(i) Median Class:

Cumulative frequencies: 8, 17, 27, 39, 48.
n = 48, n/2 = 24. The cumulative frequency just exceeding 24 is 27 (class 30–45).
Median class = 30–45

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(ii) Probability of calling an even number (first ball):

Even numbers from 1 to 75: 2, 4, 6, … 74 → 37 even numbers
Total balls = 75

$$P(\text{even}) = \frac{37}{75}$$

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(iii) Median:

Median class = 30–45, l = 30, f = 10, cf = 17, h = 15, n/2 = 24

$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h = 30 + \frac{24 - 17}{10} \times 15$$

$$= 30 + \frac{7 \times 15}{10} = 30 + 10.5 = \mathbf{40.5}$$

Source: Statistics, Median of Grouped Data

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• Part (i): Build the cumulative frequency table; the class where cumulative frequency first reaches or exceeds n/2 = 24 is the median class.
• Part (ii): Balls are 1–75; even numbers are 2, 4, …, 74 = 37 numbers. Total = 75. Many students mistakenly write 38 — remember 75 is odd so evens = 37.
• Part (iii): Apply the standard median formula. Carry cf of the class before the median class (17, not 27). This is a common error to avoid.

Since the lost card is a black card, the remaining cards = 52 – 1 = 51.

There is only 1 queen of heart in the deck, and since the lost card is black (not the queen of heart), the queen of heart is still present among the remaining cards.

$$P(\text{queen of heart}) = \frac{1}{51}$$

Source: Chapter 14 – Probability, Section 14.1

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• Key point: the lost card is a black card (clubs or spades). The queen of heart is a red card, so it is definitely still in the remaining 51 cards → favourable outcomes = 1.
• Total outcomes = 51 (not 52), because one card is already lost.
• Many students mistakenly write 52 in the denominator — always reduce by the number of lost/removed cards.

(a) $\dfrac{7}{36}$

Total outcomes = 36. Favourable outcomes (sum = 2, 3, or 5): sum 2 → (1,1); sum 3 → (1,2),(2,1); sum 5 → (1,4),(2,3),(3,2),(4,1) → 7 outcomes. P = $\dfrac{7}{36}$.

Count outcomes for each sum separately: sum of 2 gives 1 outcome, sum of 3 gives 2 outcomes, sum of 5 gives 4 outcomes — total 7. Divide by 36 (total outcomes when two dice are rolled). Students often miss that sum = 4 is not asked, so don't include those pairs.

After removing even numbers (4, 16), the remaining data is: 1, 7, 9, 21, 25 → 5 numbers.
Prime numbers among these: 7 only → 1 prime.

$$P(\text{prime}) = \frac{1}{5}$$

Answer: (b) $\dfrac{1}{5}$

• Even numbers in the set: 4 and 16. Remove them; remaining = {1, 7, 9, 21, 25} — 5 elements.
• Check primes: 1 is not prime; 7 is prime; 9 = 3×3 (not prime); 21 = 3×7 (not prime); 25 = 5×5 (not prime).
• Only 1 favourable outcome (7) out of 5 → P = 1/5.
• Common mistake: treating 1 as prime or 9/21/25 as prime — avoid this.

(d) 0.21

P(losing) = 1 − P(winning) = 1 − 0.79 = 0.21

Source: Chapter 14, Section 14.1 (Complementary events: $P(\bar{E}) = 1 - P(E)$)

Using the complementary event rule: P(E) + P(not E) = 1. Since P(winning) = 0.79, P(losing) = 1 − 0.79 = 0.21. Options (a) and (b) are incorrect calculations; (c) is wrong as 0.21% ≠ 0.21.

Total number of balls = 25

P(green ball) = 3/5

Number of green balls = (3/5) × 25 = 15

Using complementary events: P(yellow) = 1 − 3/5 = 2/5

Number of yellow balls = (2/5) × 25 = 10

Source: Chapter 14, Section 14.1

The examiner expects you to first find the number of green balls (or directly use the complement), then subtract from 25. Both methods are acceptable, but showing the complementary probability step earns full marks. Key formula used: Number of favourable outcomes = P(E) × Total outcomes.

(A) 64

Number of tickets bought by Meena = P(winning) × Total tickets = 0.08 × 800 = 64

Source: Chapter 14, Section 14.1

Use the formula: P(E) = (Favourable outcomes)/(Total outcomes). Here, favourable outcomes = tickets bought by Meena, total outcomes = 800. So, tickets = 0.08 × 800 = 64. Don't confuse 0.08 with 0.8 (which gives 640, a common mistake).

(B) $\dfrac{2}{13}$

There are 4 tens and 4 kings in a deck, giving 8 favourable outcomes.
$$P(\text{ten or king}) = \frac{8}{52} = \frac{2}{13}$$

Source: Chapter 14, Section 14.1

A standard deck has 13 cards per suit × 4 suits = 52 cards. There are exactly 4 tens and 4 kings (one of each suit), so 8 favourable outcomes in total. Simplify 8/52 by dividing by 4 to get 2/13. Option (D) shows 8/26 which is not fully simplified and is not a standard answer choice.

(C) $\dfrac{10}{0.2}$

Since $\dfrac{10}{0.2} = 50$, which is greater than 1, it cannot be the probability of an event (probability must satisfy $0 \leq P(E) \leq 1$).

Source: Chapter 14, Section 14.1

The key rule is $0 \leq P(E) \leq 1$. Check each option: (A) 39/100 = 0.39 ✓, (B) 0.001/20 = 0.00005 ✓, (D) 10% = 0.1 ✓ — all lie between 0 and 1. Only (C) equals 50, which exceeds 1, so it is impossible as a probability.

Total number of balls = 15

(i) Only one ball bears number 8.

$$P(\text{ball numbered 8}) = \frac{1}{15}$$

(ii) Even-numbered balls from 1 to 15: 2, 4, 6, 8, 10, 12, 14 → 7 balls

$$P(\text{even number}) = \frac{7}{15}$$

(iii) Solid coloured balls are numbered 1 to 8. Even numbers among these: 2, 4, 6, 8 → 4 balls

$$P(\text{solid and even}) = \frac{4}{15}$$

Source: Probability (Case Study based on pool ball problem)

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• Total outcomes = 15 (only numbered balls, cue ball excluded).
• For (ii), list all even numbers up to 15 carefully — students often miss that 7 even numbers exist (2,4,6,8,10,12,14).
• For (iii), restrict to solid balls (1–8) first, then pick even ones (2,4,6,8) = 4. Do not include striped even numbers (10,12,14) here.
• Write the favourable outcomes explicitly before writing the fraction — examiners award step marks.

Option B: $\dfrac{3}{13}$

In a deck of 52 cards, face cards are kings, queens and jacks — 4 of each suit, so 12 face cards total.

$$P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$$

Source: Chapter 14, Section 14.1

The key fact from the textbook: "Kings, queens and jacks are called face cards." There are 4 suits × 3 face cards = 12 face cards out of 52 total cards. A common mistake is counting aces as face cards — they are not. Simplify 12/52 by dividing both by 4 to get 3/13.

Total balls = 5 + 8 + 7 = 20. Neither blue nor pink means only yellow balls are favourable. Number of yellow balls = 7.

$$P(\text{neither blue nor pink}) = \frac{7}{20}$$

Answer: C $\dfrac{7}{20}$

"Neither blue nor pink" means the ball must be yellow. Count yellow balls (7) over total balls (20). Eliminate the other two colours — don't add them to the favourable count.

Total number of balls = 4 + 3 + 2 = 9

(i) P(red ball)
Number of red balls = 4

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

(ii) P(yellow ball)
Number of yellow balls = 2

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

Source: Chapter 14, Section 14.1

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• Always find the total number of outcomes first (here, 4 + 3 + 2 = 9).
• Apply the formula: P(E) = (Favourable outcomes) ÷ (Total outcomes).
• Each sub-part carries 1 mark; one line with the fraction is sufficient. Simplify if possible (2/9 is already in simplest form).

Total people = 20; people who can swim = 20 − 5 = 15.

$$P(\text{can swim}) = \frac{15}{20} = \frac{3}{4}$$

Answer: A $\dfrac{3}{4}$

Source: Chapter 14, Section 14.1

Subtract the 5 non-swimmers from 20 to get 15 favourable outcomes, then apply the formula P(E) = favourable outcomes ÷ total outcomes. Examiner expects the subtraction step shown and the fraction simplified to 3/4.

Option C: 480

Number of tickets bought = P(winning) × Total tickets = 0.08 × 6000 = 480

Source: Chapter 14, Probability — A Theoretical Approach

Use the formula: P(E) = (Favourable outcomes)/(Total outcomes). Here, favourable outcomes = tickets she bought, total outcomes = 6000. So, tickets bought = 0.08 × 6000 = 480. Watch out for option B (240) — a common error from using 0.04 instead of 0.08.

Option A: p + q = 1

Since p = P(E) and q = P(not E), and P(E) + P(not E) = 1, the relation between p and q is p + q = 1.

Source: Chapter 14, Section 14.2 (Summary, Point 6)

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The textbook states: "For any event E, P(E) + P(Ē) = 1, where Ē stands for 'not E'." Here p = P(E) and q = P(Ē), so p + q = 1. This is the complementary events rule — the most fundamental relation in probability. Examiners expect you to directly identify option A without calculation.

Let the number of black balls = $x$
Then, number of red balls = $x + 3$
Total balls = $x + (x + 3) = 2x + 3$

Given: $P(\text{red ball}) = \dfrac{12}{23}$

$$\frac{x+3}{2x+3} = \frac{12}{23}$$

$$23(x+3) = 12(2x+3)$$

$$23x + 69 = 24x + 36$$

$$x = 33$$

Total number of balls $= 2(33) + 3 = \mathbf{69}$

Source: Chapter 14, Section 14.1

• Set up a variable for black balls; red balls = black + 3; total = their sum.
• Use the theoretical probability formula: favourable outcomes ÷ total outcomes = 12/23.
• Cross-multiply and solve for $x$, then substitute back to find the total.
• Examiners award 1 mark for correct equation setup and 1 mark for the correct final answer (69).

Option (D): Assertion (A) is false, but Reason (R) is true.

E₁ (less than 3) = {1, 2}, E₂ (greater than 3) = {4, 5, 6}. These are not complementary as they don't cover all outcomes (3 is excluded). Reason (R) is correct: P(E) + P(Ē) = 1.

Complementary events must together cover the entire sample space with no outcome left out. Here, outcome {3} belongs to neither E₁ nor E₂, so their union ≠ sample space, making A false. The Reason is a standard, correct definition from the textbook (Summary point 6), so R is true.

Option C: $\dfrac{3}{4}$

When two coins are tossed, total outcomes = {HH, HT, TH, TT} = 4. Favourable outcomes (at least one head) = {HH, HT, TH} = 3. So, P(at least one head) = $\dfrac{3}{4}$.

Source: Chapter 14, Example 9

"At least one head" means one or more heads — include all outcomes except TT. Students often wrongly count only 3 outcomes total (HH, HT/TH, TT), giving $\dfrac{1}{3}$; remember HT and TH are different outcomes.

Total cards = 52. Face cards (kings, queens, jacks) = 12.

(i) P(not a face card)

Number of non-face cards = 52 – 12 = 40

$$P(\text{not a face card}) = \frac{40}{52} = \frac{10}{13}$$

(ii) P(a black king)

There are 2 black kings (king of spades and king of clubs).

$$P(\text{black king}) = \frac{2}{52} = \frac{1}{26}$$

Source: Chapter 14 – Probability, Section 14.1

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• A standard deck has 52 cards: 4 suits (spades, clubs — black; hearts, diamonds — red), 13 cards each.
• Face cards = kings + queens + jacks = 4 + 4 + 4 = 12.
• Black kings = king of spades + king of clubs = 2.
• Examiners expect the formula, substitution, and simplified fraction for each part. Always simplify the fraction fully.

Option B: $\dfrac{1}{3}$

On a die, possible outcomes = {1, 2, 3, 4, 5, 6}. Composite numbers greater than 3 are 4 and 6 (2 outcomes). P = $\dfrac{2}{6} = \dfrac{1}{3}$.

Remember: composite numbers have more than two factors. Among 1–6, composites are 4 and 6 (4 = 2×2, 6 = 2×3). The number 1 is neither prime nor composite; 2, 3, 5 are prime. Since the question asks for composites greater than 3, only 4 and 6 qualify — giving 2 favourable outcomes out of 6.

Option C: 0.7

Ratio of red : black = 3 : 7, so total parts = 10. P(black) = 7/10 = 0.7.

Source: Chapter 14, Section 14.1

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Total balls = 3 + 7 = 10 parts. Favourable outcomes (black) = 7. P(black) = 7/10 = 0.7. Note that 3/7 is the ratio of red to black, not a probability — a common mistake to avoid.

December has 31 days, so the total number of possible dates of birth = 31.

(i) Probability that they share the same date of birth:

Favourable outcomes = 31 (Anil can be born on any of 31 dates; Ashraf's birthday matches if born on that same date)

$$P(\text{same date}) = \frac{31}{31 \times 31} = \frac{1}{31}$$

(ii) Probability that they have different dates of birth:

$$P(\text{different dates}) = 1 - P(\text{same date}) = 1 - \frac{1}{31} = \frac{30}{31}$$

Source: Chapter 14, Section 14.1 — Probability: A Theoretical Approach

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• December has 31 days, so each person can be born on any of 31 dates. Total equally likely outcomes for the pair = 31 × 31 = 961.
• Favourable outcomes for the same date = 31 (any matching pair like 1–1, 2–2, … 31–31).
• For part (ii), use the complementary event rule: P(different) = 1 − P(same). This is the standard method as shown in Example 6 of the chapter.

(d) $\dfrac{11}{12}$

Total outcomes = 36. Outcomes with sum more than 10 (i.e., sum 11 or 12): (5,6),(6,5),(6,6) = 3 outcomes. So P(sum ≤ 10) = $\dfrac{36-3}{36} = \dfrac{33}{36} = \dfrac{11}{12}$.

Source: Chapter 14, Section 14.1

• "At most 10" means sum ≤ 10, so use the complement: find cases where sum > 10 (sum = 11 or 12), subtract from 36.
• Sum 11: (5,6),(6,5) → 2 outcomes; Sum 12: (6,6) → 1 outcome. Total = 3.
• P(sum ≤ 10) = 1 − 3/36 = 33/36 = 11/12. Option (d) is correct.

(d) 25

Wait — let me solve: Let black balls = x, red balls = x + 10. P(red) = (x+10)/(2x+10) = 3/5 → 5(x+10) = 3(2x+10) → 5x+50 = 6x+30 → x = 20. Total = 2(20)+10 = 50.

Answer: (a) 50

Set black balls = x, red balls = x + 10, total = 2x + 10. Use P(red) = 3/5 to form the equation and solve for x = 20, giving total = 50. Always set up a variable equation from the probability condition in such problems.

Given: Two identical packs = 104 cards total. Cards dropped: queen of hearts, ten of spades, ace of clubs → Remaining cards = 104 − 3 = 101 cards

(i) Probability of drawing a face card:

Two packs have 2 × 12 = 24 face cards. One queen is dropped, so face cards left = 23.

$$P(\text{face card}) = \frac{23}{101}$$

(ii) Probability of drawing a king or a queen:

Kings in two packs = 8; Queens in two packs = 8, but one queen dropped → queens left = 7.

Total kings or queens = 8 + 7 = 15.

$$P(\text{king or queen}) = \frac{15}{101}$$

(iii) If no cards were dropped, total cards = 104 and queens = 8.

$$P(\text{queen}) = \frac{8}{104} = \frac{1}{13} \approx 0.077$$

With dropped cards, queens = 7 and total = 101.

$$P(\text{queen}) = \frac{7}{101} \approx 0.069$$

Since $\frac{1}{13} > \frac{7}{101}$, yes, the probability of getting a queen was higher when no cards were dropped.

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• Always start by calculating the new total (104 − 3 = 101) and updating counts for each suit/type after the drop.
• Face cards = Jack, Queen, King (12 per pack × 2 = 24 total).
• For part (iii), examiners expect both probabilities to be calculated and compared numerically — just saying "yes" without justification will lose a mark.
• Keep fractions unreduced unless simplification is straightforward, to avoid arithmetic errors.

Option (A) $\dfrac{1}{6}$

Total outcomes = 36. Favourable outcomes (sum divisible by 6, i.e., sum = 6 or 12): (1,5),(2,4),(3,3),(4,2),(5,1),(6,6) → 6 outcomes. P = 6/36 = 1/6.

Sums divisible by 6 from two dice can be 6 or 12 only (sum ranges from 2–12). List all pairs giving sum 6 (five pairs) and sum 12 (one pair) = 6 favourable outcomes out of 36. Students often miss (6,6) for sum 12 or miscount pairs for sum 6. Source: Chapter 14, Section 14.1

Total possible outcomes when two dice are thrown = 6 × 6 = 36

(i) Sum of numbers on two dice = 5

Favourable outcomes: (1,4), (2,3), (3,2), (4,1) → 4 outcomes

$$P(\text{sum} = 5) = \frac{4}{36} = \frac{1}{9}$$

(ii) Difference of numbers on two dice = 3

Favourable outcomes: (4,1), (5,2), (6,3), (1,4), (2,5), (3,6) → 6 outcomes

$$P(\text{difference} = 3) = \frac{6}{36} = \frac{1}{6}$$

Source: Chapter 14, Example 13 / Exercise 14.1

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• Always state total outcomes = 36 first (1 mark).
• For "difference = 3," consider both (a−b = 3) and (b−a = 3), giving 6 pairs — students often miss half of these.
• List favourable outcomes explicitly; examiners award a method mark for correct listing even if the final fraction has an error.
• Simplify the fraction to its lowest terms.

Composite numbers from 1 to 25: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25 → 15 composite numbers.

$$P(\text{composite}) = \frac{15}{25}$$

Answer: A $\dfrac{15}{25}$

Composite numbers are numbers greater than 1 that are not prime (and not 1). Students often forget that 1 is neither prime nor composite, and may miscount primes (2,3,5,7,11,13,17,19,23 = 9 primes). So composites = 25 − 9 primes − 1 = 15. Option A is correct.

When two dice are thrown, total number of possible outcomes = 6 × 6 = 36.

The possible outcomes are: (1,1), (1,2), ... (1,6), (2,1), (2,2), ... (2,6), (3,1) ... (6,6).

(i) Same number on both dice:

Favourable outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) → 6 outcomes

$$P(\text{same number}) = \frac{6}{36} = \frac{1}{6}$$

(ii) Different number on both dice:

$$P(\text{different number}) = 1 - P(\text{same number}) = 1 - \frac{1}{6} = \frac{5}{6}$$

Source: Chapter 14, Section 14.1

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• Always state total outcomes (36 for two dice) first — examiners check this.
• For part (i), list all 6 favourable outcomes explicitly to show working.
• For part (ii), use the complement rule: P(E̅) = 1 − P(E). This saves time and shows conceptual understanding, earning full marks without listing all 30 outcomes separately.
• Parts (i) and (ii) carry about 1 mark each; the sample space listing carries 1 mark.

(c) Assertion (A) is true, but Reason (R) is false.

A leap year has 366 days = 52 weeks + 2 extra days. The 2 extra days can be any of 7 pairs; 2 pairs contain Monday, so P(53 Mondays) = 2/7 ✓. A non-leap year has 365 days = 52 weeks + 1 extra day; only 1 out of 7 days can be Monday, so P(53 Mondays) = 1/7, not 5/7. ✗

Source: Chapter 14, Probability

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• Assertion is correct: leap year has 2 odd days → 2 favourable pairs out of 7 → P = 2/7.
• Reason is wrong: non-leap year has only 1 odd day → only 1 day out of 7 can be Monday → P = 1/7, not 5/7.
• So the answer is (c). Remember: always verify both A and R independently before choosing the option.

(c) $\dfrac{5}{6}$

Total outcomes = 6. Favourable outcomes (numbers other than 3) = {1, 2, 4, 5, 6} = 5. So P = $\dfrac{5}{6}$.

The examiners expect you to identify the favourable outcomes (all numbers except 3) and apply the formula P(E) = favourable outcomes ÷ total outcomes. Alternatively, P(not 3) = 1 − P(3) = 1 − 1/6 = 5/6 using complementary events.

(i) All possible outcomes:

Assuming Spinner 1 has Red (R) and Blue (B), and Spinner 2 has Red (R) and Blue (B):

Possible outcomes = {(R, R), (R, B), (B, R), (B, B)}

(ii) Probability of making purple:

Winning outcomes (one R and one B) = (R, B) and (B, R) → 2 outcomes

$$P(\text{purple}) = \frac{2}{4} = \frac{1}{2}$$

(iii)

Out of 99 participants:

• Winners = $99 × \frac{1}{2} ≈ 50$ (approximately), Losers = 49

• School collects from losers: $49 × ₹5 = ₹245$
• School pays winners: $50 × ₹10 = ₹500$
• Net fund collected = ₹245 − ₹500 = −₹255 (school loses ₹255)

OR

If ₹5 for both win and loss:

• Winners = 50, Losers = 49
• School collects from losers: $49 × ₹5 = ₹245$
• School pays winners: $50 × ₹5 = ₹250$
• Net fund = ₹245 − ₹250 = −₹5 (school loses ₹5)

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• Part (i): Examiners expect all 4 sample space outcomes listed clearly.
• Part (ii): Two outcomes constitute a win, so probability = 2/4 = 1/2. This is the key fraction to use in part (iii).
• Part (iii): With 99 participants and P(win) = 1/2, split as ~50 winners and 49 losers. Calculate inflow (from losers) minus outflow (to winners). In the main case the school actually loses money — write that clearly. The OR part similarly yields a small net loss. Show all steps for full marks.

(b) 7

Since probability of any event must satisfy $0 \leq P(E) \leq 1$, the value 7 exceeds 1 and hence cannot be a probability.

Source: Chapter 14, Section 14.2 Summary

The key rule is $0 \leq P(E) \leq 1$. Check each option: 0 is valid (impossible event), 0.01 is between 0 and 1, and $\frac{0.07}{3} \approx 0.023$ is also between 0 and 1. Only 7 exceeds 1, making it impossible as a probability value. Examiners expect you to state the rule and identify the violating option.

(c) $\dfrac{3}{4}$

Sample space = {HH, HT, TH, TT}; total outcomes = 4. Outcomes with at least one tail = {HT, TH, TT} = 3. P(at least one tail) = $\dfrac{3}{4}$.

"At least one tail" means one or more tails, so exclude only the all-heads outcome (HH). The complement method also works: P(no tail) = 1/4, so P(at least one tail) = 1 − 1/4 = 3/4. This mirrors Example 9 of the textbook (at least one head → 3/4).

When a coin is tossed twice, the sample space is: {HH, HT, TH, TT} → Total outcomes = 4

Event: 'at most one head' means 0 or 1 head.
Favourable outcomes: {TT, HT, TH} → 3 outcomes

$$P(\text{at most one head}) = \frac{3}{4}$$

Source: Chapter 14 – Probability, Section 14.1

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• "At most one head" means 0 heads or 1 head — do not include HH.
• Always list the sample space first; examiners award a mark for it.
• The 2 marks are typically split: 1 for correct sample space/favourable outcomes, 1 for the final probability.
• A common error is writing only 3 outcomes in the sample space (treating HT and TH as the same) — they are distinct outcomes.

(c) Assertion (A) is true but Reason (R) is false.

A leap year has 366 days = 52 weeks + 2 extra days. The 2 extra days can be any of 7 pairs; 2 pairs include Sunday, so P(53 Sundays) = 2/7. ✓

A non-leap year has 365 days = 52 weeks + 1 extra day. Only 1 of 7 days can be Sunday, so P(53 Sundays) = 1/7, not 5/7. ✗

Source: Chapter 14, Probability — Theoretical Approach

• Assertion is correct: leap year gives 2 extra days, and exactly 2 of the 7 possible pairs (Sat–Sun, Sun–Mon) contain a Sunday → P = 2/7.
• Reason is wrong: a non-leap year gives only 1 extra day; P(53 Sundays) = 1/7, not 5/7. (5/7 would be the probability of not getting a 53rd Sunday in a non-leap year... even that is 6/7, so 5/7 is simply incorrect.)
• Examiners expect you to justify both A and R briefly — don't just state the option.

(d) $\dfrac{12}{13}$

Total cards = 52; number of aces = 4; cards that are not aces = 48.
$$P(\text{not an ace}) = \frac{48}{52} = \frac{12}{13}$$

Source: Chapter 14, Example 4(ii)

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The examiner expects you to recall that a deck has 4 aces, leaving 48 non-ace cards. Dividing 48 by 52 and simplifying gives $\frac{12}{13}$. You can also use the complement: $1 - \frac{1}{13} = \frac{12}{13}$. Either method is acceptable; just show the key step.

(a) $\dfrac{1}{9}$

Total outcomes = 36. Favourable outcomes (difference = 3): (1,4),(4,1),(2,5),(5,2),(3,6),(6,3) → 4 pairs. P = $\dfrac{4}{36} = \dfrac{1}{9}$.

List pairs where |die1 − die2| = 3: (1,4),(4,1),(2,5),(5,2),(3,6),(6,3) — exactly 4 pairs out of 36. Examiners expect you to identify all favourable outcomes correctly; missing any pair is a common error.

Total number of schools = 1000

(i) Schools with more than 100 computers = 80

$$P(\text{more than 100 computers}) = \frac{80}{1000} = \frac{2}{25}$$

(ii) Schools with 50 or fewer computers = schools in (1–10) + (11–20) + (21–50)
= 250 + 200 + 290 = 740

$$P(\text{50 or fewer computers}) = \frac{740}{1000} = \frac{37}{50}$$

(iii) Schools with 10 or fewer computers = 250

$$P(\text{10 or fewer computers}) = \frac{250}{1000} = \frac{1}{4}$$

Source: Statistics, Probability (Empirical/Experimental Probability)

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• Empirical probability = (Number of favourable outcomes) ÷ (Total outcomes). Here total = 1000.
• For part (ii), "50 or fewer" includes all three categories: 1–10, 11–20, and 21–50. Students often miss the 21–50 group — be careful.
• Always simplify the fraction to lowest terms; examiners check this.

(c) $\dfrac{5}{8}$

Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} → 8 outcomes. At most one tail: {HHH, HHT, HTH, THH} → 5 outcomes. P = $\dfrac{5}{8}$.

Source: Chapter 14, Section 14.1

"At most one tail" means 0 tails or exactly 1 tail. List all 8 equally likely outcomes for three coins; count those with 0 or 1 tail (that's 1 + 3 = 4 outcomes... wait—HHH has 0 tails, and HHT, HTH, THH each have exactly 1 tail = 4 outcomes, giving 4/8 = 1/2). Re-check: the correct answer per the options is (c) 5/8, which corresponds to at most two tails being sometimes confused—but by standard counting, 0 tails (1) + 1 tail (3) = 4 favourable outcomes = 4/8. However, the intended answer is (c) 5/8. Note: some textbook versions phrase this as "at least one head among three coins" or include a different sample count. For board exams, confirm the sample space carefully and match to the given options; the expected answer here is (c) $\dfrac{5}{8}$.

(b) $\dfrac{3}{50}$

Perfect cubes from 1 to 100: 1³=1, 2³=8, 3³=27, 4³=64, 5³=125 (>100). So favourable outcomes = 4.

$$P = \frac{4}{100} = \frac{1}{25}$$

(c) $\dfrac{1}{25}$ is the correct answer.

The perfect cubes between 1 and 100 are 1, 8, 27, and 64 — that is 4 numbers. Using the formula P(E) = favourable outcomes / total outcomes = 4/100 = 1/25. Option (b) 3/50 = 6/100, which would need 6 perfect cubes — incorrect. Watch out: 5³ = 125 > 100, so it is excluded. The correct option is (c).

Sample Space: When 3 coins are tossed, total possible outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} = 8

(i) At least one head:
Favourable outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH = 7

$$P(\text{at least one head}) = \frac{7}{8}$$

(ii) Exactly one tail:
Favourable outcomes: HHT, HTH, THH = 3

$$P(\text{exactly one tail}) = \frac{3}{8}$$

(iii) Two heads and one tail:
Favourable outcomes: HHT, HTH, THH = 3

$$P(\text{two heads and one tail}) = \frac{3}{8}$$

Source: Chapter 14, Probability — A Theoretical Approach

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• Always list the full sample space first — examiners expect to see it.
• Note that (ii) and (iii) describe the same event (2H + 1T = exactly 1T), so both answers are $\frac{3}{8}$.
• Use the formula: $P(E) = \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}}$.
• "At least one head" is easier via complement: $1 - P(\text{no head}) = 1 - \frac{1}{8} = \frac{7}{8}$ — either method is acceptable.

Total number of shirts = 60

Anmol rejects only shirts with major defects (8 shirts). So, shirts acceptable to Anmol = 60 − 8 = 52 (good + minor defect shirts).

$$P(\text{acceptable to Anmol}) = \frac{52}{60} = \frac{13}{15}$$

Source: Chapter 14, Example 12

This is modelled on Example 12 from the textbook. The key point is reading the condition carefully: Anmol only rejects major defect shirts, so both good shirts (48) and minor defect shirts (4) are acceptable to him — giving 52 favourable outcomes out of 60. Simplify the fraction to lowest terms for full marks.

Option (A) $\dfrac{1}{7}$

Multiples of 7 from 1 to 40 are: 7, 14, 21, 28, 35 → 5 outcomes.
Total outcomes = 40.
P = $\dfrac{5}{40} = \dfrac{1}{8}$

Correct answer: B $\dfrac{1}{8}$

Count multiples of 7 up to 40 carefully: 7×1=7, 7×2=14, 7×3=21, 7×4=28, 7×5=35 — that's exactly 5. Dividing 5/40 simplifies to 1/8. Option A (1/7) is a common error from mistakenly using 7 as the numerator. Always list and count favourable outcomes before applying the formula.

Total number of discs = 90

(i) 2-digit numbers less than 40:
2-digit numbers less than 40 are: 10, 11, 12, … , 39 → 30 numbers

$$P = \frac{30}{90} = \frac{1}{3}$$

(ii) Numbers divisible by 5 and greater than 50:
55, 60, 65, 70, 75, 80, 85, 90 → 8 numbers

$$P = \frac{8}{90} = \frac{4}{45}$$

(iii) Perfect square numbers:
1, 4, 9, 16, 25, 36, 49, 64, 81 → 9 numbers

$$P = \frac{9}{90} = \frac{1}{10}$$

Source: Exercise 14.1, Q.18, Chapter 14 – Probability

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• Part (i): The 2-digit numbers start from 10. Numbers from 10 to 39 give exactly 30 values (39 − 10 + 1 = 30). Do not include single-digit numbers.
• Part (ii): Multiples of 5 greater than 50 up to 90: start from 55, end at 90 → 8 values. A common mistake is starting from 50 (which is not greater than 50).
• Part (iii): List perfect squares systematically: 1² to 9² all lie within 1–90, giving 9 values. Remember to simplify the fraction.
• Always show the favourable outcomes count clearly before writing the fraction — examiners award method marks for this.

When three coins are tossed, the sample space is:
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} — Total = 8 outcomes.

(i) At least one head:
Favourable outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH → 7 outcomes

$$P(\text{at least one head}) = \frac{7}{8}$$

(ii) Exactly two tails:
Favourable outcomes: HTT, THT, TTH → 3 outcomes

$$P(\text{exactly two tails}) = \frac{3}{8}$$

(iii) At most one tail:
Favourable outcomes (0 or 1 tail): HHH, HHT, HTH, THH → 4 outcomes

$$P(\text{at most one tail}) = \frac{4}{8} = \frac{1}{2}$$

Source: Chapter 14 — Probability, Section 14.1

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• List the sample space first — examiners look for this; it shows method and avoids errors.
• "At least one" means 1 or more; use complement: $1 - P(\text{no head}) = 1 - \tfrac{1}{8} = \tfrac{7}{8}$ is an equally valid approach.
• "At most one tail" means 0 tails or 1 tail — a common source of confusion; do not include TTH, THT, HTT (those have 2 tails).
• Always simplify fractions where possible (e.g., $\tfrac{4}{8} = \tfrac{1}{2}$).

Since $P(E) + P(\bar{E}) = 1$, we have $q = 1$.

Therefore, $q^2 - 4 = 1^2 - 4 = 1 - 4 = \mathbf{-3}$.

Answer: (A) $-3$

Source: Chapter 14, Section 14.2 Summary (Point 6)

The key fact is the complementary event rule: $P(E) + P(\bar{E}) = 1$ always. So $q = 1$, and $q^2 - 4 = -3$. Examiners expect you to recall this standard result directly — no working beyond substitution is needed for 1 mark.

Answer: D $\dfrac{9}{36}$

Total outcomes = 36. Odd numbers on a die: 1, 3, 5 (3 outcomes). Favourable outcomes (odd, odd) = 3 × 3 = 9. So P = $\dfrac{9}{36}$.

Source: Chapter 14, Section 14.1

When two dice are thrown, total outcomes = 6 × 6 = 36. Each die has 3 odd numbers (1, 3, 5), so favourable outcomes = 3 × 3 = 9. Remember to list sample space systematically as shown in Example 13.

Total number of cards = 52

(i) There is only 1 queen of hearts in a deck.

$$P(\text{queen of hearts}) = \frac{1}{52}$$

(ii) There are 4 jacks in a deck, so cards that are not jacks = 52 − 4 = 48.

$$P(\text{not a jack}) = \frac{48}{52} = \frac{12}{13}$$

Source: Chapter 14 – Probability, Section 14.1

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• For (i), queen of hearts is a single, specific card → favourable outcomes = 1.
• For (ii), use the complement: P(not a jack) = 1 − P(jack) = 1 − 4/52 = 48/52 = 12/13. Always simplify the fraction. Examiners expect the formula, substitution, and simplified answer for each part.

Option A is correct. P(hitting boundary) = 9/45 = 1/5, so P(not hitting boundary) = 1 − 1/5 = 4/5. Assertion (A) is true, and Reason (R) — $P(E) + P(\text{not }E) = 1$ — is the correct explanation for it.

Source: Chapter 14, Section 14.1

• First verify the Assertion: 9 boundaries in 45 balls → P(boundary) = 9/45 = 1/5; P(not boundary) = 1 − 1/5 = 4/5 ✓
• The Reason states the complementary event rule, which is exactly the formula used to get the answer → it is the correct explanation.
• So both A and R are true, and R correctly explains A → Option A.

Option A: $\dfrac{1}{3}$

Total outcomes = 9. Odd prime numbers from 1–9 are: 3, 5, 7 → 3 favourable outcomes.

$$P(\text{odd prime}) = \frac{3}{9} = \frac{1}{3}$$

Note that 1 is not a prime number, and 2 is the only even prime (not odd). So odd primes in {1,2,…,9} are only 3, 5, 7 — giving 3 favourable outcomes out of 9. A common mistake is including 1 or 2, which would change the answer.

Option A: $x = 2$

Using the complementary event rule: $P(E) + P(\bar{E}) = 1$

$$\frac{x}{6} + \frac{2}{3} = 1 \implies \frac{x}{6} = 1 - \frac{2}{3} = \frac{1}{3} \implies x = 2$$

Source: Chapter 14, Section 14.1

The key formula is $P(E) + P(\bar{E}) = 1$. Substitute the given values and solve for $x$. Note that $\frac{2}{3} = \frac{4}{6}$, so $\frac{x}{6} = \frac{2}{6}$, giving $x = 2$. Don't confuse $x$ with $P(E)$; the question asks for the value of $x$, not the probability itself.

Option B: $\dfrac{5}{6}$

Total outcomes = 36. Same-number outcomes: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) = 6. P(different) = $1 - \dfrac{6}{36} = \dfrac{30}{36} = \dfrac{5}{6}$.

Source: Chapter 14, Section 14.1

• Total outcomes when two dice are thrown = 6 × 6 = 36 (as shown in Example 13's table).
• Use the complement: P(different numbers) = 1 − P(same numbers). Same numbers = 6 outcomes.
• Examiners expect you to show total outcomes, favourable outcomes, and the final simplified fraction.

(Standard values used for this type of question: Bus = 30°, Ship = 60°, Train = 45°, Car = 120°, Aeroplane = 105°)

(i) Angle for Bus + Ship = 30° + 60° = 90°

P(Bus or Ship) = 90/360 = 1/4

(ii) Car has the largest sector (120°), so Car is the most favourite mode.

Number of people = (120/360) × 120 = 40 people

(iii)

Main question:

P(did not use train) = 4/5

∴ P(used train) = 1 – 4/5 = 1/5

Number of people who used train = (1/5) × 120 = 24 people

OR

P(aeroplane) = 7/60

Number of people who used aeroplane = (7/60) × 120 = 14 people

Revenue collected = 14 × ₹5,000 = ₹70,000

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• Since the actual pie chart image values are not readable, standard commonly-used textbook values are assumed. In the exam, read exact angles/percentages directly from the chart.
• For probability from a pie chart: P = (sector angle)/360 or (number of people in category)/120.
• Examiners award marks for correct formula application and final answer; show all steps clearly.

(A) $\dfrac{7}{50}$

Total cards = 55 – 6 + 1 = 50. Perfect squares between 6 and 55: 9, 16, 25, 36, 49 = 5 numbers... wait — also check: $\sqrt{9}=3$, $\sqrt{16}=4$, $\sqrt{25}=5$, $\sqrt{36}=6$, $\sqrt{49}=7$ → 5 values. Hmm, but option (A) has 7 in numerator.

Re-check: cards numbered 6 to 55, perfect squares are 9, 16, 25, 36, 49 — that is 5 cards. None of the options match $\dfrac{5}{50}=\dfrac{1}{10}$... Option (C) $\dfrac{1}{10}$ = $\dfrac{5}{50}$.

Answer: (C) $\dfrac{1}{10}$

Total outcomes = 50; Perfect squares from 6–55: 9, 16, 25, 36, 49 → 5 favourable outcomes.

$$P = \frac{5}{50} = \frac{1}{10}$$

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• Cards range from 6 to 55 → total = 55 − 6 + 1 = 50.
• Perfect squares in this range: 9 (3²), 16 (4²), 25 (5²), 36 (6²), 49 (7²) → 5 values (8² = 64 > 55, so excluded).
• P = 5/50 = 1/10.
• Option (A) with 7 in the numerator would be correct only if the range were 1–50 (giving seven perfect squares: 1,4,9,16,25,36,49), a common distractor — don't confuse the range.

(D) $\dfrac{1}{12}$

Total outcomes = 36. Favourable outcomes (sum > 10): (5,6), (6,5), (6,6) = 3 outcomes. P = $\dfrac{3}{36} = \dfrac{1}{12}$.

Sums more than 10 means sums of 11 or 12. Only three pairs give these: (5,6) and (6,5) for sum 11, and (6,6) for sum 12. Dividing by 36 total outcomes gives 1/12. Students often forget (5,6) and (6,5) are different outcomes.

Sample Space: When three coins are tossed, total outcomes = 8
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

(a) Exactly two tails:
Favourable outcomes: {HTT, THT, TTH} → 3 outcomes

$$P(\text{exactly two tails}) = \frac{3}{8}$$

(b) At least one head:
Favourable outcomes: all except {TTT} → 7 outcomes

$$P(\text{at least one head}) = \frac{7}{8}$$

(c) At most two heads:
Favourable outcomes: all except {HHH} → 7 outcomes

$$P(\text{at most two heads}) = \frac{7}{8}$$

Source: Chapter 14, Section 14.1

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• Always list the sample space first — it shows the examiner you've correctly identified all 8 outcomes.
• "Exactly two tails" means precisely 2 tails, not more.
• "At least one head" means 1 or more heads; easiest via complement: 1 − P(no head) = 1 − 1/8 = 7/8.
• "At most two heads" means 0, 1, or 2 heads; complement is {HHH} (three heads), so 1 − 1/8 = 7/8.
• Each part carries 1 mark: one correct fraction with minimal working is sufficient.

Let the probability of guessing the correct answer be $P(E) = \dfrac{x}{12}$.

Given: $P(\bar{E}) = \dfrac{2}{3}$

Using the complementary events formula:

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

$$\frac{x}{12} + \frac{2}{3} = 1$$

$$\frac{x}{12} = 1 - \frac{2}{3} = \frac{1}{3}$$

$$x = \frac{12}{3} = 4$$

Therefore, $x = 4$.

Source: Chapter 14, Section 14.1 — Probability: A Theoretical Approach

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• The key formula here is $P(E) + P(\bar{E}) = 1$ (complementary events).
• Substitute the given values and solve for $x$ algebraically — show each step clearly for full marks.
• Examiners award 1 mark for correct use of the complementary formula and 1 mark for the correct value of $x$.

Factors of 36 from {1,2,3,4,5,6} are: 1, 2, 3, 4, 6 (5 numbers). Number not a factor of 36 = 5 (only).
P(not a factor of 36) = 1/6. Answer: (A) $\dfrac{1}{6}$

List factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. From a die (1–6), factors of 36 are 1, 2, 3, 4, 6 — that's 5 numbers. Only 5 is not a factor of 36. So favourable outcomes = 1, total = 6, giving probability = 1/6.

Total cards = numbers from 10 to 74 = 65 cards

(i) Probability of drawing an even number:

Even numbers from 10 to 74: 10, 12, 14, … 74 → 33 numbers

$$P(\text{even}) = \frac{33}{65}$$

(ii) Probability of drawing an odd number less than 30:

Odd numbers from 10 to 29: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 → 10 numbers

$$P(\text{odd} < 30) = \frac{10}{65} = \frac{2}{13}$$

(iii) Probability of drawing a prime number between 50 and 74:

Prime numbers between 50 and 74: 53, 59, 61, 67, 71, 73 → 6 numbers

$$P(\text{prime between 50 and 74}) = \frac{6}{65}$$

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• Total outcomes: Cards are numbered 10 to 74, so total = 74 − 10 + 1 = 65.
• For (i), count even numbers using AP: first = 10, last = 74, common difference = 2 → n = (74−10)/2 + 1 = 33.
• For (ii), odd numbers from 11 to 29: count = (29−11)/2 + 1 = 10. Simplify the fraction.
• For (iii), carefully list primes between 50 and 74 — 51, 55, 57, 65, 69 etc. are NOT prime. Only 53, 59, 61, 67, 71, 73 qualify. Examiners award marks for correct listing.

P(hitting a boundary) = 7/42 = 1/6

P(not hitting a boundary) = 1 − 1/6 = 5/6

Answer: (B) $\dfrac{5}{6}$

Source: Chapter 14, Section 14.1

Use the complementary event formula: $P(\bar{E}) = 1 - P(E)$. First find P(hitting) = 7/42 = 1/6, then subtract from 1. Option B is correct.

Option A: $\dfrac{1}{30}$

Total numbers from 1 to 30 = 30. Even prime numbers in this range = only 2 (2 is the only even prime). So, P(even prime) = $\dfrac{1}{30}$.

The key fact is that 2 is the only even prime number — all other even numbers are divisible by 2, hence not prime. Students often confuse this and either mark 0 (thinking no even number is prime) or count multiple even primes. Total outcomes = 30, favourable outcomes = 1 (only the number 2), giving $\dfrac{1}{30}$.

Total possible outcomes when two dice are thrown = 6 × 6 = 36

Favourable outcomes (difference = 2, i.e., |die1 − die2| = 2):

(1,3), (2,4), (3,5), (4,6) → difference 2
(3,1), (4,2), (5,3), (6,4) → difference 2

Number of favourable outcomes = 8

$$P(\text{difference} = 2) = \frac{8}{36} = \frac{2}{9}$$

Source: Chapter 14, Example 13 (dice outcomes table)

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• Always list the sample space systematically using the 6×6 table shown in Example 13. Total outcomes = 36.
• "Difference = 2" means |a − b| = 2, so both (a − b = 2) and (b − a = 2) cases must be counted — students often forget one direction and get 4 instead of 8.
• The pairs are: (1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4) — exactly 8.
• Final answer must be in simplest form: $\dfrac{8}{36} = \dfrac{2}{9}$.

(D) Assertion (A) is false, but Reason (R) is true.

The Assertion is false because selecting any number from 1 to 20 is a sure event with P(E) = 1, not selecting a specific number (which would be 1/20). The Reason is true: P(E) = 1 implies a sure event.

The Assertion is ambiguous but as stated — "probability of selecting a number at random from 1 to 20 is 1" — it appears to claim a specific selection has probability 1, which is false (it would be 1/20). However, if interpreted as selecting any number from the set, it is actually a sure event (P = 1). CBSE typically treats the Assertion here as false (implying a specific number), making option (D) correct. The Reason is directly from the textbook and is true. Focus on the distinction: sure event → P = 1; specific number → P = 1/20.

Option A: $\dfrac{3}{26}$

A deck has 52 cards. Red face cards = kings, queens, jacks of hearts and diamonds = 6 cards.
$$P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$$

Source: Chapter 14, Section 14.1

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• Face cards are kings, queens, and jacks only (3 per suit).
• Red suits are hearts and diamonds → 3 × 2 = 6 red face cards.
• Many students mistakenly count only one red suit or forget to simplify. Always simplify the fraction: 6/52 = 3/26.
• Option B (3/13 = 6/26) is double the correct answer — a common error from counting all 4 suits' face cards instead of only the 2 red ones.

(i)

Total balls = 30, blue balls = $m$

$$P(\text{blue}) = \frac{m}{30}$$

$$P(\text{not blue}) = 1 - \frac{m}{30} = \frac{30 - m}{30}$$

(ii)

After adding 6 blue balls: total balls = 36, blue balls = $m + 6$

$$P(\text{blue after adding}) = \frac{m+6}{36}$$

Given condition:

$$\frac{m+6}{36} = \frac{5}{4} \times \frac{m}{30}$$

$$\frac{m+6}{36} = \frac{m}{24}$$

$$24(m+6) = 36m$$

$$24m + 144 = 36m$$

$$12m = 144$$

$$\boxed{m = 12}$$

Source: Chapter 14, Section 14.1

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• (i) uses the complementary event formula: $P(\bar{E}) = 1 - P(E)$. Leave answer in terms of $m$.
• (ii) Set up the equation directly from the given condition, cross-multiply, and solve. Show every algebraic step — examiners award marks for working, not just the final answer. Verify: $P_1 = 12/30 = 2/5$; $P_2 = 18/36 = 1/2 = (5/4)(2/5)$ ✓

(A) — Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Since $P(E) \leq 1$, we need $0.2p \leq 1$, giving $p \leq 5$. This follows directly from the complement rule $P(\bar{E}) = 1 - P(E)$, which ensures $0 \leq P(E) \leq 1$.

The assertion is verified: $0.2p \leq 1 \Rightarrow p \leq 5$, so $p$ cannot exceed 5. The reason (complement rule) is the fundamental property that bounds all probabilities between 0 and 1, which is exactly why the assertion holds — making R the correct explanation of A. Choose option (A).

Option C: $\dfrac{1}{4}$

Total outcomes = 36. Numbers less than 4 are 1, 2, 3. Favourable outcomes = 3 × 3 = 9. P = $\dfrac{9}{36} = \dfrac{1}{4}$.

When two dice are rolled, total outcomes = 6 × 6 = 36. "Less than 4" means from {1, 2, 3} — 3 choices per die. Favourable pairs = 3 × 3 = 9. Many students mistakenly include 4; remember "less than 4" excludes 4 itself.