Question 1

P(A) = 5/9 and P(B^c) = 5/11 (so P(B) = 6/11). Assuming A and B are independent, what is the probability that at least one of A or B occurs?

Accepted Answer

Introduction / Context:The probability that at least one of two events occurs is given by P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If independence is assumed, then P(A ∩ B) = P(A)P(B). This question provides P(A) and P(B^c). Given Data / Assumptions: P(A) = 5/9. P(B^c) = 5/11, hence P(B) = 1 − 5/11 = 6/11. A and B are independent (stated in the title; needed to determine P(A ∩ B)). Concept / Approach:Use the union formula with independence: P(A ∪ B) = P(A) + P(B) − P(A)P(B). Compute exactly as a rational number to avoid rounding. Step-by-Step Solution: P(A ∪ B) = 5/9 + 6/11 − (5/9)(6/11).Common denominator 99: 55/99 + 54/99 − 30/99 = 79/99.Therefore, P(at least one) = 79/99. Verification / Alternative check:Compute P(neither) = P(A^c)P(B^c) = (4/9)(5/11) = 20/99. Then 1 − 20/99 = 79/99, confirming the result. Why Other Options Are Wrong:6/11 and 5/9 are individual event probabilities, not the union. 4/9 is P(A^c). 0.8 is a rounded approximation (79/99 ≈ 0.79798) and is not exact. Common Pitfalls:Forgetting to subtract the intersection or neglecting the independence assumption; adding probabilities directly can exceed 1. Final Answer:79/99

Question 2

Two fair dice are thrown. What is the probability that the absolute difference between the two numbers is 2 or 3?

Accepted Answer

Introduction / Context:We count ordered pairs (a,b) with a,b ∈ {1,…,6} such that |a − b| ∈ {2,3}. Counting favorable outcomes directly is efficient for small dice problems. Given Data / Assumptions: Sample space size = 36 equally likely ordered pairs. We need |a − b| = 2 or 3. Concept / Approach:Enumerate the pairs for |a − b| = 2 and for |a − b| = 3. Sum them, divide by 36. Step-by-Step Solution: |a − b| = 2 pairs: (1,3),(3,1),(2,4),(4,2),(3,5),(5,3),(4,6),(6,4) ⇒ 8.|a − b| = 3 pairs: (1,4),(4,1),(2,5),(5,2),(3,6),(6,3) ⇒ 6.Total favorable = 8 + 6 = 14; probability = 14/36 = 7/18. Verification / Alternative check:Symmetry around the main diagonal implies for each a there are at most two b values satisfying |a − b| = k; tabulation matches the counts above. Why Other Options Are Wrong:5/18 corresponds to only one of the difference values; 1/2 greatly overestimates the event; 3/11 is not tied to a 36-sized space. Common Pitfalls:Forgetting that outcomes are ordered, or double-counting/omitting symmetric pairs. Final Answer:7/18

Question 3

From cards numbered 1 to 17, one card is drawn at random. What is the probability that the number is divisible by 3 or by 7?

Accepted Answer

Introduction / Context:We apply inclusion–exclusion for divisibility conditions over a finite range. Count multiples of 3, multiples of 7, then subtract overlaps (if any). Given Data / Assumptions: Numbers 1,2,…,17 (uniformly likely). Event: divisible by 3 or by 7. Concept / Approach:Count: ⌊17/3⌋ = 5 (3,6,9,12,15), ⌊17/7⌋ = 2 (7,14), and overlap ⌊17/21⌋ = 0 since 21 > 17. Total favorable = 5 + 2 = 7, total outcomes 17. Step-by-Step Solution: Favorable = 7.Total = 17.Probability = 7/17. Verification / Alternative check:Explicit list confirms the count: {3,6,7,9,12,14,15} has size 7. Why Other Options Are Wrong:10/17 would require 10 favorable numbers; 2/17 or 1/7 are not supported by counts. Common Pitfalls:Forgetting to subtract overlaps (not applicable here but critical in general) or miscounting the edge near 17. Final Answer:7/17

Question 4

One card is drawn at random from 100 cards numbered 1 to 100. What is the probability that the number is a perfect square?

Accepted Answer

Introduction / Context:Perfect squares ≤ 100 occur at n^2 for n = 1,…,10. The sample space contains 100 equally likely outcomes. We count and divide. Given Data / Assumptions: Numbers 1 to 100 equiprobable. Squares: 1,4,9,16,25,36,49,64,81,100. Concept / Approach:Count favorable outcomes, then compute probability = favorable / total. Step-by-Step Solution: Favorable squares = 10.Total outcomes = 100.Probability = 10/100 = 1/10. Verification / Alternative check:Because 100 = 10^2, there are exactly 10 perfect squares up to 100. The list confirms the count. Why Other Options Are Wrong:1/5 (= 0.2) and 2/5 (= 0.4) overestimate; “None of these” is unnecessary as 1/10 is exact. Common Pitfalls:Including 0 (not in range) or omitting 100 (which is 10^2). Final Answer:1/10

Question 5

In shuffling a standard deck, three cards are accidentally dropped. What is the probability that the three missing cards are all of different suits?

Accepted Answer

Introduction / Context:We draw 3 cards from 52 without order. We want all three to come from three distinct suits. Counting can be done by choosing suits first and then one card from each suit. Given Data / Assumptions: Standard deck: 4 suits, 13 cards per suit. Drawn cards are unordered (combinations). Concept / Approach:Favorable outcomes: choose 3 suits out of 4 (C(4,3)) and choose 1 card from each chosen suit (13 choices independently per suit). Total outcomes: choose any 3 cards (C(52,3)). The ratio gives the probability. Step-by-Step Solution: Favorable = C(4,3) * 13^3 = 4 * 2197 = 8788.Total = C(52,3) = 525150/6 = 22100.Probability = 8788 / 22100 = (13^2)/(425) = 169/425. Verification / Alternative check:Reduction: 8788/22100 divide numerator and denominator by 52? Instead, observe 8788 = 413^3 and 22100 = 42552; simplifying yields 169/425 exactly. Why Other Options Are Wrong:261/425 and 104/425 correspond to different suit-pattern counts (allowing repetitions); “None of these” is unnecessary as the exact simplified fraction is available. Common Pitfalls:Overcounting by permuting the three cards even though combinations are used; or forcing an order among suits (unnecessary), which would multiply by 3! incorrectly. Final Answer:169/425

Question 6

From a standard 52-card deck, what is the probability that a randomly drawn card is a diamond or a king?

Accepted Answer

Introduction / Context:We compute a simple union of two card events: “diamond” or “king”. Since one card (king of diamonds) lies in both sets, inclusion–exclusion is needed to avoid double-counting. Given Data / Assumptions: Diamonds = 13 cards. Kings = 4 cards. Overlap = 1 card (king of diamonds). Total cards = 52. Concept / Approach:P(diamond ∪ king) = (13 + 4 − 1)/52 = 16/52. Reduce the fraction to simplest terms. Step-by-Step Solution: Favorable = 16.Probability = 16/52 = 4/13. Verification / Alternative check:Complementary event is “not diamond and not king” = 39/52; thus union is 1 − 39/52 = 13/52? Careful: “not diamond and not king” equals 52 − (diamonds ∪ kings) = 36; hence union is 16/52 = 4/13, matching above. Why Other Options Are Wrong:4/52 is only diamonds OR only kings if miscounted; 2/13 undercounts; 1/52 is just a single card. Common Pitfalls:Double-counting the king of diamonds by simply adding 13 + 4 without subtracting the overlap. Final Answer:4/13

Question 7

Four machines are known to contain exactly two faulty ones. Machines are tested one by one in random order until both faulty machines are identified. What is the probability that only two tests are needed?

Accepted Answer

Introduction / Context:We randomly order four machines containing exactly two faulty units. We test until both faulty ones are found. “Only two tests” means the first two tested are precisely the two faulty machines (in any order). Given Data / Assumptions: Total machines: 4, faulty: 2, good: 2. Random order of testing; each permutation equally likely. Success event: the first two positions are the two faulty machines. Concept / Approach:Compute the probability that position 1 is faulty and position 2 is faulty given the first. Multiply sequential conditional probabilities. Step-by-Step Solution: P(first is faulty) = 2/4 = 1/2.Given first faulty, remaining: 1 faulty out of 3. P(second faulty | first faulty) = 1/3.Therefore P(both first two faulty) = (1/2) * (1/3) = 1/6. Verification / Alternative check:Permutation counting: number of orders with the two faulty first equals 2! * 2! (arrangements of faulty then good) out of 4! total? More directly, probability that a specific 2-set occupies the first two slots is C(2,2)C(2,0)/C(4,2) = 1/6; same result. Why Other Options Are Wrong:1/3, 1/2, 1/4 mis-handle sequential conditioning or overcount permutations. Common Pitfalls:Assuming independence between first and second picks; the second probability is conditional because sampling is without replacement. Final Answer:1/6

Question 8

Let E be the set of integers whose unit digit is 1. If a number is chosen uniformly from {2, 3, 4, …, 50}, what is the probability that it belongs to E?

Accepted Answer

Introduction / Context:We count how many integers between 2 and 50 inclusive end with digit 1. Divide by the total count to obtain the probability. Given Data / Assumptions: Range: {2,3,…,50} has 49 numbers. Numbers with unit digit 1 in this range: 11, 21, 31, 41. Concept / Approach:Simple counting; all outcomes equally likely. Step-by-Step Solution: Favorable = 4 (11, 21, 31, 41).Total = 49.Probability = 4/49. Verification / Alternative check:Note that 1 and 51 lie just outside the range; hence there are exactly four valid entries spaced by 10. Why Other Options Are Wrong:5/49, 3/49, 2/49 are inconsistent with the explicit list. Common Pitfalls:Accidentally including 1 or 51, which are not in the specified set. Final Answer:4/49

Question 9

Three students are picked at random from a school with 1000 students. What is the probability that these three have the same birthday (same day and month)? Assume 365 equally likely birthdays and independence.

Accepted Answer

Introduction / Context:This is a variant of the birthday problem. We want the probability that all three selected students share the same day and month of birth. The school size (1000) only ensures that sampling without replacement effects are negligible compared to assuming independent birthdays across students. Given Data / Assumptions: 365 possible birthdays (ignoring leap day), equally likely. Three students chosen; independence of birthdays. Concept / Approach:Fix the first student’s birthday arbitrarily. For the other two to match, each must equal the first student’s day and month. Therefore multiply the two independent matching probabilities: (1/365)·(1/365) = 1/365^2. Step-by-Step Solution: Let student 1 have any date (probability 1).P(student 2 matches) = 1/365.P(student 3 matches) = 1/365 (independent of student 2).Thus P(all three same) = 1/365^2. Verification / Alternative check:If we insisted on unordered counting across 365 categories, this matches the multinomial perspective: the only favorable case has all three in the same category, giving the same probability. Why Other Options Are Wrong:3/365 or 3/1000 are unrelated heuristics; 1/365 is the two-student match probability, not three. Common Pitfalls:Accidentally computing the probability that at least two match; or using 366 (leap year) when the question specifies a non-leap model. Final Answer:1/(365)^2

Question 10

You correctly dial the first four digits of a 7-digit telephone number but have forgotten the last three. If you dial the last three digits at random, what is the probability of dialing the correct number on a single attempt?

Accepted Answer

Introduction / Context:The last three digits are unknown and assumed equally likely across 000–999. A single random attempt has success probability equal to 1 divided by the number of possibilities. Given Data / Assumptions: Three unknown digits, each 0–9. Total combinations = 10^3 = 1000. Exactly one correct triple. Concept / Approach:Uniform discrete probability over 1000 equally likely outcomes, with one favorable outcome. Step-by-Step Solution: Total possibilities = 1000.Favorable = 1.Probability = 1/1000. Verification / Alternative check:Whether the digits can begin with 0 does not change the count; most telephone numbering schemes allow any 3-digit tail in this abstract setting. Why Other Options Are Wrong:1/999, 1/990, 1/1001 are based on ad-hoc exclusions or inclusions that the problem statement does not impose. Common Pitfalls:Assuming digits cannot be zero or must be distinct; no such constraints are given. Final Answer:1/1000

Question 11

In a non-leap year (365 days), what is the probability that the year contains exactly 52 Tuesdays (i.e., not 53 Tuesdays)?

Accepted Answer

Introduction / Context:A non-leap year has 365 days = 52 full weeks + 1 extra day. Thus each weekday occurs 52 times, and one weekday (the weekday of January 1) occurs a 53rd time. We need exactly 52 Tuesdays, so the extra day must NOT be Tuesday. Given Data / Assumptions: Non-leap year: 365 days. Exactly one extra day beyond 52 weeks. Start weekday is equally likely among 7 days (assuming a random model across years). Concept / Approach:There are 7 equally likely choices for which weekday occurs 53 times. Exactly 52 Tuesdays means Tuesday is not the extra weekday. Therefore probability = 6/7. Step-by-Step Solution: Total possible “extra weekdays” = 7.Favorable choices (extra day is not Tuesday) = 6.Probability = 6/7. Verification / Alternative check:If the extra day is Tuesday, there are 53 Tuesdays; in the other six cases there are exactly 52 Tuesdays. Hence 6 of 7 outcomes are favorable. Why Other Options Are Wrong:1/7 is the probability of 53 Tuesdays; 2/7 and 3/7 are not consistent with the single extra-day structure. Common Pitfalls:Confusing “at least 52” with “exactly 52”, or applying leap-year logic (366 days) by mistake. Final Answer:6/7

Question 12

Two cards are drawn together at random from a 52-card deck. What is the probability that both cards are kings?

Accepted Answer

Introduction / Context:We draw two cards without replacement. The number of ways to choose two kings from four over the number of ways to choose any two cards yields the probability. Given Data / Assumptions: Deck: 52 cards; kings: 4. Draw two cards simultaneously (order irrelevant). Concept / Approach:Use combinations: favorable = C(4,2); total = C(52,2). The ratio simplifies to a neat reduced fraction. Step-by-Step Solution: Favorable = C(4,2) = 6.Total = C(52,2) = 1326.Probability = 6 / 1326 = 1 / 221. Verification / Alternative check:Sequential multiplication: P(first king) = 4/52, P(second king | first king) = 3/51 ⇒ product = 12/2652 = 1/221. Same value. Why Other Options Are Wrong:1/15, 25/57, 35/256 arise from incorrect denominators or mixing with replacement. Common Pitfalls:Forgetting that order does not matter in combinations, or not reducing the fraction fully. Final Answer:1/221

Question 13

A bag initially contains three unknown balls. We add one white ball, then draw one ball at random. If all hypotheses about the initial composition (0,1,2,3 whites) are equally likely, what is the probability that the drawn ball is white?

Accepted Answer

Introduction / Context:This problem uses a simple Bayesian averaging over hypotheses. Before adding a white ball, the bag has three balls but we do not know how many are white. We assume four hypotheses are equally likely: initially 0, 1, 2, or 3 whites. After adding one white ball, a single draw is made; we want the overall chance the drawn ball is white. Given Data / Assumptions: Prior over k ∈ {0,1,2,3} whites initially: each 1/4. After adding one white, the bag has 4 balls with (k+1) white. On a single draw: P(white | k) = (k+1)/4. Concept / Approach:Law of total probability: average the conditional probabilities over the four equally likely hypotheses. Step-by-Step Solution: P(white) = (1/4) * Σ_{k=0}^3 (k+1)/4.Compute Σ (k+1) from k=0 to 3: 1 + 2 + 3 + 4 = 10.Thus P(white) = (1/4) * (10/4) = 10/16 = 5/8. Verification / Alternative check:Explicit mixture: If initially 0W ⇒ 1/4; 1W ⇒ 2/4; 2W ⇒ 3/4; 3W ⇒ 4/4; average of {1/4, 1/2, 3/4, 1} is (2.5)/4 = 0.625 = 5/8. Why Other Options Are Wrong:3/4 (0.75) would require biasing toward larger k; 1/2 (0.5) or 3/8 (0.375) undervalue the influence of adding the extra white ball. Common Pitfalls:Assuming the initial composition is uniformly random over individual colorings rather than over k; the statement says “all hypotheses concerning the colour” by count, i.e., each k equally likely. Final Answer:5/8

Question 14

Raja, Karan, and Ashwini each fire one shot at a target. Their hit probabilities are: Raja = 3/6, Karan = 2/6, Ashwini = 4/4. What is the probability that at least two shots hit the target?

Accepted Answer

Introduction / Context:We have three independent Bernoulli trials with different success probabilities: p_R = 1/2, p_K = 1/3, p_A = 1. We want P(at least two hits). Because Ashwini always hits, the total hits are 1 (from Ashwini) plus the number of hits from Raja and Karan. Hence, “at least two” reduces to “Raja or Karan (or both) hits”. Given Data / Assumptions: p_R = 3/6 = 1/2. p_K = 2/6 = 1/3. p_A = 1. Independence of shooters. Concept / Approach:Compute P(exactly 2 hits) and P(exactly 3 hits) and sum them. Exactly 2 occurs when Ashwini hits and exactly one of {Raja, Karan} hits. Exactly 3 occurs when all three hit. Step-by-Step Solution: P(exactly 2) = p_A * [p_R(1 − p_K) + (1 − p_R)p_K] = 1 * [(1/2)(2/3) + (1/2)(1/3)] = 1/3 + 1/6 = 1/2.P(exactly 3) = p_R * p_K * p_A = (1/2)(1/3)(1) = 1/6.Therefore P(≥ 2) = 1/2 + 1/6 = 2/3. Verification / Alternative check:Shortcut: Because p_A = 1, at least two hits ⇔ (R or K hits). P(R ∪ K) = 1/2 + 1/3 − (1/2)(1/3) = 5/6 − 1/6 = 2/3. Why Other Options Are Wrong:1/2 ignores triple-hit scenarios; 5/6 overcounts; 1/3 is too small. Common Pitfalls:Forgetting independence or misinterpreting “at least two” as “exactly two”. Final Answer:2/3

Question 15

Wardrobe selection in the dark: Girdhari has 3 trousers (grey, blue, brown) and 4 shirts (1 grey, 3 white). He randomly picks one shirt–trouser pair without seeing the colors. What is the probability that neither the shirt nor the trouser is grey?

Accepted Answer

Introduction / Context:This problem is a direct application of independent selection with simple fractions. We have a small wardrobe with categories (shirts, trousers) and want the probability that neither selected item is grey when picked at random in the dark. Given Data / Assumptions: Trousers: 3 total (1 grey, 2 non-grey). Shirts: 4 total (1 grey, 3 non-grey). One shirt and one trouser are chosen independently and uniformly at random. Concept / Approach:For independence, multiply the probabilities of the required events in each category. The desired event is “non-grey shirt” AND “non-grey trouser.” Step-by-Step Solution: P(non-grey trouser) = 2/3P(non-grey shirt) = 3/4P(neither grey) = (2/3) * (3/4) = 1/2 Verification / Alternative check:List outcomes: trousers (G, B, Br), shirts (G, W, W, W). Favorable pairs exclude any with a G. Count = 2 * 3 = 6 favorable out of 3*4 = 12 total ⇒ 6/12 = 1/2. Why Other Options Are Wrong: 1/12 or 1/6 severely undercount favorable outcomes. 1/4 double-filters incorrectly or ignores independence. Common Pitfalls:Treating the combined choice as dependent in the wrong way, or forgetting to multiply across categories for independent events. Final Answer:1/2

Question 16

Hiring scenario with two candidates: A brother and sister apply for two vacant posts. P(brother selected) = 1/5 and P(sister selected) = 1/3, independently. What is the probability that exactly one of them is selected?

Accepted Answer

Introduction / Context:We calculate the probability that exactly one of two independent events occurs. This is a common inclusion–exclusion style computation for Bernoulli outcomes (selected vs not selected). Given Data / Assumptions: P(Brother selected) = 1/5 ⇒ P(B̄) = 4/5. P(Sister selected) = 1/3 ⇒ P(S̄) = 2/3. Selections are independent (as stated). Concept / Approach:“Exactly one” happens if the brother is selected and the sister is not, OR the sister is selected and the brother is not. Add these disjoint probabilities. Step-by-Step Solution: P(exactly one) = P(B ∩ S̄) + P(B̄ ∩ S)= (1/5)*(2/3) + (4/5)*(1/3)= 2/15 + 4/15 = 6/15 = 2/5 Verification / Alternative check:Use complement: 1 − P(both) − P(neither) = 1 − (1/5)*(1/3) − (4/5)*(2/3) = 1 − 1/15 − 8/15 = 1 − 9/15 = 6/15 = 2/5. Why Other Options Are Wrong:1/5 or 1/3 capture only one piece; 2/3 is the complement of “at most one,” not “exactly one.” Common Pitfalls:Confusing “exactly one” with “at least one,” or forgetting independence while multiplying. Final Answer:2/5

Question 17

Two full decks mixed: Draw two cards in sequence without replacement. What is the probability both drawn cards are jacks?

Accepted Answer

Introduction / Context:When two standard decks (2 × 52 = 104 cards) are thoroughly shuffled, there are 8 jacks total. We draw two cards sequentially without replacement and want both to be jacks. Given Data / Assumptions: Total cards = 104; total jacks = 8. Sampling without replacement; order matters only in probability multiplication. Concept / Approach:Use sequential probability: P(first jack) × P(second jack | first jack). Multiply numerators and denominators carefully to avoid arithmetic slips. Step-by-Step Solution: P(first jack) = 8/104 = 1/13P(second jack | first jack) = 7/103P(both jacks) = (1/13) * (7/103) = 7/1339 Verification / Alternative check:Combinatorial form: C(8,2)/C(104,2) = [28]/[5356] = 7/1339, same result. Why Other Options Are Wrong: 1/13 or 2/13 ignore the second conditional step. 1/169 assumes replacement or independence that does not hold here. Common Pitfalls:Using 52 instead of 104; forgetting there are 8 jacks; assuming replacement. Final Answer:7/1339

Question 18

Three fair coins tossed together: What is the probability that all three coins show the same face (either all heads or all tails)?

Accepted Answer

Introduction / Context:Tossing three fair coins creates 2^3 equally likely outcomes. We want the probability of the event where all faces match, i.e., either HHH or TTT. Given Data / Assumptions: Coins are fair and independent. Sample space size = 8 outcomes. Concept / Approach:Count favorable outcomes and divide by total outcomes. Favorable = {HHH, TTT} = 2 outcomes. Step-by-Step Solution: Total outcomes = 8Favorable outcomes = 2P(all same) = 2/8 = 1/4 Verification / Alternative check:Compute P(HHH) + P(TTT) = (1/2)^3 + (1/2)^3 = 1/8 + 1/8 = 1/4. Why Other Options Are Wrong:1/3 results from averaging; 1/6 confuses with “exactly two heads”; 1/8 counts only one of the two symmetric cases. Common Pitfalls:Forgetting to include both all-heads and all-tails; mixing up “all same” with “all heads.” Final Answer:1/4

Question 19

Cards numbered 1 to 200: A card is drawn uniformly at random. What is the probability the number on it is a perfect cube?

Accepted Answer

Introduction / Context:We need the density of perfect cubes among the integers 1 through 200. Recognize that cubes grow quickly, so only a few lie within this range. Given Data / Assumptions:Numbers are 1,2,3,…,200 with equal probability; perfect cubes are n^3 for integer n ≥ 1. Concept / Approach:Find the largest n with n^3 ≤ 200. Then probability = (count of cubes)/200. Step-by-Step Solution: 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216 > 200Cubes ≤ 200 are 1, 8, 27, 64, 125 ⇒ 5 numbersProbability = 5/200 = 1/40 Verification / Alternative check:Use floor of cube root: ⌊∛200⌋ = 5 ⇒ count = 5; divide by 200. Why Other Options Are Wrong:3/40 and 7/40 miscount; 19/40 is not plausible given rarity of cubes. Common Pitfalls:Including 216 (6^3) erroneously or forgetting 1 is a cube. Final Answer:1/40

Question 20

Two unbiased dice thrown together: What is the probability that the sum of the two faces is divisible by 3?

Accepted Answer

Introduction / Context:We classify die faces by residues modulo 3 and use independence. A 1–6 die has two faces of each residue class 0,1,2 (since 3 and 6 are 0 mod 3; 1,4 are 1 mod 3; 2,5 are 2 mod 3). Given Data / Assumptions:Die outcomes are independent and uniform on {1,…,6}. Concept / Approach:A sum is divisible by 3 when residues add to 0 mod 3: (0,0), (1,2), or (2,1). Each residue probability per die is 2/6. Step-by-Step Solution: P(0,0) = (2/6)*(2/6) = 4/36P(1,2) = (2/6)*(2/6) = 4/36P(2,1) = (2/6)*(2/6) = 4/36Total = 12/36 = 1/3 Verification / Alternative check:Explicit counting yields 12 favorable ordered pairs out of 36 total, confirming 1/3. Why Other Options Are Wrong:11/36 and 13/36 are off by ±1 pair; 17/36 is too large. Common Pitfalls:Forgetting that there are two faces in each residue class; double-counting permutations. Final Answer:12/36

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