
• Step 3: Use complement rule.
• Step 4: Check mutual exclusivity condition.

Concept Visualization

Solution

Given:

\[ P(E' \cup F') = 0.25 \]

Using De Morgan’s law:

\[ E' \cup F' = (E \cap F)' \]

Therefore:

\[ P((E \cap F)') = 0.25 \]

Using complement rule:

\[ P(E \cap F) = 1 - P((E \cap F)') \]

Substitute:

\[ \begin{aligned} P(E \cap F) &= 1 - 0.25 \\ &= 0.75 \end{aligned} \]

Check condition for mutually exclusive events:

\[ \begin{aligned} P(E \cap F) = 0 \\ \text{(required)} \end{aligned} \]

But:

\[ P(E \cap F) = 0.75 \ne 0 \]

Hence, events overlap significantly.

Final Conclusion: E and F are NOT mutually exclusive.

Significance for Exams

• Tests understanding of De Morgan’s law in probability.
• Common trap: confusing union of complements with intersection.
• Very important for JEE logical reasoning MCQs.
• Strengthens concept of event overlap vs exclusivity.

Theory

Two fundamental rules are used:

• Complement Rule: \(P(A') = 1 - P(A)\)
• Addition Rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

The subtraction in the addition rule avoids double counting of the overlap.

Solution Roadmap

• Step 1: Use complement rule for (i) and (ii).
• Step 2: Apply addition law for (iii).
• Step 3: Perform arithmetic carefully.

Concept Visualization

Solution

Given:

\[ P(A)=0.42,\quad P(B)=0.48,\quad P(A \cap B)=0.16 \]

(i) Find \(P(A')\)

Using complement rule:

\[ \begin{aligned} P(A') &= 1 - P(A) \\ &= 1 - 0.42 \\ &= 0.58 \end{aligned} \]

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

\[ \begin{aligned} P(B') &= 1 - P(B) \\ &= 1 - 0.48 \\ &= 0.52 \end{aligned} \]

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

Using addition rule:

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

Substitute values:

\[ \begin{aligned} P(A \cup B) &= 0.42 + 0.48 - 0.16 \\ &= (0.42 + 0.48) - 0.16 \\ &= 0.90 - 0.16 \\ &= 0.74 \end{aligned} \]

Final Answers: (i) \(0.58\), (ii) \(0.52\), (iii) \(0.74\)

Significance for Exams

• Direct application of complement and addition rules.
• Common in CBSE 2–3 mark questions.
• Important for JEE MCQ elimination techniques.
• Builds base for conditional probability and Bayes theorem.

Theory

When dealing with real-life data (percentages), probabilities are obtained by converting percentages into decimals.

The addition rule of probability:

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

This subtraction avoids double counting of common elements.

Solution Roadmap

• Step 1: Convert percentages into probabilities.
• Step 2: Apply addition rule.
• Step 3: Simplify carefully.

Concept Visualization

Solution

Given:

• 40% study Mathematics ⇒ \(P(M) = \frac{40}{100} = 0.4\)
• 30% study Biology ⇒ \(P(B) = \frac{30}{100} = 0.3\)
• 10% study both ⇒ \(P(M \cap B) = \frac{10}{100} = 0.1\)

Using addition rule:

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

Substitute values:

\[ \begin{aligned} P(M \cup B) &= 0.4 + 0.3 - 0.1 \\ &= (0.4 + 0.3) - 0.1 \\ &= 0.7 - 0.1 \\ &= 0.6 \end{aligned} \]

Final Answer: \(0.6\)

Significance for Exams

• Classic real-life probability application.
• Very common in CBSE case-based questions.
• Important for JEE data interpretation + probability.
• Key concept: avoid double counting.

Theory

The addition rule of probability connects individual events, their union, and their intersection:

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

Rearranging:

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

This is used when “at least one” probability is given.

Solution Roadmap

• Step 1: Assign events properly.
• Step 2: Substitute values into formula.
• Step 3: Simplify step-by-step.

Concept Visualization

Solution

Let:

\[ F = \text{passing first exam}, \quad S = \text{passing second exam} \]

Given:

\[ P(F)=0.8,\quad P(S)=0.7,\quad P(F \cup S)=0.95 \]

Find \(P(F \cap S)\)

Using formula:

\[ P(F \cap S) = P(F) + P(S) - P(F \cup S) \]

Substitute values:

\[ \begin{aligned} P(F \cap S) &= 0.8 + 0.7 - 0.95 \\ &= (0.8 + 0.7) - 0.95 \\ &= 1.5 - 0.95 \\ &= 0.55 \end{aligned} \]

Final Answer: \(0.55\)

Significance for Exams

• Classic application of addition rule (reverse use).
• Common in CBSE case-based problems.
• Very important for JEE MCQ elimination.
• Key insight: “at least one” ⇒ use union formula.

Theory

Key probability relations used:

• Complement Rule: \(P(A') = 1 - P(A)\)
• Addition Rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

Also:

\[ P(\text{neither }A\text{ nor }B) = 1 - P(A \cup B) \]

Solution Roadmap

• Step 1: Convert “neither” into union probability.
• Step 2: Apply addition rule.
• Step 3: Solve for unknown probability.

Concept Visualization

Solution

Let:

\[ E = \text{passing English}, \quad H = \text{passing Hindi} \]

Given:

\[ P(E)=0.75,\quad P(E \cap H)=0.5,\quad P(\text{neither})=0.1 \]

Step 1: Find \(P(E \cup H)\)

Using complement:

\[ P(E \cup H) = 1 - P(\text{neither}) \] \[ \begin{aligned} P(E \cup H) &= 1 - 0.1 \\ &= 0.9 \end{aligned} \]

Step 2: Use addition rule

\[ P(E \cup H) = P(E) + P(H) - P(E \cap H) \]

Substitute values:

\[ \begin{aligned} 0.9 &= 0.75 + P(H) - 0.5 \end{aligned} \]

Simplify RHS:

\[ 0.75 - 0.5 = 0.25 \] \[ 0.9 = 0.25 + P(H) \]

Solve:

\[ \begin{aligned} P(H) &= 0.9 - 0.25 \\ &= 0.65 \end{aligned} \]

Final Answer: \(0.65\)

Significance for Exams

• Combines complement + addition rule in one problem.
• Very common in CBSE case-based questions.
• Important for JEE logical probability problems.
• Key idea: “neither” ⇒ use complement of union.

Theory

This is a standard application of set theory in probability.

• \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
• \(P(\text{neither}) = 1 - P(A \cup B)\)
• \(P(\text{only }B) = P(B) - P(A \cap B)\)

Solution Roadmap

• Step 1: Convert numbers into probabilities.
• Step 2: Use addition rule for union.
• Step 3: Use complement for “neither”.
• Step 4: Use subtraction for “only” case.

Concept Visualization

Solution

Total students:

\[ n(S) = 60 \]

Given:

\[ n(NCC)=30,\quad n(NSS)=32,\quad n(NCC \cap NSS)=24 \]

(i) Probability of NCC or NSS

Using addition rule:

\[ \begin{aligned} n(NCC \cup NSS) &= n(NCC) + n(NSS) - n(NCC \cap NSS) \\ &= 30 + 32 - 24 \\ &= 62 - 24 \\ &= 38 \end{aligned} \]

Therefore:

\[ \begin{aligned} P(NCC \cup NSS) &= \frac{38}{60} \\ &= \frac{19}{30} \end{aligned} \]

(ii) Probability of neither NCC nor NSS

Using complement:

\[ \begin{aligned} P(\text{neither}) &= 1 - P(NCC \cup NSS) \\ &= 1 - \frac{19}{30} \end{aligned} \]

Convert 1 into fraction:

\[ 1 = \frac{30}{30} \] \[ \begin{aligned} P(\text{neither}) &= \frac{30}{30} - \frac{19}{30} \\ &= \frac{11}{30} \end{aligned} \]

(iii) Probability of NSS but not NCC

Students in NSS but not NCC:

\[ \begin{aligned} n(\text{only NSS}) &= n(NSS) - n(NCC \cap NSS) \\ &= 32 - 24 \\ &= 8 \end{aligned} \]

Therefore:

\[ \begin{aligned} P(\text{NSS but not NCC}) &= \frac{8}{60} \\ &= \frac{2}{15} \end{aligned} \]

Final Answers: (i) \(\frac{19}{30}\), (ii) \(\frac{11}{30}\), (iii) \(\frac{2}{15}\)

Significance for Exams

• Classic set theory + probability integration.
• Very common in CBSE case-based and data interpretation questions.
• Important for JEE Main logical elimination.
• Builds foundation for Venn diagram-based reasoning.