Question 1

A bag has 4 white, 5 red, and 6 blue balls. Three are drawn at random. What is the probability all three are red?

Accepted Answer

Given data Total balls = 4 + 5 + 6 = 15. Red balls = 5. Concept/Approach Use combinations without replacement: P(all red) = C(5,3) / C(15,3). Step-by-step calculation Favourable = C(5,3) = 10Total = C(15,3) = 455Probability = 10/455 = 2/91 Verification/Alternative Sequential draw: (5/15) × (4/14) × (3/13) = 60/2730 = 2/91. Common pitfalls Using permutations or treating draws as independent with replacement. Final Answer 2/91

Question 2

Tickets numbered 1 to 20 are mixed. What is the probability the drawn ticket is a multiple of 3 or 5?

Accepted Answer

Given data Sample space size = 20. Multiples of 3: 3,6,9,12,15,18 (6 numbers). Multiples of 5: 5,10,15,20 (4 numbers). Concept/Approach Apply inclusion–exclusion: count(3 ∪ 5) = count(3) + count(5) − count(15). Step-by-step calculation Common multiples (15): 1 (i.e., 15)Favourable = 6 + 4 − 1 = 9Probability = 9/20 = 9/20 Verification/Alternative Explicit list: {3,5,6,9,10,12,15,18,20} (9 numbers). Common pitfalls Double-counting 15 or forgetting it entirely. Final Answer 9/20

Question 3

Two dice are tossed. What is the probability that the total score is a prime number?

Accepted Answer

Given data Sums possible: 2 to 12 (36 equally likely outcomes). Prime sums: 2, 3, 5, 7, 11. Concept/Approach Count outcomes for each prime sum using standard dice-sum counts. Step-by-step calculation Sum 2: 1 way; 3: 2 ways; 5: 4 ways; 7: 6 ways; 11: 2 waysTotal favourable = 1 + 2 + 4 + 6 + 2 = 15Probability = 15/36 = 5/12 Verification/Alternative Non-prime sums (4,6,8,9,10,12) account for 21 outcomes; 36 − 21 = 15. Common pitfalls Including 1 or 13 as prime sums (not possible with two dice). Final Answer 5/12

Question 4

One card is drawn at random from 52. What is the probability it is a face card (Jack, Queen, or King)?

Accepted Answer

Given data Face cards per suit: J, Q, K ⇒ 3 per suit. Number of suits = 4. Concept/Approach Total face cards = 3 × 4 = 12; equally likely draws from 52. Step-by-step calculation Probability = 12/52 = 3/13 Verification/Alternative Non-face (A, 2–10) count: 13 − 3 = 10 per suit ⇒ 40 non-face; 12 face; 12/52 = 3/13. Common pitfalls Including Aces as face cards; in this context they are not. Final Answer 3/13

Question 5

Two cards are drawn together from a pack of 52. What is the probability that one is a spade and the other is a heart?

Accepted Answer

Given data Spades = 13, Hearts = 13. Total two-card selections = C(52,2) = 1326. Concept/Approach Choose 1 spade and 1 heart: C(13,1) × C(13,1); order does not matter because the draw is simultaneous. Step-by-step calculation Favourable = 13 × 13 = 169Probability = 169/1326 = 13/102 Verification/Alternative Sequential approach without order: (13/52)×(13/51)×2 = 338/2652 = 169/1326 = 13/102. Common pitfalls Forgetting to treat order properly (double-counting or undercounting). Final Answer 13/102

Question 6

A card is drawn from 52. What is the probability of getting the queen of clubs or the king of hearts?

Accepted Answer

Given data Two specific, distinct cards: Q♣ and K♥. Single draw from 52 equally likely cards. Concept/Approach Mutually exclusive events: P(Q♣ or K♥) = P(Q♣) + P(K♥). Step-by-step calculation Probability = 1/52 + 1/52 = 1/26 Verification/Alternative Count favourable outcomes = 2; total = 52 ⇒ 2/52 = 1/26. Common pitfalls Treating as non-exclusive or counting all queens and all kings. Final Answer 1/26

Question 7

What is the probability of getting a total of 9 from two throws of a die?

Accepted Answer

Given data Two throws (equivalently, two dice) ⇒ 36 ordered outcomes. Target sum = 9. Concept/Approach List the ordered pairs that sum to 9. Step-by-step calculation Favourable pairs: (3,6), (4,5), (5,4), (6,3) ⇒ 4 outcomesProbability = 4/36 = 1/9 Verification/Alternative Counts for sums follow the standard 1,2,3,4,5,6,5,4,3,2,1 pattern; sum 9 has 4 ways. Common pitfalls Missing order (e.g., counting (4,5) and (5,4) as one). Final Answer 1/9

Question 8

Two dice are thrown simultaneously. What is the probability that the product of the two numbers is even?

Accepted Answer

Given data Each die shows 1–6 with equal probability. Concept/Approach The product is even unless both outcomes are odd. Compute P(both odd) and subtract from 1. Step-by-step calculation P(odd on one die) = 3/6 = 1/2P(both odd) = (1/2) × (1/2) = 1/4P(product even) = 1 − 1/4 = 3/4 Verification/Alternative Complementary counting among 36 outcomes: 9 odd-odd pairs (3×3) ⇒ 36 − 9 = 27 even-product outcomes ⇒ 27/36 = 3/4. Common pitfalls Counting only pairs with at least one even face but forgetting total outcomes are ordered. Final Answer 3/4

Question 9

In a class with 15 boys and 10 girls, three students are chosen at random. What is the probability that exactly 1 girl and 2 boys are chosen?

Accepted Answer

Given data Boys = 15, Girls = 10, Total = 25. Selection size = 3; event = exactly 1 girl and 2 boys. Concept/Approach Use combinations without replacement: favorable ways ÷ total ways. Step-by-Step calculation Total ways = C(25, 3) = 25 × 24 × 23 ÷ 6 = 2300 Favorable ways = C(10, 1) × C(15, 2) = 10 × 105 = 1050 Probability = 1050 ÷ 2300 = 105 ÷ 230 = 21 ÷ 46 Verification/Alternative Hypergeometric check: P = (10÷25) × (15÷24) × (14÷23) × 3 (for the three placements of the girl) simplifies to 21÷46. Common pitfalls Choosing 2 girls and 1 boy by mistake. Using permutations instead of combinations. Final Answer 21/46

Question 10

From a standard deck of 52 cards, two cards are drawn together at random. What is the probability that both are kings?

Accepted Answer

Given data Total cards = 52, Kings = 4. Draw two cards together without replacement. Concept/Approach Probability = C(4, 2) ÷ C(52, 2). Step-by-Step calculation C(4, 2) = 6 C(52, 2) = 52 × 51 ÷ 2 = 1326 P = 6 ÷ 1326 = 1 ÷ 221 Verification/Alternative Sequential view: (4÷52) × (3÷51) = 12÷2652 = 1÷221. Common pitfalls Treating as with replacement (which would give 4÷52 × 4÷52). Final Answer 1/221

Question 11

A bag has 6 black and 8 white balls. One ball is drawn at random. What is the probability that it is white?

Accepted Answer

Given data Black = 6, White = 8, Total = 14. Concept/Approach Favorable outcomes ÷ total outcomes. Step-by-Step calculation P(white) = 8 ÷ 14 = 4 ÷ 7 Verification/Alternative Complement: 1 − P(black) = 1 − 6÷14 = 8÷14. Final Answer 4/7

Question 12

A box has 8 red, 7 blue, and 6 green balls. One ball is drawn at random. What is the probability that it is neither red nor green?

Accepted Answer

Given data Red = 8, Blue = 7, Green = 6; Total = 21. Concept/Approach Neither red nor green means blue. Step-by-Step calculation P(blue) = 7 ÷ 21 = 1 ÷ 3 Verification/Alternative Complement: 1 − P(red or green) = 1 − (8+6)÷21 = 1 − 14÷21 = 7÷21. Final Answer 1/3

Question 13

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

Accepted Answer

Given data Number of coins = 3; each fair (unbiased). Concept/Approach At most two heads = 1 − P(three heads). Step-by-Step calculation Total outcomes = 23 = 8 P(3 heads) = 1 ÷ 8 P(at most 2 heads) = 1 − 1÷8 = 7÷8 Verification/Alternative Direct sum: P(0H) + P(1H) + P(2H) = 1÷8 + 3÷8 + 3÷8 = 7÷8. Final Answer 7/8

Question 14

A bag contains 2 red, 3 green, and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

Accepted Answer

Given data Red balls = 2 Green balls = 3 Blue balls = 2 Total balls = 2 + 3 + 2 = 7 Draws = 2 (without replacement) Event: none of the drawn balls is blue (i.e., both drawn from only red/green set) Concept/Approach Use combinations (sampling without replacement). The favorable outcomes are all 2-ball selections from the non-blue balls. The total outcomes are all 2-ball selections from the full set. Probability = favorable ÷ total. Step-by-Step calculation Non-blue balls = red + green = 2 + 3 = 5 Favorable ways = C(5, 2) = 5 × 4 ÷ 2 = 10 Total ways = C(7, 2) = 7 × 6 ÷ 2 = 21 P(no blue) = Favorable ÷ Total = 10 ÷ 21 Verification/Alternative Sequential method: First draw non-blue with probability 5÷7; then, without replacement, 4 non-blue remain out of 6 total, so probability 4÷6. Multiply: (5÷7) × (4÷6) = 20÷42 = 10÷21 (order of the two non-blue balls does not matter, and the calculation already accounts for that because draws are successive without replacement). Complement check: P(at least one blue) = 1 − P(no blue) = 1 − 10÷21 = 11÷21. Common pitfalls Using permutations instead of combinations; order does not matter for selection probabilities. Treating draws as independent with replacement; here they are dependent without replacement. Forgetting that the event is both draws non-blue (not just the first). Final Answer 10/21

Question 15

In a lottery with 10 prizes and 25 blanks, a single ticket is drawn at random. What is the probability of getting a prize?

Accepted Answer

Given data Prizes = 10, Blanks = 25, Total tickets = 35. Concept/Approach Probability = favorable ÷ total. Step-by-Step calculation P(prize) = 10 ÷ 35 = 2 ÷ 7 Final Answer 2/7

Question 16

From the integers 1 to 16 (inclusive), a single number is selected uniformly at random. What is the probability that the chosen number is a prime?

Accepted Answer

Introduction / Context: Prime-number selection questions are standard in basic probability. We pick one number uniformly from a finite set and ask for the probability that it satisfies a property (being prime). The solution reduces to counting favourable outcomes over total outcomes. Given Data / Assumptions: Sample space: integers 1 through 16 inclusive. Each integer is equally likely. Primes are integers greater than 1 with exactly two positive divisors: 1 and itself. Concept / Approach: Probability = (number of favourable outcomes) / (total outcomes). We identify all primes in 1..16 and divide their count by 16. Step-by-Step Solution: List 1..16: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16.Primes in this range are 2, 3, 5, 7, 11, 13 → 6 primes.Total outcomes = 16; favourable = 6 → probability = 6/16 = 3/8. Verification / Alternative check: Double-check the non-primes: 1 (neither prime nor composite); 4,6,8,9,10,12,14,15,16 are composite. This confirms exactly 6 primes remain. Why Other Options Are Wrong: 1/16 assumes only one prime; 5/8 = 10/16 overcounts; 7/16 = 7 primes (incorrect, we have 6). Common Pitfalls: Erroneously treating 1 as prime; forgetting 9 and 15 are composite (divisible by 3). Final Answer: 3/8

Question 17

An urn has 9 red, 7 white, and 4 black balls (total 20). If one ball is drawn at random, what is the probability that it is not red?

Accepted Answer

Introduction / Context: This is a direct single-draw probability problem with a finite equally likely sample space. We want the chance that the colour of the drawn ball is not red. Given Data / Assumptions: Red: 9, White: 7, Black: 4. Total balls = 9 + 7 + 4 = 20. Single draw, each ball equally likely. Concept / Approach: P(not red) = 1 − P(red) = (non-red count) / (total count). Non-red balls are white + black. Step-by-Step Solution: Non-red balls = 7 + 4 = 11.Total = 20.Probability = 11/20. Verification / Alternative check: Compute P(red) = 9/20, then 1 − 9/20 = 11/20, consistent. Why Other Options Are Wrong: 1/11 and 2/11 invert counts; 9/20 is P(red); 9/20 ≠ P(not red). Common Pitfalls: Forgetting to add both white and black when counting non-red outcomes. Final Answer: 11/20

Question 18

A lottery has 20 prizes and 15 blanks (total 35 tickets). If one ticket is drawn at random, what is the probability of getting a prize?

Accepted Answer

Introduction / Context: We compute a basic probability from a finite set of equally likely outcomes. Each ticket is equally likely, and we count favourable outcomes (prizes). Given Data / Assumptions: Prizes: 20, blanks: 15. Total tickets = 35. Single random draw. Concept / Approach: Probability = favourable/total = 20/35. Reduce the fraction to simplest form. Step-by-Step Solution: Favourable outcomes (prizes) = 20.Total outcomes = 35.Probability = 20/35 = 4/7 after dividing numerator and denominator by 5. Verification / Alternative check: Compute P(blank) = 15/35 = 3/7. Then P(prize) = 1 − 3/7 = 4/7, consistent. Why Other Options Are Wrong: 1/10 and 2/7 are too small; 2/5 = 14/35 undercounts prizes. Common Pitfalls: Not simplifying the fraction or mixing up the counts for prizes and blanks. Final Answer: 4/7

Question 19

If the odds in favour of an event are 3:5, what is the probability that the event occurs?

Accepted Answer

Introduction / Context: “Odds in favour” and “probability” are related but not identical. Converting between the two is a common basic skill in probability. Given Data / Assumptions: Odds in favour = a:b means success : failure = a : b. Here a = 3, b = 5. Concept / Approach: If odds in favour are a:b, then P(success) = a / (a + b) and P(failure) = b / (a + b). Step-by-Step Solution: P = 3 / (3 + 5) = 3/8. Verification / Alternative check: Corresponding odds against would be 5:3; then P = 1 / (1 + (5/3)) = 3/8, same result. Why Other Options Are Wrong: 3/5 uses only the “for” part; 1/3 and 1/5 are unrelated to 3:(something) totals. Common Pitfalls: Confusing 3/5 with probability; odds are a ratio of counts, not a direct probability. Final Answer: 3/8

Question 20

A fair coin is tossed twice. What is the probability of getting heads on both tosses?

Accepted Answer

Introduction / Context: Independent trials with a fair coin lead to a product rule for probabilities: multiply the probabilities of independent events to get the probability of their joint occurrence. Given Data / Assumptions: Coin is fair: P(H) = 1/2, P(T) = 1/2. Two independent tosses. Concept / Approach: P(H on toss 1 and H on toss 2) = P(H) * P(H) because tosses are independent. Step-by-Step Solution: P = (1/2) * (1/2) = 1/4. Verification / Alternative check: Enumerate outcomes: {HH, HT, TH, TT}. Exactly one outcome (HH) out of four gives two heads → 1/4. Why Other Options Are Wrong: 1/2 is for “at least one head” in two tosses (not exactly both); 3/4 is “not both tails”. Common Pitfalls: Adding probabilities instead of multiplying for independent joint events. Final Answer: 1/4

Probability Questions