Question 1

Eight fair coins tossed simultaneously: What is the probability of getting at least 6 heads (i.e., 6, 7, or 8 heads)?

Accepted Answer

Introduction / Context: “At least k heads” suggests summing binomial probabilities. For 8 fair coins, we sum for 6, 7, and 8 heads. Each outcome has probability (1/2)^8. Given Data / Assumptions: n = 8, p(head) = 1/2, independence across coins. Concept / Approach: Use binomial coefficients C(8,6), C(8,7), C(8,8) and divide by 2^8 = 256. Step-by-Step Solution: C(8,6) = 28C(8,7) = 8C(8,8) = 1Total favorable = 28 + 8 + 1 = 37Probability = 37/256 Verification / Alternative check: By symmetry, P(X ≥ 6) = P(X ≤ 2); indeed C(8,0)+C(8,1)+C(8,2) = 1+8+28 = 37 ⇒ same result. Why Other Options Are Wrong: Values like 25/57 or 1/13 are unrelated to 2^8 denominators and correct counts. Common Pitfalls: Forgetting to include all three counts (6,7,8) or miscomputing combinations. Final Answer: 37/256

Question 2

Target practice with two shooters: P(A hits) = 5/7 and P(B hits) = 7/10, independently. What is the probability exactly one of them hits?

Accepted Answer

Introduction / Context: We need the probability of the exclusive OR (XOR) event: exactly one out of two independent shooters hits the target. Given Data / Assumptions: P(A) = 5/7 ⇒ P(Ā) = 2/7; P(B) = 7/10 ⇒ P(B̄) = 3/10; independence assumed. Concept / Approach: Compute P(A ∩ B̄) + P(Ā ∩ B). Multiply due to independence and sum the disjoint cases. Step-by-Step Solution: P(A ∩ B̄) = (5/7)*(3/10) = 15/70P(Ā ∩ B) = (2/7)*(7/10) = 14/70Total = 29/70 Verification / Alternative check: Complement method: P(exactly one) = P(A ∪ B) − 2P(A ∩ B) = [5/7 + 7/10 − (5/7)(7/10)] − 2*(5/7)(7/10) = 29/70. Why Other Options Are Wrong: 31/70 or 33/70 come from arithmetic slips; 21/10 is impossible (>1). Common Pitfalls: Using “at least one” instead of “exactly one,” or failing to apply independence correctly. Final Answer: 29/70

Question 3

Urn with replacement: A bag has 5 white, 7 red, and 8 black balls (20 total). Four balls are drawn one by one with replacement. What is the probability all four are white?

Accepted Answer

Introduction / Context: With replacement, each draw is independent and identically distributed (i.i.d.). We want the probability all four outcomes are white from a fixed proportion. Given Data / Assumptions: White balls = 5, total = 20 ⇒ P(white on any draw) = 1/4; 4 i.i.d. draws. Concept / Approach: P(all white) = (P(white))^4 because of independence with replacement. Step-by-Step Solution: P(all white) = (1/4)^4 = 1/256 Verification / Alternative check: Simulated reasoning: each draw 25% chance, the joint event multiplies four times. Why Other Options Are Wrong: 1/16 corresponds to only two draws; fractions like 4/20 or 4/8 are single-draw proportions, not fourfold joint probabilities. Common Pitfalls: Forgetting that “with replacement” makes draws independent, leading to exponentiation of a single probability. Final Answer: 1/256

Question 4

Urn without replacement: A bag has 8 blue and 4 white balls (12 total). Two balls are drawn without replacement. What is the probability both are blue?

Accepted Answer

Introduction / Context: Sampling without replacement changes the denominator on the second draw. We compute the sequential probability of drawing two blue balls in a row from a finite urn. Given Data / Assumptions: Blue = 8, White = 4, Total = 12; draws are without replacement. Concept / Approach: Use sequential multiplication with conditional probability for the second draw. Step-by-Step Solution: P(first blue) = 8/12 = 2/3P(second blue | first blue) = 7/11P(both blue) = (2/3) * (7/11) = 14/33 Verification / Alternative check: Combinatorial: C(8,2)/C(12,2) = 28/66 = 14/33. Why Other Options Are Wrong: 7/33 is half the correct value; 11/33 or 2/33 do not align with either method. Common Pitfalls: Using 12 in the denominator for both draws (as if with replacement) or ignoring the reduced blue count after the first draw. Final Answer: 14/33

Question 5

Two dice thrown together: What is the probability that neither die shows a 4 (i.e., the outcome on every die is not 4)?

Accepted Answer

Introduction / Context: We seek the probability both dice avoid a particular face. With independence, multiply the single-die avoidance probability across the two dice. Given Data / Assumptions: Each die: P(not 4) = 5/6; two independent dice. Concept / Approach: P(neither shows 4) = (5/6) * (5/6) by independence. Step-by-Step Solution: P = (5/6)^2 = 25/36 Verification / Alternative check: Complement: 1 − P(at least one 4) = 1 − [1 − (5/6)^2] = (5/6)^2 = 25/36. Why Other Options Are Wrong: 2/3 or 1/3 reflect single-die logic or complement misapplied; 17/36 is off by arithmetic. Common Pitfalls: Using 1 − 2*(1/6) which double-counts the “both 4” case. Final Answer: 25/36

Question 6

Tokens numbered 1 to 25: Draw one token uniformly at random. What is the probability the number is divisible by both 2 and 3?

Accepted Answer

Introduction / Context: Divisibility by both 2 and 3 means divisibility by 6. Count multiples of 6 up to 25 and divide by total tokens. Given Data / Assumptions: Numbers: 1,…,25; uniform random draw. Concept / Approach: Use floor division to count: ⌊25/6⌋ = 4. Step-by-Step Solution: Multiples of 6: 6, 12, 18, 24 (four numbers)Probability = 4/25 Why Other Options Are Wrong: 7/25 or 9/25 overcount; 2/25 undercounts. Common Pitfalls: Forgetting that 0 is not in the set; miscounting multiples. Final Answer: 4/25

Question 7

Traffic light cycle of 60 s: Green = 25 s, Yellow = 5 s, Red = 30 s. At a random time, what is the probability the light is not green?

Accepted Answer

Introduction / Context: Over a uniform 60-second cycle, the fraction of time in each color equals its probability at a random instant. We are asked for “not green.” Given Data / Assumptions: Total cycle = 60 s; Green = 25 s; Yellow = 5 s; Red = 30 s; random time uniformly across the cycle. Concept / Approach: P(not green) = (Yellow + Red) / Total. Step-by-Step Solution: Yellow + Red = 5 + 30 = 35 sProbability = 35/60 = 7/12 Verification / Alternative check: P(green) = 25/60 = 5/12 ⇒ complement 1 − 5/12 = 7/12. Why Other Options Are Wrong: 1/3 or 1/4 do not match the given durations; 5/12 is P(green), not “not green.” Common Pitfalls: Mixing up complement or misreading the seconds per color. Final Answer: 7/12

Question 8

Independent survival over 10 years: P(man lives 10y) = 1/4, P(wife lives 10y) = 1/3. What is the probability that after 10 years neither is alive?

Accepted Answer

Introduction / Context: We are given survival probabilities over a fixed horizon for two people and asked for the probability that neither survives. Assume independence unless stated otherwise. Given Data / Assumptions: P(M survives) = 1/4 ⇒ P(M dies) = 3/4; P(W survives) = 1/3 ⇒ P(W dies) = 2/3; independence assumed. Concept / Approach: P(neither alive) = P(M dies) * P(W dies). Step-by-Step Solution: P(neither) = (3/4) * (2/3) = 1/2 Verification / Alternative check: Complement perspective: 1 − P(at least one survives) = 1 − [1 − P(neither)] ⇒ consistent when computing directly. Why Other Options Are Wrong: 5/12 and 7/12 arise from mixing survival vs failure; 11/12 is impossible here. Common Pitfalls: Confusing given survival with failure probabilities; overlooking independence. Final Answer: 1/2

Question 9

Random permutation of letters: The letters B, G, I, N, R are rearranged uniformly at random. What is the probability the arrangement spells BRING?

Accepted Answer

Introduction / Context: We ask for the probability that a random permutation of five distinct letters equals one specific word. This is a classic counting problem using factorials. Given Data / Assumptions: Letters: B, R, I, N, G (5 distinct). Each permutation is equally likely. Concept / Approach: Total permutations = 5! = 120. Exactly one of those permutations is BRING. Step-by-Step Solution: Total outcomes = 120Favorable outcomes = 1Probability = 1/120 Verification / Alternative check: Enumerating permutations conceptually reinforces 5 positions × decreasing choices framework. Why Other Options Are Wrong: 1/54 or 1/24 do not correspond to 5! counting; the expression “5/5 x 42” is malformed. Common Pitfalls: Using 4! accidentally (treating one letter as fixed) or miscounting distinctness. Final Answer: 1/120

Question 10

Fair die rolled twice: What is the probability the die shows the same number on both rolls (i.e., a double)?

Accepted Answer

Introduction / Context: On two independent fair die rolls, there are 36 ordered outcomes. We want those where both faces match (1–1, 2–2, …, 6–6). Given Data / Assumptions: Uniform outcomes on pairs (1,…,6) × (1,…,6). Concept / Approach: Count matching pairs and divide by total pairs. Step-by-Step Solution: Favorable = 6 (the doubles)Total = 36Probability = 6/36 = 1/6 Verification / Alternative check: Alternatively, P(match) = sum over i of P(first=i)*P(second=i) = 6*(1/6)*(1/6) = 1/6. Why Other Options Are Wrong: 1/12 or 1/24 undercount; 1/8 corresponds to 4.5 favorable outcomes (nonsense). Common Pitfalls: Confusing ordered vs unordered outcomes; both methods agree on 1/6. Final Answer: 1/6

Question 11

100 identical biased coins: Each coin shows heads with probability p (0 < p < 1). If P(50 heads) equals P(51 heads), find p.

Accepted Answer

Introduction / Context: For a Binomial(n, p) with n = 100, we are told two adjacent probabilities match: P(X = 50) = P(X = 51). This determines p via the ratio of consecutive binomial terms. Given Data / Assumptions: X ~ Binomial(100, p); 0 < p < 1; equality of probabilities at 50 and 51. Concept / Approach: Use the identity P(k+1)/P(k) = [(n − k)/(k + 1)] * [p/(1 − p)] and set it equal to 1 at k = 50. Step-by-Step Solution: P(51)/P(50) = [(100 − 50)/(50 + 1)] * (p/(1 − p))= (50/51) * (p/(1 − p)) = 1⇒ p/(1 − p) = 51/50 ⇒ p = 51/101 Verification / Alternative check: Numerically, p ≈ 0.50495 is slightly > 1/2, which makes P(51) slightly larger than P(50) when p exceeds this threshold, matching equality at p = 51/101. Why Other Options Are Wrong: 1/2 is close but not exact; 49/101 and 50/101 arise from flipping or off-by-one ratios. Common Pitfalls: Setting p = 1/2 by symmetry without checking the discrete correction embodied in the ratio formula. Final Answer: 51/101

Question 12

Last digit behavior under multiplication: Four whole numbers are chosen independently at random. What is the probability that the last digit of their product is in {1, 3, 7, 9}?

Accepted Answer

Introduction / Context: The last digit of a product depends only on the last digits of the factors modulo 10. A product will end in 1, 3, 7, or 9 precisely when it has no factor 2 or 5 in its decimal factorization, i.e., every factor has a last digit coprime to 10. Given Data / Assumptions: Each whole number’s last digit is assumed independent and uniformly distributed over {0,…,9}. Allowed last digits (coprime to 10): {1, 3, 7, 9} (four choices). Concept / Approach: For the product to end with 1, 3, 7, or 9, every chosen last digit must be from {1, 3, 7, 9}. If any factor contributes a 0, 2, 4, 5, 6, or 8, the product gains a factor 2 and/or 5, making last digit even or 0/5. Step-by-Step Solution: Single number: P(last digit ∈ {1,3,7,9}) = 4/10 = 2/5Four independent choices ⇒ P = (2/5)^4 = 16/625 Verification / Alternative check: Complement counts any occurrence of a digit with a factor 2 or 5; the direct approach above is simpler and exact. Why Other Options Are Wrong: Values like 4/25 or 8/125 correspond to fewer than four constraints; 1/210 is unrelated to base-10 residue counting. Common Pitfalls: Allowing last digit 5 or an even digit in any factor, which invalidates membership in {1,3,7,9} for the product. Final Answer: 16/625

Question 13

Standard 52-card deck: What is the probability that a card drawn at random is either a king or a spade?

Accepted Answer

Introduction / Context: We need P(King ∪ Spade) from a standard deck. Use inclusion–exclusion to avoid double counting the king of spades. Given Data / Assumptions: There are 4 kings total and 13 spades total; the king of spades is in both sets. Concept / Approach: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Step-by-Step Solution: P(king) = 4/52P(spade) = 13/52P(king of spades) = 1/52P(either) = (4 + 13 − 1)/52 = 16/52 = 4/13 Verification / Alternative check: Counting: favorable cards are 3 non-spade kings + 13 spades = 16 distinct cards. Why Other Options Are Wrong: 17/52 double-counts; 13/52 is spades only; 3/13 undercounts. Common Pitfalls: Forgetting to subtract the overlap (king of spades). Final Answer: 4/13

Question 14

Single card from a well-shuffled deck: What is the probability of drawing a queen of clubs or a king of hearts?

Accepted Answer

Introduction / Context: We consider two specific distinct cards in a standard deck. The probability of drawing one of them is the count of favorable cards divided by 52. Given Data / Assumptions: Two distinct favorable cards: Q♣ and K♥; deck size = 52; uniform randomness. Concept / Approach: Since the events are disjoint singletons, add their probabilities. Step-by-Step Solution: P(Q♣ or K♥) = 1/52 + 1/52 = 2/52 = 1/26 Verification / Alternative check: Counting directly: 2 favorable outcomes out of 52. Why Other Options Are Wrong: 1/52 counts only one; 1/13 treats a whole rank or suit; 1/39 mixes numerator/denominator incorrectly. Common Pitfalls: Accidentally assuming overlap; these are different cards. Final Answer: 1/26

Question 15

Three unbiased coins are tossed simultaneously. What is the probability of getting exactly two heads?

Accepted Answer

Introduction / Context: Coin-toss probability questions test counting of outcomes under independence. With unbiased coins, each toss has outcomes H or T with equal likelihood, and combined outcomes are equally likely. Given Data / Assumptions: Three independent, fair coins. Sample space size = 2^3 = 8 equiprobable outcomes. We want exactly two heads. Concept / Approach: Count favorable outcomes using combinations: choose which 2 of the 3 tosses are heads. Probability = favorable / total. Step-by-Step Solution: Total outcomes = 2^3 = 8Favorable outcomes (exactly two H) = C(3,2) = 3Required probability = 3 / 8 Verification / Alternative check: List outcomes: HHT, HTH, THH are the only three with exactly two heads; count = 3 out of 8 total outcomes; probability = 3/8. Why Other Options Are Wrong: 1/8 corresponds to exactly three heads (only HHH). 2/8 is not the correct count; it simplifies to 1/4. 4/8 (= 1/2) counts “at least two heads,” not “exactly two heads.” Common Pitfalls: Confusing “exactly two” with “at least two.” Forgetting outcomes are equally likely only if coins are unbiased. Final Answer: 3/8

Question 16

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

Accepted Answer

Introduction / Context: This question checks counting with simple modular (last-digit) patterns over a finite range. We pick uniformly from consecutive integers and count those ending with 1. Given Data / Assumptions: Set S = {2, 3, 4, …, 50} has 49 numbers. Event E: unit digit = 1. Concept / Approach: Within any block of ten consecutive integers, exactly one ends with digit 1. Enumerate within the given finite range. Step-by-Step Solution: Eligible numbers in S ending with 1: 11, 21, 31, 41Count(E) = 4; Count(S) = 50 − 2 + 1 = 49Probability = 4 / 49 Verification / Alternative check: Check endpoints: 1 is not included; 51 is outside the range; thus exactly four hits (every ten steps) remain: 11, 21, 31, 41. Why Other Options Are Wrong: 5/49, 3/49, 2/49 miscount how many candidates end with 1. “None of these” is false because 4/49 is attainable. Common Pitfalls: Accidentally including 1 or 51 which are outside the closed interval. Final Answer: 4/49

Question 17

A number is chosen uniformly at random from {1, 2, 3, …, 50}. What is the probability that the chosen number is prime?

Accepted Answer

Introduction / Context: Prime-counting over a small range is straightforward by enumeration or recalling standard lists up to 50. Probability equals (number of primes up to 50)/50 for uniform selection from 1–50. Given Data / Assumptions: Sample space: integers 1 through 50 (size 50). Primes in this range are to be counted exactly once. Concept / Approach: List primes ≤ 50. Compute probability = count / 50. Step-by-Step Solution: Primes ≤ 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (15 numbers)Probability = 15 / 50 = 3 / 10 = 0.30 Verification / Alternative check: Cross-check by excluding composites and 1; the remaining count remains 15, a standard fact set for small prime tables. Why Other Options Are Wrong: 0.18, 0.24, 0.36 reflect incorrect prime counts. “None of these” is wrong because 0.30 is correct. Common Pitfalls: Forgetting that 1 is not prime and that 49 (= 7^2) is composite. Final Answer: 0.30

Question 18

From a group of 3 men and 2 women, a committee of 3 is to be selected uniformly at random. What is the probability that the committee has at least one woman?

Accepted Answer

Introduction / Context: “At least one” events are often faster via the complement rule: P(at least one woman) = 1 − P(no women). Counting committees uses combinations since only membership matters, not order. Given Data / Assumptions: Total people = 3 men + 2 women = 5. Committee size = 3. All committees are equally likely. Concept / Approach: Total committees = C(5,3). Committees with no women = choose all from the 3 men: C(3,3). Probability = 1 − [C(3,3)/C(5,3)]. Step-by-Step Solution: Total committees = C(5,3) = 10No-woman committees = C(3,3) = 1P(at least one woman) = 1 − 1/10 = 9/10 Verification / Alternative check: Direct count of “at least one woman” committees = total 10 minus the single all-men committee = 9 favorable; probability 9/10. Why Other Options Are Wrong: 1/10 is the complement (no woman), not the wanted event. 9/20 and 1/20 come from incorrect denominators or double counting. Common Pitfalls: Overcounting when splitting into cases (exactly one, two, or three women) instead of using the quick complement. Final Answer: 9/10

Question 19

An urn contains 3 red and 4 green marbles (7 total). If three marbles are picked at random without replacement, what is the probability that two are green and one is red?

Accepted Answer

Introduction / Context: Sampling without replacement from colored marbles is modeled by combinations. We count favorable color-compositions and divide by total equally likely 3-combinations from the urn. Given Data / Assumptions: R = 3 (red), G = 4 (green), total N = 7. Draw k = 3 without replacement. Target composition: 2 green, 1 red. Concept / Approach: Total ways = C(7,3). Favorable = C(4,2) * C(3,1). Probability = favorable / total. Step-by-Step Solution: Total ways = C(7,3) = 35Favorable ways = C(4,2) * C(3,1) = 6 * 3 = 18Probability = 18 / 35 Verification / Alternative check: Hypergeometric model: HG(N=7, success-type G=4, draws k=3, want x=2) gives the same ratio 18/35 after multiplying by C(3,1) for the red pick. Why Other Options Are Wrong: 3/7, 5/14, 4/21 result from mixing replacement assumptions or miscounting combinations. Common Pitfalls: Treating order as distinct (it is not; combinations handle orderless draws). Final Answer: 18/35

Question 20

Two cards are drawn at random from a standard pack of 52 playing cards without replacement. What is the probability that both drawn cards are aces?

Accepted Answer

Introduction / Context: Card-draw probabilities often use combinations since order does not matter for the set of drawn cards. There are 4 aces in a 52-card deck; we want both drawn cards to be aces without replacement. Given Data / Assumptions: Deck size = 52, aces = 4. Draw 2 cards without replacement. Concept / Approach: Total unordered draws = C(52,2). Favorable draws (both aces) = C(4,2). Probability = C(4,2)/C(52,2). Step-by-Step Solution: C(4,2) = 6C(52,2) = 52*51/2 = 1326Probability = 6/1326 = 1/221 Verification / Alternative check: Sequential method: P(first ace) = 4/52; P(second ace | first ace) = 3/51; product = (4/52)*(3/51) = 12/2652 = 1/221. Same result. Why Other Options Are Wrong: 2/13 and 2/26 ignore shrinking deck or use wrong counting. “None of these” is false because 1/221 is exact. Common Pitfalls: Using replacement probabilities or confusing ordered vs unordered counting. Final Answer: 1/221

Probability Questions