Question 1

A single card is drawn uniformly at random from a standard 52-card deck. What is the probability that it is neither a heart nor a king?

Accepted Answer

Introduction / Context: Use inclusion–exclusion to avoid double-counting the king of hearts when excluding hearts and kings from a deck. Then divide by total cards for the probability. Given Data / Assumptions: 52-card standard deck. Hearts = 13; Kings = 4; overlap (king of hearts) = 1. Concept / Approach: Cards that are heart or king = 13 + 4 − 1 = 16 (by inclusion–exclusion). “Neither” count = 52 − 16 = 36. Probability = 36/52 = 9/13. Step-by-Step Solution: Count(forbidden) = 13 + 4 − 1 = 16Count(allowed) = 52 − 16 = 36Probability = 36/52 = 9/13 Verification / Alternative check: Complement method directly: P(neither) = 1 − P(heart ∪ king) = 1 − 16/52 = 36/52 = 9/13. Why Other Options Are Wrong: 4/13 and 2/13 correspond to the complement or partial counts. Duplicate 4/13 remains incorrect; “None of these” is false because 9/13 is correct. Common Pitfalls: Double-counting the king of hearts if adding 13 and 4 without subtracting 1. Final Answer: 9/13

Question 2

Three boxes contain balls as follows: • Box 1: 3 white, 1 black • Box 2: 2 white, 2 black • Box 3: 1 white, 3 black One ball is drawn at random from each box. What is the probability that exactly two whites and one black are drawn in total?

Accepted Answer

Introduction / Context: We draw one ball independently from each of three differently composed boxes. The overall event “exactly two whites and one black” occurs via three mutually exclusive cases depending on which box contributes the black ball. Sum those case probabilities. Given Data / Assumptions: Box 1: P(W) = 3/4, P(B) = 1/4. Box 2: P(W) = 2/4 = 1/2, P(B) = 1/2. Box 3: P(W) = 1/4, P(B) = 3/4. Independent draws across boxes. Concept / Approach: Casework by which single box yields the black ball. Add the disjoint case probabilities (law of total probability). Step-by-Step Solution: Case A (black from Box 1): (1/4)*(1/2)*(1/4) = 1/32Case B (black from Box 2): (3/4)*(1/2)*(1/4) = 3/32Case C (black from Box 3): (3/4)*(1/2)*(3/4) = 9/32Total probability = 1/32 + 3/32 + 9/32 = 13/32 Verification / Alternative check: Complement check is longer here; the direct case-sum is clean and matches repeated simulation intuition (Box 3 is most likely to give black). Why Other Options Are Wrong: 1/4 (= 8/32) and 3/16 (= 6/32) undercount; 1/32 counts only one case. “None of these” is false because 13/32 matches the exact calculation. Common Pitfalls: Double-counting cases or forgetting that exactly one box must contribute black. Final Answer: 13/32

Question 3

A box contains 3 white balls and 2 red balls (total 5). Two draws are made without replacement. What is the probability that the second ball drawn is red (no information about the first draw is used)?

Accepted Answer

Introduction / Context: When two balls are drawn without replacement from a small urn, the colour of the second ball can be analysed either conditionally on the first draw or by symmetry arguments. The question explicitly asks for the unconditional probability that the second ball is red, i.e., we do not observe the first draw. Given Data / Assumptions: Total balls = 5 (3 white, 2 red). Two draws are made without replacement. No information about the first draw is used; draws are from a well-mixed box. Concept / Approach: Key fact: In simple random sampling without replacement, each position in the draw sequence is equally likely to be any given ball. Therefore, the marginal probability that the kth draw is red equals the overall red proportion in the urn: red_count / total_count. Alternatively, use the law of total probability conditioning on the first draw. Step-by-Step Solution: Red proportion initially = 2/5.Therefore, P(second is red) = 2/5 by symmetry of positions.Check by conditioning: P(2nd R) = P(1st W)*P(2nd R|1st W) + P(1st R)*P(2nd R|1st R) = (3/5)*(2/4) + (2/5)*(1/4) = (6/20) + (2/20) = 8/20 = 2/5. Verification / Alternative check: The symmetry argument applies for any position: at each draw, without conditioning on previous outcomes, the chance of red remains the initial fraction. Why Other Options Are Wrong: 3/5 and 21/25 overstate the likelihood; 8/25 confuses a conditional product with the unconditional calculation. Common Pitfalls: Confusing the unconditional probability with probabilities conditional on the first draw being white or red. Final Answer: 2/5

Question 4

A box holds 3 mangoes and 3 apples (6 fruits total). Two fruits are chosen uniformly at random without replacement. What is the probability that one fruit is a mango and the other is an apple?

Accepted Answer

Introduction / Context: This is a classic two-draw mixture question. With equal counts of two fruit types, we need the probability of selecting a mixed pair (one of each) when drawing two without replacement. Given Data / Assumptions: Mangoes = 3, Apples = 3. Total fruits = 6. Two draws without replacement. Concept / Approach: Count favourable unordered pairs and divide by the total number of unordered pairs. Alternatively, compute sequentially: P(mango then apple) + P(apple then mango). Step-by-Step Solution (combinatorial): Total unordered pairs = C(6,2) = 15.Favourable pairs (one of each) = 3 * 3 = 9 (choose 1 mango and 1 apple).Probability = 9/15 = 3/5. Verification / Alternative check (sequential): P(M then A) = (3/6)*(3/5) = 3/10.P(A then M) = (3/6)*(3/5) = 3/10.Total = 3/10 + 3/10 = 3/5. Why Other Options Are Wrong: 2/3 and 1/3 are typical but incorrect guesses; 1/5 is far too small. Common Pitfalls: Forgetting that order does not matter for the event “one of each” but it matters in sequential computation (hence we add both orders). Final Answer: 3/5

Question 5

The probability a man survives 10 more years is 1/4 and the probability his wife survives 10 more years is 1/3. Assuming independence, what is the probability that neither of them is alive after 10 years?

Accepted Answer

Introduction / Context: Given two independent survival probabilities over a fixed horizon, we are asked for the probability that neither survives. This is a direct complement multiplication problem. Given Data / Assumptions: P(man survives 10 yrs) = 1/4 → P(man does not survive) = 3/4. P(wife survives 10 yrs) = 1/3 → P(wife does not survive) = 2/3. Independence of the two events. Concept / Approach: For independent events X and Y, P(X and Y) = P(X)*P(Y). Here X = “man does not survive,” Y = “wife does not survive.” Step-by-Step Solution: P(neither survives) = (3/4) * (2/3) = 6/12 = 1/2. Verification / Alternative check: Compute P(at least one survives) = 1 − 1/2 = 1/2. Alternatively, use inclusion–exclusion on survival events and complement. Why Other Options Are Wrong: 5/12 and 7/12 are near but result from incorrect sums/products; 11/12 is far too large. Common Pitfalls: Adding the “non-survival” probabilities instead of multiplying for the joint event. Final Answer: 1/2

Question 6

A lottery sells 10,000 tickets and awards 10 prizes. If you buy one ticket, what is the probability that you do not win a prize?

Accepted Answer

Introduction / Context: In a finite uniform lottery, the probability of winning equals the number of prize tickets divided by total tickets. The probability of not winning is the complement. Given Data / Assumptions: Total tickets = 10,000. Prizes = 10. One ticket purchased; each ticket is equally likely to be drawn as a prize. Concept / Approach: P(no prize) = 1 − P(prize) = 1 − (prize tickets / total tickets). Step-by-Step Solution: P(prize) = 10 / 10000 = 1 / 1000.Therefore, P(no prize) = 1 − 1/1000 = 999/1000. Verification / Alternative check: Non-prize tickets = 10000 − 10 = 9990. For one ticket chosen uniformly, P(no prize) = 9990/10000 = 999/1000 after simplification. Why Other Options Are Wrong: 9999/10000 would correspond to 1 prize; 9/10 ignores the precise counts; 1/1000 is P(win), not P(no win). Common Pitfalls: Mixing up the probability of winning with not winning; not simplifying to lowest terms. Final Answer: 999/1000

Question 7

Two candidates A and B interview for two vacancies. The probabilities that A and B are selected are 1/3 and 1/6 respectively. Assuming independence, what is the probability that neither is selected?

Accepted Answer

Introduction / Context: Given marginal selection probabilities for two candidates and assuming independence of their outcomes, the chance that neither is selected is the product of their non-selection probabilities. Given Data / Assumptions: P(A selected) = 1/3 → P(A not selected) = 2/3. P(B selected) = 1/6 → P(B not selected) = 5/6. Selections are independent (standard assumption unless stated otherwise). Concept / Approach: P(neither) = P(A not selected) * P(B not selected). Step-by-Step Solution: P(neither) = (2/3) * (5/6) = 10/18 = 5/9. Verification / Alternative check: P(at least one selected) = 1 − 5/9 = 4/9. Inclusion–exclusion also yields P(A ∪ B) = 1/3 + 1/6 − (1/3)(1/6) = 1/2 − 1/18 = 8/18 = 4/9. Why Other Options Are Wrong: 5/12 and smaller values underestimate; independence requires multiplying complements, not adding/subtracting naïvely. Common Pitfalls: Assuming mutual exclusivity or that exactly two positions force dependence; the statement provides independent success chances for each candidate. Final Answer: 5/9

Question 8

From 13 applicants (5 women, 8 men), two persons are selected at random. What is the probability that at least one selected person is a woman?

Accepted Answer

Introduction / Context: “At least one” probability questions are usually simpler via complement: subtract the probability that no one satisfies the property from 1. Here, “no woman” means both selected are men. Given Data / Assumptions: Total applicants = 13 (5 women, 8 men). Two are selected uniformly without order. Concept / Approach: P(≥1 woman) = 1 − P(both men) = 1 − C(8,2)/C(13,2). Step-by-Step Solution: C(8,2) = 28; C(13,2) = 78.P(≥1 woman) = 1 − 28/78 = 50/78 = 25/39. Verification / Alternative check: Direct count of favourable outcomes: total pairs (78) minus all-men pairs (28) = 50 favourable → 50/78 = 25/39. Why Other Options Are Wrong: 10/13 and 5/13 arise from incorrect linear approximations; 14/35 is unrelated. Common Pitfalls: Forgetting to use combinations (unordered selection) or to apply the complement method. Final Answer: 25/39

Question 9

Five fair coins are tossed simultaneously. What is the probability of obtaining at least one tail?

Accepted Answer

Introduction / Context: For multiple independent coin tosses, it is often fastest to compute the complement of the desired event. “At least one tail” is the complement of “no tails,” i.e., all heads. Given Data / Assumptions: Five independent fair coin tosses. P(H) = P(T) = 1/2. Concept / Approach: P(≥1 T) = 1 − P(all heads) = 1 − (1/2)^5. Step-by-Step Solution: P(all heads) = (1/2)^5 = 1/32.Therefore, P(≥1 tail) = 1 − 1/32 = 31/32. Verification / Alternative check: Enumerating outcomes confirms that only one of the 32 equally likely outcomes has zero tails. Why Other Options Are Wrong: 1/32 is the complement event; 5/32 is the probability of exactly one tail, not at least one; 1/5 is unrelated. Common Pitfalls: Misreading “at least one” as “exactly one.” Final Answer: 31/32

Question 10

Events A and B satisfy P(A) = 0.3 and P(A ∪ B) = 0.8. Assuming independence, find P(B).

Accepted Answer

Introduction / Context: This problem combines the union formula with independence. When A and B are independent, P(A ∩ B) = P(A)P(B), which plugs into the union identity P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Given Data / Assumptions: P(A) = 0.3. P(A ∪ B) = 0.8. A and B are independent. Concept / Approach: Let p = P(B). Then 0.8 = 0.3 + p − 0.3p = 0.3 + 0.7p, so solve for p. Step-by-Step Solution: 0.8 = 0.3 + 0.7p ⇒ 0.5 = 0.7p ⇒ p = 5/7 ≈ 0.7142857. Verification / Alternative check: Compute P(A ∩ B) = 0.3 * (5/7) = 1.5/7. Then P(A) + P(B) − P(A ∩ B) = 0.3 + 5/7 − 1.5/7 = 0.3 + 3.5/7 = 0.3 + 0.5 = 0.8, consistent. Why Other Options Are Wrong: 3/8, 2/7, 4/7 do not satisfy 0.8 = 0.3 + p − 0.3p under independence. Common Pitfalls: Using the addition rule for mutually exclusive events (A ∪ B = A + B) instead of the correct union formula with intersection. Final Answer: 5/7

Question 11

A box contains 24 marbles. The probability that a randomly drawn marble is green is 2/3. How many marbles in the box are blue?

Accepted Answer

Introduction / Context: This is a direct conversion between probability and counts. If a colour has probability 2/3 in a finite uniform selection, then 2/3 of the items must be of that colour. Given Data / Assumptions: Total marbles = 24. P(green) = 2/3 → green count = (2/3)*24. All non-green marbles are blue. Concept / Approach: Compute the number of green marbles and subtract from the total to get the number of blue marbles. Step-by-Step Solution: Green marbles = (2/3)*24 = 16.Blue marbles = 24 − 16 = 8. Verification / Alternative check: Check probability: 16/24 = 2/3 is consistent with the given probability. Why Other Options Are Wrong: 12 and 13 are incorrect residual counts; 16 is the green count, not blue. Common Pitfalls: Applying the 2/3 to the wrong colour or forgetting that only two colours are present. Final Answer: 8

Question 12

A bag contains 3 red, 4 white, and 7 black balls (14 total). If one ball is drawn at random, what is the probability that it is red or black?

Accepted Answer

Introduction / Context: The probability of a union of disjoint simple events (red or black) equals the sum of their individual probabilities because the events do not overlap on a single draw. Given Data / Assumptions: Red = 3, White = 4, Black = 7, Total = 14. One random draw. Concept / Approach: P(red or black) = (red + black) / total = (3 + 7) / 14. Step-by-Step Solution: (3 + 7) / 14 = 10/14 = 5/7. Verification / Alternative check: P(white) = 4/14 = 2/7; 5/7 + 2/7 = 1 as expected. Why Other Options Are Wrong: 4/7 omits some favourable outcomes; 1/7 and 3/7 undercount severely. Common Pitfalls: Accidentally double-counting mutually exclusive colour events (not applicable on a single draw). Final Answer: 5/7

Question 13

Nitish has 3 trousers (black, blue, brown) and 4 shirts (1 black, 3 white). In the dark he randomly picks one shirt–trouser pair. What is the probability that neither the shirt nor the trouser is black?

Accepted Answer

Introduction / Context: This is a simple independent choice problem: one trouser is chosen from 3, and one shirt from 4, with equal likelihood. We need the probability that both chosen items are non-black. Given Data / Assumptions: Trousers: {black, blue, brown} → 2 non-black out of 3. Shirts: {black, white, white, white} → 3 non-black out of 4. Choices are independent (one trouser and one shirt). Concept / Approach: P(neither black) = P(non-black trouser) * P(non-black shirt) = (2/3) * (3/4). Step-by-Step Solution: (2/3) * (3/4) = 6/12 = 1/2. Verification / Alternative check: Total outfit combinations = 3*4 = 12. Non-black trouser (2 choices) with non-black shirt (3 choices) = 6 combinations → 6/12 = 1/2. Why Other Options Are Wrong: 1/12 and 1/6 are much too small; 1/4 ignores the independence product. Common Pitfalls: Counting only white shirts as non-black but forgetting there are three of them. Final Answer: 1/2

Question 14

Murari has 9 pairs of dark-blue socks and 9 pairs of black socks (all loose in one bag). If he picks three socks at random, what is the probability that he gets at least one matching pair?

Accepted Answer

Introduction / Context: With only two colours available (dark blue and black), any selection of three socks must contain at least two of the same colour by the pigeonhole principle. Therefore, a matching colour pair is guaranteed. Given Data / Assumptions: Two colours: dark blue and black. Three socks drawn without replacement. Pairs do not matter; colour matches are sufficient for a “pair.” Concept / Approach: Pigeonhole Principle: placing 3 items into 2 categories ensures at least one category receives ≥2 items. Step-by-Step Reasoning: Possible colour-count splits for 3 socks with 2 colours are 3–0 or 2–1; both contain a colour appearing at least twice.Hence the event “at least one matching pair by colour” occurs with certainty. Verification / Alternative check: Trying to avoid a pair would require 3 socks all of different colours—impossible with only two colours available. Why Other Options Are Wrong: Combinatorial expressions given in other options do not evaluate to 1 and misrepresent the structure of the sample space. Common Pitfalls: Confusing identical socks within a pair with “colour” matching; here, only colour match is needed to constitute a pair. Final Answer: 1

Question 15

An anti-aircraft gun fires up to four shots at a receding plane. The hit probabilities on the 1st, 2nd, 3rd, and 4th shots are 0.4, 0.3, 0.2, and 0.1 respectively (independent). What is the probability the plane is hit at least once when all four sh…

Accepted Answer

Introduction / Context: When several independent attempts can cause success, the probability of at least one success is 1 minus the probability of zero successes. Here each shot has a different success probability. Given Data / Assumptions: P(hit on 1st) = 0.4, P(hit on 2nd) = 0.3, P(hit on 3rd) = 0.2, P(hit on 4th) = 0.1. Shot results are independent. Concept / Approach: P(at least one hit) = 1 − Π P(miss on shot i). Miss probabilities are the complements: 0.6, 0.7, 0.8, 0.9. Step-by-Step Calculation: P(no hit) = 0.6 * 0.7 * 0.8 * 0.9 = 0.3024.Therefore, P(at least one hit) = 1 − 0.3024 = 0.6976. Verification / Alternative check: Any inclusion–exclusion computation must yield the same value; using the complement avoids lengthy term expansions. Why Other Options Are Wrong: Values like 0.6872 or 0.4379 arise from arithmetic slips or partial sums of inclusion–exclusion. Common Pitfalls: Adding hit probabilities directly (invalid due to overlaps) instead of using the complement of “no hit.” Final Answer: 0.6976

Question 16

Four boys and three girls line up for an interview. What is the probability that they stand in strictly alternating gender order (B–G–B–G–B–G–B)?

Accepted Answer

Introduction / Context: Alternation constraints are classic permutation-with-pattern problems. With 4 boys and 3 girls, a perfect alternation is only possible in the pattern B–G–B–G–B–G–B (starting and ending with boys). Given Data / Assumptions: Distinct individuals: 4 boys, 3 girls. Total permutations without restriction = 7!. Alternation pattern forced: B–G–B–G–B–G–B. Concept / Approach: Place boys in the 4 boy slots in 4! ways, and girls in the 3 girl slots in 3! ways. Divide by 7! to get the probability. Step-by-Step Solution: Favourable arrangements = 4! * 3! = 24 * 6 = 144.Total arrangements = 7! = 5040.Probability = 144 / 5040 = 1 / 35. Verification / Alternative check: Since the counts differ by 1, only a single alternation pattern is feasible; the computed probability matches that unique pattern count. Why Other Options Are Wrong: 1/34 and 1/17 are not reducible forms from factorial counts; 1/68 would correspond to an extra symmetry that does not exist. Common Pitfalls: Allowing a starting girl pattern (impossible with 4 boys and 3 girls) or double-counting placements. Final Answer: 1/35

Question 17

An elevator starts with 5 passengers and stops independently at each of 8 floors. Assuming each passenger independently chooses a floor uniformly among the 8, what is the probability that all 5 alight at different floors?

Accepted Answer

Introduction / Context: We model each passenger’s floor choice as an independent uniform pick among 8 floors. We seek the probability that the 5 chosen floors are all distinct. Given Data / Assumptions: Passengers choose floors independently and uniformly from {1,…,8}. Choices are with replacement (conceptually), since multiple people could choose the same floor. Concept / Approach: Favourable outcomes: permutations of choosing 5 distinct floors from 8 → P(8,5) = 8*7*6*5*4. Total outcomes: 8^5 ordered choices. Probability = P(8,5)/8^5. Step-by-Step Solution: P = (8*7*6*5*4) / 8^5 = (8/8)*(7/8)*(6/8)*(5/8)*(4/8).= 1 * (7/8) * (3/4) * (5/8) * (1/2) = (105)/(512). Verification / Alternative check: Compute numerators and denominators explicitly: 8*7*6*5*4 = 6720; 8^5 = 32768; 6720/32768 reduces to 105/512. Why Other Options Are Wrong: Other fractions correspond to nearby but incorrect numerators in the product (e.g., replacing 6/8 by 1/2 too early or arithmetic slips). Common Pitfalls: Using combinations instead of permutations; order matters because passengers are distinguishable. Final Answer: 105/512

Question 18

A fair coin is tossed 100 times. What is the probability that the number of heads obtained is an odd number?

Accepted Answer

Introduction / Context: When a fair coin is tossed repeatedly, the binomial distribution with parameters n and p applies. Here, n = 100 independent tosses and p = 1/2 for heads. We are asked for the probability that the count of heads is odd. Given Data / Assumptions: Number of trials n = 100 (even). Per-trial head probability p = 1/2; tail probability q = 1/2. Outcomes across trials are independent. Concept / Approach: For a symmetric Bernoulli trial with p = q = 1/2 and an even number of trials n, the binomial mass is perfectly split between “even” and “odd” counts. This follows from the binomial theorem applied to (p + q)^n and (p − q)^n. Step-by-Step Solution: Let S_even = P(even heads) and S_odd = P(odd heads).(p + q)^n = 1 = S_even + S_odd.(p − q)^n = 0 (since p = q and n is even) = S_even − S_odd.Solving gives S_even = S_odd = 1/2. Verification / Alternative check: This parity-splitting result holds for any even n with p = 1/2 and is a standard identity in probability. Why Other Options Are Wrong: 1/4 and 3/4 would imply asymmetry; 2/3 does not arise from binomial parity symmetry; 0 is impossible. Common Pitfalls: Forgetting that the parity-splitting property requires p = 1/2 and n even; it does not generally hold for p ≠ 1/2 or odd n. Final Answer: 1/2

Question 19

From a class of 15 students (7 from Maharashtra, 5 from Karnataka, 3 from Goa), four students are selected at random. What is the probability that at least one selected student is from Karnataka?

Accepted Answer

Introduction / Context: This is a classical “at least one” selection problem from disjoint groups. The cleanest approach is to compute the complement (no one from Karnataka) and subtract from 1. Given Data / Assumptions: Total students = 15 (7 Maharashtra, 5 Karnataka, 3 Goa). We choose 4 students uniformly at random without replacement. Event of interest: at least one from Karnataka among the selected four. Concept / Approach: Use complement: P(≥1 from Karnataka) = 1 − P(none from Karnataka). “None from Karnataka” means all four are from the remaining 10 students (Maharashtra + Goa). Step-by-Step Solution: Total ways = C(15, 4) = 1365.Ways with none from Karnataka = C(10, 4) = 210.Therefore, probability = 1 − 210/1365 = 1155/1365 = 11/13. Verification / Alternative check: Arithmetic check: 1365 − 210 = 1155; divide numerator and denominator by 105 to get 11/13. Why Other Options Are Wrong: 12/13 overstates; 10/15 is not a valid hypergeometric probability; 1/15 is the complement of something else, not this event; 9/13 undercounts. Common Pitfalls: Trying to sum “exactly k” cases directly; the complement method is faster and less error-prone here. Final Answer: 11/13

Question 20

A box contains 12 oranges, of which one-third are bad (i.e., 4 bad and 8 good). If three oranges are drawn at random without replacement, what is the probability that at least one of the three is good?

Accepted Answer

Introduction / Context: “At least one good” is most efficiently computed via the complement event: “all drawn are bad.” The composition is small, making the computation straightforward with combinations. Given Data / Assumptions: Total oranges = 12. Bad = 4; Good = 8. Three are drawn uniformly at random without replacement. Concept / Approach: P(≥1 good) = 1 − P(all 3 are bad). The “all bad” event exists only if we draw all three from the 4 bad oranges. Step-by-Step Solution: Total ways to choose 3 = C(12, 3) = 220.Ways to choose 3 all bad = C(4, 3) = 4.P(all bad) = 4 / 220 = 1 / 55.Therefore P(≥1 good) = 1 − 1/55 = 54/55. Verification / Alternative check: Direct enumeration of mixed cases is longer and error-prone compared to the clean complement method. Why Other Options Are Wrong: 1/55 is the probability that all three are bad, not “at least one good”; 45/55 and 3/55 are arithmetic mismatches; 53/55 still undercounts. Common Pitfalls: Confusing “one-third bad” as 1/3 probability per draw without replacement; combinatorics must reflect finite counts. Final Answer: 54/55

Probability Questions