Question 1

A number is chosen uniformly at random from the first 120 natural numbers. What is the probability that it is a multiple of 5 or 15?

Accepted Answer

Introduction / Context:We must count numbers divisible by 5 or 15 among 1..120. Since “multiple of 15” is a subset of “multiple of 5,” the union simplifies to just “multiple of 5.” Given Data / Assumptions: Range: 1 to 120 inclusive. Multiples of 15 are included in multiples of 5. Concept / Approach:Count multiples of 5: floor(120/5). Divide by 120 to get probability. Step-by-Step Solution:Multiples of 5 in 1..120: 120/5 = 24 numbers.Probability = 24 / 120 = 1/5. Verification / Alternative check:Explicit list (5,10,...,120) has 24 terms; inclusion–exclusion not needed since 15-subset is contained within 5-multiples. Why Other Options Are Wrong:1/6 and 1/8 mismatch the correct count; “None of these” is false because 1/5 is available. Common Pitfalls:Double counting with inclusion–exclusion when one set is wholly contained in the other. Final Answer:1/5

Question 2

A bag contains 7 white and 9 red balls (total 16). One ball is drawn at random. What is the probability that it is white?

Accepted Answer

Introduction / Context:Single-draw colour problems require counting favourable outcomes (white balls) and dividing by the total number of balls, assuming all are equally likely to be drawn. Given Data / Assumptions: White = 7, Red = 9, Total = 16. Uniform random draw. Concept / Approach:P(white) = white/total = 7/16. Step-by-Step Solution:Favourable = 7 white balls.Total = 16 balls.Probability = 7/16. Verification / Alternative check:P(red) = 9/16; 7/16 + 9/16 = 1 confirms consistency. Why Other Options Are Wrong:1/16 and 7/52 confuse totals or a 52-card deck; 1/52 is irrelevant. Common Pitfalls:Transferring fractions from card problems to urn problems. Final Answer:7/16

Question 3

From a standard 52-card deck, what is the probability of drawing either the queen of clubs or the king of hearts?

Accepted Answer

Introduction / Context:We calculate the probability of drawing one of two distinct, non-overlapping single cards from a standard deck: the queen of clubs (unique) or the king of hearts (unique). Given Data / Assumptions: Standard deck: 52 cards, 4 suits, 13 ranks per suit. Queen of clubs (1 card) and king of hearts (1 card) are distinct events. Concept / Approach:Since the two favourable outcomes cannot occur simultaneously on a single draw, add their probabilities: 1/52 + 1/52. Step-by-Step Solution:P(Q♣) = 1/52.P(K♥) = 1/52.P(Q♣ or K♥) = 1/52 + 1/52 = 2/52 = 1/26. Verification / Alternative check:Counting favourable cards directly: 2 out of 52 → 2/52 → 1/26. Why Other Options Are Wrong:1/52 counts only one specific card; 1/13 equals probability of any one rank in a suit or any one suit for a rank, not this combined two-card event. Common Pitfalls:Accidentally counting extra queens or kings; here only the specified two single cards are favourable. Final Answer:1/26

Question 4

From the five English vowels {a, e, i, o, u}, one vowel is selected uniformly at random. What is the probability that it is the letter i?

Accepted Answer

Introduction / Context:A uniform choice from a set with equally likely outcomes yields a probability equal to 1 divided by the set size for any particular specified element. Given Data / Assumptions: Vowel set: {a, e, i, o, u}. Each vowel is equally likely. Concept / Approach:P(chosen vowel is i) = 1 / 5 because there is exactly one favourable element among five equally likely vowels. Step-by-Step Solution:Total outcomes = 5.Favourable outcomes (letter i) = 1.Probability = 1/5. Verification / Alternative check:Symmetry: each vowel has identical probability → each 1/5; summing five such equal probabilities gives 1. Why Other Options Are Wrong:4/5 would be “not i”; 1/2 and 1/3 do not match the equally likely 5-way split. Common Pitfalls:Overthinking the selection—this is a direct uniform-choice scenario. Final Answer:1/5

Question 5

If P(A lives 15 years) = 7/8 and P(B lives 15 years) = 9/10, assuming independence, what is the probability that both A and B are alive after 15 years?

Accepted Answer

Introduction / Context:For independent events, the probability that both occur equals the product of their individual probabilities. Survival-type questions often include this assumption unless stated otherwise. Given Data / Assumptions: P(A survives 15 years) = 7/8. P(B survives 15 years) = 9/10. Independence of the two survival events (as typical in basic probability exercises). Concept / Approach:P(A ∩ B) = P(A) * P(B) for independent A and B. Step-by-Step Solution:P = (7/8) * (9/10) = 63/80. Verification / Alternative check:63/80 = 0.7875 lies between individual probabilities (0.875 and 0.9), as expected for a conjunction. Why Other Options Are Wrong:1/20 and 1/5 are far too small; “None of these” is incorrect since 63/80 is present. Common Pitfalls:Adding probabilities instead of multiplying for independent “both happen” scenarios. Final Answer:63/80

Question 6

From the word PROBABILITY (11 letters), one letter is chosen uniformly at random. What is the probability that the chosen letter is a vowel?

Accepted Answer

Introduction / Context:We select one character uniformly from a word and ask for the chance it is a vowel. Repetitions matter in counting because each occurrence is a distinct favourable outcome. Given Data / Assumptions: Word: PROBABILITY. Total letters = 11. Vowels among English letters: a, e, i, o, u. Concept / Approach:Count total characters and the number of vowel occurrences (including duplicates). Probability = vowel count / 11. Step-by-Step Solution:Letters with multiplicities: P(1), R(1), O(1), B(2), A(1), I(2), L(1), T(1), Y(1).Vowels present (with counts): A(1), O(1), I(2) → 4 vowels.Probability = 4/11. Verification / Alternative check:Consonant count = 7; 4 + 7 = 11 confirms total. Why Other Options Are Wrong:3/11 or 2/11 undercount vowel multiplicities; 0 ignores vowels entirely. Common Pitfalls:Counting distinct vowel types instead of occurrences; duplicates (I appears twice) must be included. Final Answer:4/11

Question 7

From a standard 52-card deck, what is the probability of drawing a red card?

Accepted Answer

Introduction / Context:A standard deck has equal numbers of red and black cards, making this a symmetry-based probability. Given Data / Assumptions: Standard deck: 52 cards. Red suits: hearts and diamonds (26 cards total). Concept / Approach:P(red) = number of red cards / total cards = 26 / 52. Step-by-Step Solution:Probability = 26/52 = 1/2. Verification / Alternative check:P(black) = 26/52 = 1/2; the two are complementary and sum to 1. Why Other Options Are Wrong:1/13 corresponds to a specific rank in a suit; 1/4 is a single suit; 2/18 is nonsensical for a 52-card deck. Common Pitfalls:Confusing suit probabilities (1/4) with colour probabilities (1/2). Final Answer:1/2

Question 8

When an unbiased six-faced die is tossed, what is the probability that the outcome is a composite number?

Accepted Answer

Introduction / Context:The sample space for a single fair die roll is {1,2,3,4,5,6}. We classify which outcomes are composite and compute the probability accordingly. Given Data / Assumptions: Composite numbers have more than two positive divisors. 1 is neither prime nor composite. Concept / Approach:Identify composite outcomes, count them, and divide by 6. Step-by-Step Solution:Outcomes: 1,2,3,4,5,6.Composite among these: 4 (2*2) and 6 (2*3).Count = 2; probability = 2/6 = 1/3. Verification / Alternative check:Prime outcomes: 2,3,5 (3 outcomes). 3 primes + 2 composites + 1 neither = 6 total. Why Other Options Are Wrong:1/2 would require 3 composites; 1/4 would require only 1 composite; 1 is impossible for a proper subset event. Common Pitfalls:Including 1 as composite or missing that 4 and 6 are the only composites on a standard die. Final Answer:1/3

Question 9

Two fair dice are rolled simultaneously. What is the probability that the sum of the numbers shown is 11?

Accepted Answer

Introduction / Context:When two dice are rolled, there are 36 equally likely ordered outcomes. We identify those with sum 11 and divide by 36. Given Data / Assumptions: Ordered pairs (d1, d2) with d1, d2 ∈ {1..6}. Each pair equally likely with probability 1/36. Concept / Approach:Find all pairs (x, y) such that x + y = 11. Step-by-Step Solution:Pairs summing to 11: (5,6) and (6,5) → 2 favourable outcomes.Probability = 2/36 = 1/18. Verification / Alternative check:Neighbouring sums: 10 has 3 pairs; 12 has 1 pair. The count for 11 (=2) fits the symmetric distribution of sums. Why Other Options Are Wrong:1/6 implies 6 favourable outcomes; 1/9 implies 4; 1 is impossible for a specific sum event. Common Pitfalls:Forgetting ordered pairs are distinct; (5,6) and (6,5) are different outcomes. Final Answer:1/18

Question 10

Tickets numbered 1 to 20 are mixed and one ticket is drawn at random. What is the probability that the number on the ticket is a multiple of 3 or 7?

Accepted Answer

Introduction / Context:We need the probability that the chosen number is in the union of two sets: multiples of 3 and multiples of 7. Inclusion–exclusion prevents double counting when overlaps exist. Given Data / Assumptions: Range: 1..20 inclusive. Multiples of 3: ⌊20/3⌋ = 6. Multiples of 7: ⌊20/7⌋ = 2. LCM(3,7) = 21 → overlap count in 1..20 is 0. Concept / Approach:Count = count(3) + count(7) − count(21). Probability = count / 20. Step-by-Step Solution:Count(3) = 6 (3,6,9,12,15,18).Count(7) = 2 (7,14).Overlap = 0; total favourable = 6 + 2 = 8.Probability = 8/20 = 2/5. Verification / Alternative check:Listing confirms 8 valid numbers; fraction reduces to 2/5. Why Other Options Are Wrong:1/2 would require 10 favourable numbers; 7/20 equals 7/20 = 0.35 (undercount). Common Pitfalls:Forgetting to apply inclusion–exclusion when overlaps are possible; here the overlap happens at 21 which is outside the range, so it is 0. Final Answer:2/5

Question 11

If the odds against an event are 5:4, what is the probability that the event occurs?

Accepted Answer

Introduction / Context:“Odds against” a:b means failure:success = a:b. We convert this ratio into a probability of success by dividing the success count by the total a + b. Given Data / Assumptions: Odds against = 5:4. Concept / Approach:If odds against are a:b, then P(success) = b / (a + b). We simply plug in a = 5, b = 4. Step-by-Step Solution:P = 4 / (5 + 4) = 4/9. Verification / Alternative check:Corresponding odds in favour would be 4:5, giving P = 4/(4+5) = 4/9, consistent. Why Other Options Are Wrong:4/5 and 1/4 invert relationships; 1/5 uses only the “against” numerator as probability. Common Pitfalls:Confusing odds with probabilities; always divide by total a + b. Final Answer:4/9

Question 12

A bag contains 8 red and 5 white balls (13 total). Two balls are drawn without replacement. What is the probability that both balls drawn are white?

Accepted Answer

Introduction / Context:For draws without replacement, probabilities change after each draw. The exact probability that both draws are white can be computed by sequential multiplication or combinatorics. Given Data / Assumptions: Red = 8, White = 5, Total = 13. Two draws without replacement. Concept / Approach:Method 1 (sequential): P(white first) * P(white second | white first). Method 2 (combinatorial): C(5,2) / C(13,2). Step-by-Step Solution (combinatorial):Favourable ways = C(5,2) = 10.Total ways = C(13,2) = 78.Probability = 10/78 = 5/39. Verification / Alternative check (sequential):P = (5/13) * (4/12) = 20/156 = 5/39, same result. Why Other Options Are Wrong:5/16 assumes replacement or independence; 2/13 and 3/26 do not match either correct computation. Common Pitfalls:Treating draws as independent when they are without replacement; forgetting to reduce the denominator after the first draw. Final Answer:5/39

Question 13

A bag contains 6 black and 8 white balls (14 total). One ball is drawn at random. What is the probability that it is white?

Accepted Answer

Introduction / Context:Single-draw colour probability with equally likely balls is computed by favourable/total. Given Data / Assumptions: White = 8, Black = 6, Total = 14. Concept / Approach:P(white) = 8/14, then reduce the fraction. Step-by-Step Solution:P = 8/14 = 4/7 after dividing by 2. Verification / Alternative check:P(black) = 6/14 = 3/7; 4/7 + 3/7 = 1 confirms complementarity. Why Other Options Are Wrong:3/4 and 4/5 overstate the probability; 1/8 confuses with the inverse count. Common Pitfalls:Not simplifying the fraction; mixing counts for colours. Final Answer:4/7

Question 14

Three fair coins are tossed simultaneously. What is the probability of getting at most two heads?

Accepted Answer

Introduction / Context:“At most two heads” means 0, 1, or 2 heads. It is often fastest to use the complement: 1 minus the probability of 3 heads. Given Data / Assumptions: Three independent fair coin tosses. Each toss: P(H) = 1/2. Concept / Approach:P(at most 2H) = 1 − P(3H). Since P(3H) = (1/2)^3, subtract from 1. Step-by-Step Solution:P(3 heads) = (1/2)^3 = 1/8.Therefore P(at most two heads) = 1 − 1/8 = 7/8. Verification / Alternative check (enumeration):Outcomes with ≤2 heads: 7 out of 8 ({TTT, HTT, THT, TTH, HHT, HTH, THH}). Only HHH is excluded. Why Other Options Are Wrong:1/4 and 3/8 undercount; 1/2 is the probability of exactly 1 or exactly 2 heads combined for two coins, not three. Common Pitfalls:Counting mistakes when listing outcomes; using “at least” instead of “at most”. Final Answer:7/8

Question 15

Three unbiased coins are tossed. What is the probability of obtaining exactly two heads (and therefore one tail) among the three outcomes?

Accepted Answer

Introduction / Context:When fair coins are tossed, each outcome in the sample space is equally likely. Here, three unbiased coins are tossed, and we seek the probability of getting exactly two heads (and one tail). Given Data / Assumptions: Number of coins = 3. Each coin is fair: P(H) = 1/2, P(T) = 1/2. All 2^3 outcomes are equiprobable. We require exactly two heads. Concept / Approach:Use counting with combinations. The number of outcomes with exactly k successes (heads) in n independent Bernoulli trials with p = 1/2 is C(n, k). The probability is C(n, k) / 2^n for fair coins. Step-by-Step Solution:Total outcomes = 2^3 = 8.Favorable outcomes for exactly two heads = C(3, 2) = 3 (e.g., HHT, HTH, THH).Probability = 3 / 8. Verification / Alternative check:Enumerate all eight outcomes and count the three with two heads to confirm. Why Other Options Are Wrong:1/3 and 2/3 confuse counts with probabilities; 3/4 corresponds to “at least two heads” (not correct here); 1/4 corresponds to “all heads” (HHH). Common Pitfalls:Forgetting that order matters when counting sequences, but combinations already capture how many distinct placements of heads exist. Final Answer:3/8

Question 16

Two unbiased coins are tossed simultaneously. What is the probability of getting at least one head?

Accepted Answer

Introduction / Context:“At least one” events are conveniently handled using complements. For two fair coins, we want the probability of at least one head in the pair of tosses. Given Data / Assumptions: Two unbiased coins; outcomes are HH, HT, TH, TT. Each outcome has probability 1/4. Event: At least one head occurs. Concept / Approach:P(at least one head) = 1 − P(no heads). “No heads” means both coins show tails, i.e., TT. Step-by-Step Solution:P(TT) = 1/4.Therefore P(at least one head) = 1 − 1/4 = 3/4. Verification / Alternative check:Direct counting: favorable outcomes are HH, HT, TH — 3 outcomes out of 4 gives 3/4. Why Other Options Are Wrong:1/2 and 2/3 underestimate; 1/3 is the probability of exactly one tail, not the target event; 1/4 is P(no heads), i.e., the complement. Common Pitfalls:Confusing “at least one head” with “exactly one head”. The former includes HH as well. Final Answer:3/4

Question 17

Single unbiased coin is tossed once. What is the probability of obtaining a head?

Accepted Answer

Introduction / Context:An unbiased (fair) coin has equally likely faces: head (H) and tail (T). We want the probability of a head on a single toss. Given Data / Assumptions: One toss of a fair coin. Sample space: {H, T}. Each outcome has probability 1/2. Concept / Approach:For a fair coin, symmetry implies P(H) = P(T) = 1/2. No further computation is required. Step-by-Step Solution:P(H) = 1/2. Verification / Alternative check:Long-run frequency under many tosses approaches 1/2 for heads (law of large numbers) for a fair coin. Why Other Options Are Wrong:1 implies certainty; 1/4 is for two successive events with probability 1/2 each; “None of these” is incorrect because 1/2 is a valid option; 0 is impossible for a fair coin. Common Pitfalls:Overthinking a basic fairness assumption; unless otherwise stated, “unbiased” means equal probabilities. Final Answer:1/2

Question 18

Surprise test on one of five class meetings; student absent twice. Assuming exactly one test is equally likely on any of the five meetings, what is the probability the student misses the test if they are absent on two meetings?

Accepted Answer

Introduction / Context:The original wording is ambiguous. By the Recovery-First Policy, we adopt the standard model: there is exactly one surprise test during the week, and each of the five class meetings is equally likely to host it. A student is absent on two of those five meetings. We want the probability they miss the (only) test. Given Data / Assumptions (clarified): Exactly one test in the week. The test day is uniformly random among 5 meetings. The student is absent on exactly 2 specific meetings. Concept / Approach:The event “misses the test” occurs iff the test falls on one of the 2 absent days. With a uniform choice among 5 days, the probability is favorable days / total days. Step-by-Step Solution:Favorable = 2 days (absences).Total = 5 days.Probability = 2 / 5. Verification / Alternative check:Complement: Probability of not missing the test (i.e., test on one of the 3 present days) is 3/5. Then 1 − 3/5 = 2/5, consistent. Why Other Options Are Wrong:4/15, 1/15, 16/125 arise from different (independent-per-day) interpretations not implied after clarification; 3/5 is the complement. Common Pitfalls:Mixing the “exactly one test” model with a “test happens with probability 1/5 independently each day.” We use the standard competitive-exam model of a single, uniformly placed surprise test. Final Answer:2/5

Question 19

Two successive tosses of a fair coin. Find the probabilities of (1) exactly one head and (2) at least one head (express as ordered pair).

Accepted Answer

Introduction / Context:With two independent fair coin tosses, we compute probabilities for two events: exactly one head, and at least one head. Given Data / Assumptions: Outcomes: HH, HT, TH, TT with equal probability 1/4 each. Event A: exactly one head (HT or TH). Event B: at least one head (HH, HT, TH). Concept / Approach:Count favorable outcomes for each event and divide by 4 (total outcomes). Use complement for “at least one head” if desired. Step-by-Step Solution:P(exactly one head) = 2 / 4 = 1/2.P(at least one head) = 1 − P(TT) = 1 − 1/4 = 3/4. Verification / Alternative check:Direct counting confirms 2 favorable for A and 3 favorable for B. Why Other Options Are Wrong:Other ordered pairs do not match the correct counts; 2/3 or 4/5 are not attainable with a four-element uniform sample space in this context. Common Pitfalls:Including HH in “exactly one head” or excluding it from “at least one head”. Final Answer:1/2 , 3/4

Question 20

Two standard dice are thrown simultaneously. What is the probability that the sum of the two upper faces equals 8?

Accepted Answer

Introduction / Context:Rolling two fair six-sided dice yields 36 equally likely ordered pairs. We want the chance that the sum is exactly 8. Given Data / Assumptions: Die outcomes are 1–6. Sample space size = 6 × 6 = 36. Sum target = 8. Concept / Approach:Enumerate the pairs whose coordinates sum to 8 and divide by 36. Step-by-Step Solution:Favorable pairs: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes.Probability = 5 / 36. Verification / Alternative check:Check neighboring sums: sums of 7 have 6 outcomes (symmetric peak), sums of 8 have 5, consistent with standard sum distribution. Why Other Options Are Wrong:2/9 ≈ 8/36 (too large); 1/6 = 6/36 (wrong count); “Data Inadequate” is inappropriate since the model is standard. Common Pitfalls:Forgetting that ordered pairs (3,5) and (5,3) are distinct outcomes. Final Answer:5/36

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