Question 1

Ordinary (non-leap) year picked uniformly at random. What is the probability that it contains 53 Sundays?

Accepted Answer

Introduction / Context:An ordinary year has 365 days, i.e., 52 weeks plus 1 extra day. Each weekday normally appears 52 times, and one weekday appears a 53rd time. We ask the probability that Sunday is the extra occurrence. Given Data / Assumptions: Ordinary year length = 52 weeks + 1 day. The extra day of the year is equally likely to be any of the seven weekdays for a uniformly random start day. We assume no bias in start-of-year weekdays. Concept / Approach:The only way to have 53 Sundays is for the extra day to land on a Sunday. Step-by-Step Solution:Extra day takes one of 7 weekdays with probability 1/7 each.Thus P(53 Sundays) = 1/7. Verification / Alternative check:A leap year (366 days) has two extra days, yielding 53 of two adjacent weekdays; but here the year is ordinary, so exactly one weekday reaches 53. Why Other Options Are Wrong:53/365 is not a valid probability model here; 2/7 would correspond to leap years, not ordinary years; other fractions do not reflect the uniform weekday assumption. Common Pitfalls:Confusing ordinary and leap years, or assuming the calendar always starts on a fixed weekday. Final Answer:1/7

Question 2

Single draw from a standard 52-card deck. What is the probability that the card is a face card (J, Q, or K of any suit)?

Accepted Answer

Introduction / Context:Face cards are Jacks, Queens, and Kings in each of the four suits. We compute the probability of drawing any face card in one draw from a well-shuffled standard deck. Given Data / Assumptions: Total cards = 52. Face cards per suit = 3 (J, Q, K). Total face cards = 4 × 3 = 12. Concept / Approach:Probability = favorable / total = 12 / 52, simplified by dividing numerator and denominator by 4. Step-by-Step Solution:Favorable = 12.Total = 52.Probability = 12/52 = 3/13. Verification / Alternative check:Counting per rank: there are 4 jacks, 4 queens, and 4 kings, summing to 12. Why Other Options Are Wrong:1/4 = 13/52 overcounts; 1/13 counts a single rank; 9/52 is not the count of any standard subset; 4/13 is too large. Common Pitfalls:Including Aces as face cards (they are not); miscounting the total number of face cards. Final Answer:3/13

Question 3

Single draw from a 52-card deck. What is the probability that the card is either red or a king (count red kings only once)?

Accepted Answer

Introduction / Context:The event “red or king” requires inclusion–exclusion to avoid double-counting the red kings. Compute the probability in one draw from a standard deck. Given Data / Assumptions: Red cards = 26 (hearts + diamonds). Kings = 4 total. Overlap (red kings) = 2 (K♥, K♦). Concept / Approach:Use P(A ∪ B) = P(A) + P(B) − P(A ∩ B) with card counts. Step-by-Step Solution:Count-based: favorable = 26 + 4 − 2 = 28.Probability = 28/52 = 7/13. Verification / Alternative check:Fraction approach: 1/2 + 1/13 − 1/26 = 7/13. Why Other Options Are Wrong:1/2 ignores kings of black suits; 6/13 or 27/52 misapply inclusion–exclusion; 5/13 undercounts. Common Pitfalls:Double-counting red kings; forgetting there are two red suits. Final Answer:7/13

Question 4

Single draw from a 52-card deck. What is the probability of drawing a king or a queen?

Accepted Answer

Introduction / Context:We seek the probability that a single draw yields either a king or a queen. Given Data / Assumptions: Kings = 4. Queens = 4. Total favorable = 8; deck size = 52. Concept / Approach:Events K and Q are disjoint in one draw, so P(K ∪ Q) = P(K) + P(Q) = 4/52 + 4/52. Step-by-Step Solution:Probability = 8/52 = 2/13. Verification / Alternative check:Simplify 8/52 by dividing by 4 to obtain 2/13. Why Other Options Are Wrong:1/26 counts only one rank; 1/13 is for a specific named rank; 3/26 is 6/52, undercounting. Common Pitfalls:Accidentally including jacks or aces; double-counting is not an issue here because the sets are disjoint. Final Answer:2/13

Question 5

Single draw from a 52-card deck. What is the probability that the card is neither a heart nor a king?

Accepted Answer

Introduction / Context:We want the complement of “heart or king”. Use inclusion–exclusion to find the union first, then subtract from 1. Given Data / Assumptions: Hearts = 13 cards. Kings = 4 cards. Overlap (king of hearts) = 1 card. Concept / Approach:P(neither) = 1 − P(heart ∪ king). Compute the union via counts. Step-by-Step Solution:Favorable for union = 13 + 4 − 1 = 16.P(heart ∪ king) = 16/52 = 4/13.Therefore P(neither) = 1 − 4/13 = 9/13. Verification / Alternative check:Direct count of “neither” = 52 − 16 = 36 → 36/52 = 9/13. Why Other Options Are Wrong:4/13 is the union, not the complement; 2/13 and 5/13 do not match the correct count. Common Pitfalls:Double-counting the king of hearts or forgetting to subtract overlap. Final Answer:9/13

Question 6

Seven white balls and three black balls are arranged randomly in a row. What is the probability that no two black balls are adjacent?

Accepted Answer

Introduction / Context:With indistinguishable balls of two colors, the number of distinct linear arrangements equals the number of ways to choose positions for the black balls among 10 slots. We require that no two black balls be adjacent. Given Data / Assumptions: Whites W = 7 (identical), blacks B = 3 (identical). Total arrangements (distinguishing only by color) = C(10, 3) by choosing positions for B's among 10 slots. Constraint: B's are non-adjacent. Concept / Approach:Place 7 W first, creating 8 gaps (including ends). To avoid adjacency, place each B in a distinct gap. Choose 3 of the 8 gaps to host one B each. Step-by-Step Solution:Favorable = C(8, 3) (choose which gaps receive a B).Total = C(10, 3) (choose any 3 positions for B among 10).Probability = C(8, 3) / C(10, 3) = 56 / 120 = 7/15. Verification / Alternative check:Compute combinations: C(8, 3) = 56; C(10, 3) = 120; reduce the fraction. Why Other Options Are Wrong:1/2 and 1/3 are rough guesses; 2/15 undercounts; 1/5 ignores the combinatorial structure. Common Pitfalls:Forgetting to include end gaps or attempting to “permute” identical balls. Final Answer:7/15

Question 7

Two fair dice are thrown. What is the probability that the sum is 3, 5, or 11 (any of the three totals)?

Accepted Answer

Introduction / Context:We aggregate favorable outcomes for three different target sums and divide by 36. Given Data / Assumptions: Two fair dice → 36 equally likely ordered pairs. Target sums: 3, 5, 11. Concept / Approach:Count outcomes for each sum and add them. Step-by-Step Solution:Sum 3: (1,2), (2,1) → 2 outcomes.Sum 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes.Sum 11: (5,6), (6,5) → 2 outcomes.Total favorable = 2 + 4 + 2 = 8.Probability = 8 / 36 = 2 / 9. Verification / Alternative check:Check that counts align with standard sum-distribution table for two dice. Why Other Options Are Wrong:5/36 applies to sum 8; 1/9 is 4/36 (too small); 19/36 is much too large; 7/36 undercounts by one outcome. Common Pitfalls:Missing symmetric pairs (e.g., both (2,3) and (3,2)). Final Answer:2/9

Question 8

Independent survival over 10 years. If P(man alive after 10 years) = 1/4 and P(wife alive after 10 years) = 1/3, what is P(neither alive after 10 years)?

Accepted Answer

Introduction / Context:Assuming independence of survival between spouses, the probability that neither is alive equals the product of their individual death probabilities. Given Data / Assumptions: P(man alive) = 1/4 ⇒ P(man dies) = 3/4. P(wife alive) = 1/3 ⇒ P(wife dies) = 2/3. Independence of outcomes. Concept / Approach:P(neither alive) = P(man dies) × P(wife dies) under independence. Step-by-Step Solution:P(neither) = (3/4) × (2/3) = 1/2. Verification / Alternative check:Complement: P(at least one alive) = 1 − 1/2 = 1/2. This equals P(man alive) + P(wife alive) − P(both alive) = 1/4 + 1/3 − (1/4 × 1/3) = 7/12 − 1/12 = 1/2. Why Other Options Are Wrong:1/12 is the probability both are alive; 3/4 and 7/12 are not the required joint death probability. Common Pitfalls:Confusing survival with death probabilities or assuming dependence without evidence. Final Answer:1/2

Question 9

Target practice with hit probability 0.3 per shot and 10 shots. What is the probability of at least one hit?

Accepted Answer

Introduction / Context:With independent trials, the probability of “at least one success” is one minus the probability of “no success at all.” Here, success = hit, p = 0.3 per shot, n = 10 shots. Given Data / Assumptions: Per-shot hit probability p = 0.3, miss probability q = 0.7. Number of shots n = 10 (independent). Concept / Approach:P(≥1 hit) = 1 − P(0 hits) = 1 − q^n with q = 0.7. Step-by-Step Solution:P(0 hits) = (0.7)^{10}.Therefore P(≥1 hit) = 1 − (0.7)^{10}. Verification / Alternative check:As p increases or n grows, the probability approaches 1, consistent with intuition. Why Other Options Are Wrong:(0.7)^10 is the probability of zero hits; (0.3)^10 is “all hits” if inverted (actually all-hit probability is (0.3)^{10}); 1 is only true in the limit, not exactly; 1 − (0.3)^10 is not the correct complement. Common Pitfalls:Confusing “at least one” with “exactly one,” or complementing the wrong event. Final Answer:1 - (0.7)^10

Question 10

Marksman with hit probability 1/5 per shot and 10 shots. What is the probability of at least one hit?

Accepted Answer

Introduction / Context:This is the same “at least one success” structure with p = 1/5 (hit) and q = 4/5 (miss), repeated over 10 independent trials. Given Data / Assumptions: Hit probability p = 1/5; miss probability q = 4/5. Number of shots n = 10; independence assumed. Concept / Approach:Complement rule: P(≥1 hit) = 1 − q^n = 1 − (4/5)^{10}. Step-by-Step Solution:Compute (4/5)^{10} if a numeric value is needed; algebraic form is acceptable. Verification / Alternative check:As n increases, (4/5)^n decreases, so the at-least-one probability increases, consistent with intuition. Why Other Options Are Wrong:(4/5)10 is missing an exponent; 1/(5)10 is not a probability expression; 1 − 1/(5)10 is unrelated; (1/5)10 would be all-hits if it had an exponent. Common Pitfalls:Dropping exponents or misreading (4/5)^10 as 4/(5×10). Final Answer:1 - (4/5)^10

Question 11

How many fair-coin tosses are required so that the probability of at least one head is at least 99%?

Accepted Answer

Introduction / Context:We seek the smallest n such that the probability of seeing at least one head in n fair-coin tosses is ≥ 0.99. Given Data / Assumptions: P(head) = 1/2 per toss; independence between tosses. Target probability ≥ 0.99 for “≥ 1 head”. Concept / Approach:P(≥1 head) = 1 − P(no heads) = 1 − (1/2)^n. Set 1 − (1/2)^n ≥ 0.99 and solve for the minimal integer n. Step-by-Step Solution:1 − (1/2)^n ≥ 0.99 ⇒ (1/2)^n ≤ 0.01.Take logs: n ≥ log(0.01) / log(0.5) ≈ 6.6439.Minimal integer n = 7. Verification / Alternative check:For n = 6: 1 − (1/2)^6 = 1 − 1/64 ≈ 0.9844 < 0.99. For n = 7: 1 − 1/128 ≈ 0.9922 ≥ 0.99. Why Other Options Are Wrong:5 or 6 give probabilities below 0.99; 8 or 9 work but are not minimal. Common Pitfalls:Rounding the logarithmic threshold the wrong way (must take the ceiling to ensure ≥ 0.99). Final Answer:7

Question 12

Integer chosen uniformly from 1 to 100. What is the probability that it is relatively prime to 100?

Accepted Answer

Introduction / Context:“Relatively prime to 100” means gcd(n, 100) = 1. Euler’s totient φ(100) counts such integers in {1, …, 100}. Given Data / Assumptions: 100 = 2^2 × 5^2. φ(n) = n × ∏(1 − 1/p) over distinct prime factors p of n. Concept / Approach:Compute φ(100) and divide by 100 to obtain the desired probability. Step-by-Step Solution:φ(100) = 100 × (1 − 1/2) × (1 − 1/5) = 100 × (1/2) × (4/5) = 40.Probability = φ(100)/100 = 40/100 = 2/5. Verification / Alternative check:Numbers divisible by 2 or 5 are not coprime to 100; inclusion–exclusion gives the same count. Why Other Options Are Wrong:61/100 overcounts; 17/100 and 19/100 undercount; 3/5 is the complement of 2/5 among multiples of 2 or 5, not the coprime share. Common Pitfalls:Confusing “not divisible by 2 or 5” with “coprime to 100” (these notions coincide here but must be handled carefully with inclusion–exclusion). Final Answer:2/5

Question 13

Arrange n distinct books in a row, of which m are volumes of one science series (m < n). What is the probability that those m volumes appear in ascending volume order (not necessarily consecutively)?

Accepted Answer

Introduction / Context:We randomly arrange n distinct books, of which m belong to a numbered series. We only care about the relative order among the m volumes; they need not be adjacent. What is the chance their left-to-right order is ascending by volume number? Given Data / Assumptions: All n! permutations are equally likely. The m special volumes are distinct and labeled (by volume numbers). We examine their relative order induced by any permutation. Concept / Approach:Among the m books of the series, all m! relative orders are equally likely when one looks only at their induced order in any full permutation. Exactly one of these m! orders is ascending. Step-by-Step Solution:Number of possible relative orders for the m volumes = m!.Favorable relative orders (ascending) = 1.Probability = 1 / m!. Verification / Alternative check:This is a standard symmetry argument: conditioning on the positions occupied by the m volumes, all m! permutations of those m among the chosen positions are equally likely. Why Other Options Are Wrong:m!/n! involves absolute positions, not just relative order; 1/(n−m)! is unrelated; 1/n! is far too small. Common Pitfalls:Assuming the volumes must be consecutive (not required). Only relative increasing order matters. Final Answer:1/m!

Question 14

n independent men, each of age x years. For a given man, the probability of dying within a years is p. What is the probability that a particular man M_k is the first to die (symmetry assumption for ties)?

Accepted Answer

Introduction / Context:Each of n men independently dies within a years with probability p and survives that period with probability (1 − p). We want the probability that a particular individual M_k is the first to die. Because the model is discrete over a fixed period, “first” is interpreted via symmetry: if at least one death occurs, all n individuals are equally likely to be the one identified as “first.” Given Data / Assumptions: Independence across individuals. P(death within a years) = p; P(survival) = 1 − p. At least one death event in the cohort has probability 1 − (1 − p)^n. Ties (multiple deaths) are assigned uniformly by symmetry as per standard exam convention. Concept / Approach:Condition on the event “there is at least one death.” Under symmetry among the n men, the chance that M_k is selected as the first equals 1/n given that at least one death happens. Unconditioning multiplies by the probability that at least one death occurs. Step-by-Step Solution:P(at least one death) = 1 − (1 − p)^n.By symmetry, P(M_k is the identified first | at least one death) = 1/n.Therefore P(M_k is first) = (1/n) × [1 − (1 − p)^n]. Verification / Alternative check:The n such probabilities sum to 1 − (1 − p)^n across all k, matching the total probability that someone dies, as required. Why Other Options Are Wrong:1 − (1 − p)^n is the probability that at least one death occurs (not tied to a specific person); 1/n^2 variants are dimensionally incorrect; p(1 − p)^{n−1} would be the probability that exactly one specific person dies and all others survive, not accounting for scenarios where multiple die. Common Pitfalls:Ignoring the conditional symmetry or misinterpreting “first” in a discrete-period setup. The adopted convention aligns with many exam problems. Final Answer:1/n[1 - (1 - p)^n]

Question 15

A machine has a daily failure probability of 0.95 (so its daily working probability is 0.05). What is the probability that it works for four consecutive days without failing?

Accepted Answer

Introduction / Context:This question tests independent-trial probability. If a device has a given chance to function (or fail) each day, the probability of a multi-day run is the product of day-wise probabilities, assuming independence from day to day. Given Data / Assumptions: Daily failure probability = 0.95. Hence daily working probability = 1 − 0.95 = 0.05. Days are independent (standard assumption in such problems). We require “works for four consecutive days”. Concept / Approach:For independent events A, B, C, D, P(A∩B∩C∩D) = P(A) * P(B) * P(C) * P(D). Here, each day has the same success probability p = 0.05. Step-by-Step Solution: Let p_work = 0.05.We need P(works 4 days) = (0.05)^4.Compute: 0.05^2 = 0.0025; then 0.05^4 = (0.0025)^2 = 0.00000625. Verification / Alternative check:Express 0.05 as 5 × 10^−2. Then (5 × 10^−2)^4 = 5^4 × 10^−8 = 625 × 10^−8 = 6.25 × 10^−6 = 0.00000625. Matches the direct computation. Why Other Options Are Wrong:0.0625 is (0.25)^2, unrelated here; 0.16548375 and 0.000625 come from incorrect powers or mixing failure vs success probability; “None of these” is not needed because the exact value appears among the options. Common Pitfalls:Using the failure probability instead of the working probability, or raising to the wrong power (e.g., three days) will yield incorrect values. Independence across days must also be assumed explicitly. Final Answer:0.00000625

Question 16

Seven white balls and three black balls are placed randomly in a row. What is the probability that no two black balls are adjacent?

Accepted Answer

Introduction / Context:This is a classic “no two alike adjacent” arrangement probability. Treat same-colored balls as indistinguishable, and count ways to position the black balls among the whites so that adjacency is avoided. Given Data / Assumptions: 7 white (W) balls, 3 black (B) balls. Balls of the same color are indistinguishable for counting positions. Random linear arrangement among 10 positions. Concept / Approach:First arrange the 7 W’s. They create 8 “gaps” (including ends). To avoid adjacency, place the 3 B’s in 3 distinct gaps. Total ways to choose 3 gaps from 8 is C(8,3). Total ways to place 3 B’s among 10 slots (without restriction) is choosing 3 positions for B out of 10: C(10,3). The desired probability is C(8,3) / C(10,3). Step-by-Step Solution: Number of safe gap choices = C(8,3) = 56.Total B-placement choices = C(10,3) = 120.Probability = 56 / 120 = 7 / 15. Verification / Alternative check:Direct small enumeration for a reduced case (e.g., 3 W and 2 B) gives identical “gap” logic, supporting the method. The gap method is a standard combinatorial approach for adjacency constraints. Why Other Options Are Wrong:1/2, 1/3, and 2/15 do not match the ratio of admissible to total placements; they stem from counting errors or treating balls as distinct when they are not. Common Pitfalls:Forgetting that black balls must occupy distinct gaps (placing two in the same gap creates adjacency), or incorrectly counting permutations of indistinguishable balls. Final Answer:7/15

Question 17

Two fair dice are thrown together. Find the probability that one die shows a multiple of 2 and the other die shows a multiple of 3 (order doesn’t matter).

Accepted Answer

Introduction / Context:We are asked for the probability that, in a single toss of two fair dice, the pair of outcomes has one value that is a multiple of 2 and the other value that is a multiple of 3. Order does not matter, so pairs like (6,6) should be counted once overall, not twice. Given Data / Assumptions: Each die has faces {1,2,3,4,5,6} with equal probability. Multiples: of 2 are {2,4,6}; of 3 are {3,6}. Sample space size = 36 ordered pairs. Concept / Approach:Use inclusion–exclusion over the two ordered scenarios: (Die1 multiple of 2 & Die2 multiple of 3) OR (Die1 multiple of 3 & Die2 multiple of 2). Subtract the overlap where both dice show a common multiple of 2 and 3, namely 6. Step-by-Step Solution: P(Die1 mult 2, Die2 mult 3) = (3/6)(2/6) = 1/6.P(Die1 mult 3, Die2 mult 2) = (2/6)(3/6) = 1/6.Overlap P(Die1=6, Die2=6) = (1/6)*(1/6) = 1/36.Total = 1/6 + 1/6 − 1/36 = 12/36 − 1/36 = 11/36. Verification / Alternative check:Counting favorable ordered pairs directly: {2,3},{2,6},{4,3},{4,6},{6,3},{6,6} and the symmetric ones where roles swap; careful de-duplication yields 11 favorable outcomes out of 36. Why Other Options Are Wrong:1/3 double-counts (6,6) without inclusion–exclusion. 1/4 and 7/26 come from ad-hoc approximations, not exact counting. Common Pitfalls:Double-counting (6,6) because it satisfies both role assignments; forgetting that order is part of the 36 equally likely ordered outcomes. Final Answer:11/36

Question 18

Two numbers are chosen without replacement from S = {1, 2, 3, 4, 5, 6}. What is the probability that the minimum of the two numbers is less than 4?

Accepted Answer

Introduction / Context:This is a complementary counting problem. Instead of directly counting outcomes where the minimum is less than 4, it is often easier to count the opposite event and subtract from 1. Given Data / Assumptions: Set S = {1,2,3,4,5,6}. Two distinct numbers are drawn without replacement. Total unordered pairs = C(6,2) = 15 (order does not matter for “minimum”). Concept / Approach:Let E be the event “min < 4”. Its complement E^c is “both numbers ≥ 4”. The subset {4,5,6} contains 3 numbers; choosing both from it gives pairs with minimum ≥ 4. Count pairs in E^c, then compute 1 − P(E^c). Step-by-Step Solution: Total pairs = C(6,2) = 15.Pairs with both ≥ 4: choose from {4,5,6} ⇒ C(3,2) = 3.P(E^c) = 3/15 = 1/5.Therefore P(E) = 1 − 1/5 = 4/5. Verification / Alternative check:Direct count for min < 4: any pair that includes at least one of {1,2,3}. Counting shows 12 out of 15 pairs, i.e., 4/5, consistent with the complement approach. Why Other Options Are Wrong:14/15 would require only one “bad” pair, but there are 3; 1/15 or 1/5 reflect counting the complement or partial cases only. Common Pitfalls:Confusing ordered vs unordered counting; the event depends only on the set of two numbers, not their order, so combinations are appropriate. Final Answer:4/5

Question 19

A bag has 5 blue and 4 black balls. Three balls are drawn at random without replacement. What is the probability that exactly two are blue and one is black?

Accepted Answer

Introduction / Context:This is a standard hypergeometric probability: sampling without replacement from a finite population containing two types (blue and black). We compute the probability of drawing a specified composition. Given Data / Assumptions: Total balls = 9 (5 blue, 4 black). Draw 3 balls without replacement. Desired outcome: exactly 2 blue and 1 black. Concept / Approach:Hypergeometric probability: P = [C(blue, needed_blue) * C(black, needed_black)] / C(total, draws). Here needed_blue = 2 and needed_black = 1. No order factor is needed because combinations already account for unordered draws. Step-by-Step Solution: Numerator = C(5,2) * C(4,1) = 10 * 4 = 40.Denominator = C(9,3) = 84.Probability = 40 / 84 = 10 / 21. Verification / Alternative check:Decimal check: 10/21 ≈ 0.4762. A quick simulation mindset (blue fraction ≈ 0.556) suggests the most likely triple roughly contains 2 blues, supporting plausibility of ≈0.48. Why Other Options Are Wrong:2/5 (=0.4) and 1/3 (=0.333…) underestimate because two-blue outcomes are relatively frequent given more blues than blacks; 1/6 is far too small. Common Pitfalls:Accidentally multiplying by permutations (overcounting), or using replacement formulas; also mixing numerator terms (e.g., C(4,2) instead of C(5,2)). Final Answer:10/21

Question 20

A six-faced die is biased so that an even number is twice as likely as an odd number. It is thrown twice. What is the probability that the sum of the two outcomes is even?

Accepted Answer

Introduction / Context:The die favors even results. The sum of two throws is even if both outcomes are even or both are odd. We must first compute the marginal probabilities of even and odd under the bias, then combine them for two throws (independent trials). Given Data / Assumptions: Even outcome is twice as likely as an odd outcome. There are 3 even faces (2,4,6) and 3 odd faces (1,3,5). All even faces share equal probability; all odd faces share equal probability. Throws are independent. Concept / Approach:Assign weight 2 to each even face and weight 1 to each odd face. Total weight = 3×2 + 3×1 = 9. Thus P(single even) = 6/9 = 2/3 and P(single odd) = 3/9 = 1/3. For two throws, P(sum even) = P(both even) + P(both odd) = (2/3)^2 + (1/3)^2. Step-by-Step Solution: P(E) = 2/3, P(O) = 1/3.P(sum even) = (2/3)^2 + (1/3)^2 = 4/9 + 1/9 = 5/9. Verification / Alternative check:Because the die is symmetric within parity classes, only parity probabilities matter, not specific face values; independence across throws validates the product rule. Why Other Options Are Wrong:1/12, 1/6, and 1/3 ignore the elevated chance of evens; the correct value exceeds 1/2 because evens are frequent. Common Pitfalls:Assigning double probability to the set of evens but forgetting there are three even faces (leading to mis-normalization). Final Answer:5/9

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