Probability Solved Examples 2026 (40+ Problems with Solutions)

Last Updated: June 2026

This page collects 40+ solved probability problems at placement-exam difficulty, organized from basics through dice, coins, cards, balls, and conditional cases. Each problem has a full step-by-step solution. Candidates report probability as a recurring quantitative topic in TCS, Infosys, and Wipro rounds, and one where careful counting of the sample space decides the marks.

Formula Reference

Section A: Dice (Problems 1-10)

Problem 1. A die is rolled. Probability of getting an even number? Solution: Favourable 2, 4, 6 = 3 of 6 = 1/2.

Problem 2. A die is rolled. Probability of a number greater than 4? Solution: 5, 6 = 2 of 6 = 1/3.

Problem 3. Two dice are rolled. Probability the sum is 7? Solution: 6 outcomes of 36 = 1/6.

Problem 4. Two dice are rolled. Probability the sum is 9? Solution: (3,6),(4,5),(5,4),(6,3) = 4 of 36 = 1/9.

Problem 5. Two dice are rolled. Probability of a doublet? Solution: 6 doublets of 36 = 1/6.

Problem 6. Two dice are rolled. Probability the sum is at least 10? Solution: Sums 10, 11, 12 have 3 + 2 + 1 = 6 outcomes of 36 = 1/6.

Problem 7. A die is rolled. Probability of a prime number? Solution: 2, 3, 5 = 3 of 6 = 1/2.

Problem 8. Two dice are rolled. Probability the sum is a prime number? Solution: Prime sums 2,3,5,7,11 have 1+2+4+6+2 = 15 of 36 = 5/12.

Problem 9. Two dice are rolled. Probability that both show the same number? Solution: 6 of 36 = 1/6.

Problem 10. Two dice are rolled. Probability the product is even? Solution: Product is odd only when both are odd: 9 outcomes. Even product = 36 - 9 = 27 of 36 = 3/4.

Section B: Coins (Problems 11-18)

Problem 11. A coin is tossed. Probability of a head? Solution: 1/2.

Problem 12. Two coins are tossed. Probability of exactly one head? Solution: HT, TH = 2 of 4 = 1/2.

Problem 13. Two coins are tossed. Probability of at least one head? Solution: 1 - P(no head) = 1 - 1/4 = 3/4.

Problem 14. Three coins are tossed. Probability of exactly two heads? Solution: HHT, HTH, THH = 3 of 8.

Problem 15. Three coins are tossed. Probability of at least two heads? Solution: Two heads (3) plus three heads (1) = 4 of 8 = 1/2.

Problem 16. Three coins are tossed. Probability of no head? Solution: TTT = 1 of 8.

Problem 17. Four coins are tossed. Probability of all heads? Solution: 1/16.

Problem 18. Three coins are tossed. Probability of at least one tail? Solution: 1 - P(all heads) = 1 - 1/8 = 7/8.

Section C: Cards (Problems 19-28)

Problem 19. A card is drawn from 52. Probability it is a king? Solution: 4 of 52 = 1/13.

Problem 20. A card is drawn. Probability it is a heart? Solution: 13 of 52 = 1/4.

Problem 21. A card is drawn. Probability it is a face card? Solution: 12 of 52 = 3/13.

Problem 22. A card is drawn. Probability it is a king or a queen? Solution: 8 of 52 = 2/13.

Problem 23. A card is drawn. Probability it is a red card? Solution: 26 of 52 = 1/2.

Problem 24. A card is drawn. Probability it is a king or a heart? Solution: 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

Problem 25. A card is drawn. Probability it is an ace of spades? Solution: 1 of 52.

Problem 26. Two cards are drawn without replacement. Probability both are aces? Solution: (4/52) × (3/51) = 12/2652 = 1/221.

Problem 27. A card is drawn. Probability it is neither a king nor a queen? Solution: 1 - 8/52 = 44/52 = 11/13.

Problem 28. A card is drawn. Probability it is a number card (2 to 10)? Solution: 9 ranks × 4 suits = 36 of 52 = 9/13.

Section D: Balls and Conditional (Problems 29-42)

Problem 29. A bag has 5 red and 3 blue balls. One is drawn. Probability it is red? Solution: 5 of 8 = 5/8.

Problem 30. A bag has 4 red and 6 green balls. Two are drawn. Probability both are red? Solution: (4/10) × (3/9) = 12/90 = 2/15.

Problem 31. A bag has 4 red and 6 green. Two are drawn. Probability both are green? Solution: (6/10) × (5/9) = 30/90 = 1/3.

Problem 32. A bag has 3 red, 4 blue, 5 green. One is drawn. Probability it is not blue? Solution: Not blue = 8 of 12 = 2/3.

Problem 33. A bag has 5 white and 5 black. Two drawn. Probability one of each colour? Solution: (5/10 × 5/9) + (5/10 × 5/9) = 25/90 + 25/90 = 50/90 = 5/9.

Problem 34. A bag has 6 red and 4 blue. One is drawn and replaced, then another drawn. Probability both red? Solution: With replacement: (6/10) × (6/10) = 36/100 = 9/25.

Problem 35. A bag has 7 red and 3 blue. Probability of drawing a blue? Solution: 3 of 10 = 3/10.

Problem 36. Two balls drawn from 3 red and 2 blue. Probability at least one blue? Solution: 1 - P(no blue) = 1 - (3/5 × 2/4) = 1 - 6/20 = 14/20 = 7/10.

Problem 37. A number is chosen from 1 to 20. Probability it is a multiple of 3? Solution: 3, 6, 9, 12, 15, 18 = 6 of 20 = 3/10.

Problem 38. A number from 1 to 50. Probability it is a perfect square? Solution: 1, 4, 9, 16, 25, 36, 49 = 7 of 50.

Problem 39. A letter is chosen from the word PROBABILITY. Probability it is a vowel? Solution: Vowels O, A, I, I = 4 of 11.

Problem 40. Two dice are rolled. Given the sum is even, probability the sum is 8? Solution: Even sums total 18 outcomes; sum 8 has 5 outcomes. P = 5/18.

Problem 41. A bag has 2 red, 3 blue, 5 green. Probability the drawn ball is red or blue? Solution: (2 + 3) of 10 = 5/10 = 1/2.

Problem 42. Three coins are tossed. Given at least one head, probability of exactly two heads? Solution: At least one head = 7 outcomes; exactly two heads = 3 outcomes. P = 3/7.

The Three Probability Methods You Must Master

Most placement probability questions yield to one of three methods, and recognizing which to apply is the core skill.

The first method is direct counting: count the favourable outcomes and divide by the total sample space. This works for single events like one die roll or one card draw. The discipline here is counting the denominator correctly. For two dice the sample space is 36, not 12; for two cards drawn the total combinations are 52 times 51 if order matters or 52 choose 2 if it does not. Getting the denominator right is where most marks are won or lost.

The second method is the complement: for any "at least one" question, compute the probability of none happening and subtract from one. Listing all the "at least one" cases is slow and error-prone, while the "none" case is usually a single product. For at least one head in three coin tosses, the none case is all tails with probability one-eighth, so the answer is seven-eighths in one step.

The third method is the multiplication rule for sequential events. When two or more events must all happen, multiply their probabilities, reducing the sample space after each draw if there is no replacement. For two aces drawn without replacement, the first ace has probability four-fifty-seconds and the second three-fifty-firsts, multiplying to one over 221. The key judgment is whether the draws are independent (with replacement) or dependent (without replacement), since that changes the second factor.

Common Probability Traps

The most frequent error is miscounting the sample space, especially undercounting two-dice outcomes as 12 instead of 36 or forgetting that order can double the count. The second trap is adding probabilities of events that overlap without subtracting the intersection, which double counts the shared outcomes; the king-or-heart question requires subtracting the king of hearts once. The third trap is treating dependent draws as independent by forgetting to reduce the denominator after a without-replacement draw. The fourth trap is brute-forcing an "at least one" question instead of using the complement. Internalizing these four traps, alongside the three methods above, makes most placement probability questions routine.

Key Takeaways

Candidates report that the most common probability error is miscounting the total sample space, especially with two dice (36 outcomes) and card overlaps. Count the denominator carefully, reach for the complement whenever "at least one" appears, and most probability questions resolve cleanly.

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Frequently Asked Questions

What is the basic probability formula?

Probability of an event equals the number of favourable outcomes divided by the total number of equally likely outcomes. The value always lies between 0 and 1. Candidates report that correctly counting the total sample space is where most errors happen, so listing outcomes carefully is the key habit.

How do I find the probability of at least one event?

Use the complement: the probability of at least one occurrence equals one minus the probability of no occurrence. This is almost always faster than adding the separate cases. Candidates report the complement method is the single most useful probability shortcut in placement tests.

When do I add probabilities versus multiply them?

Add when you want the probability of either of two mutually exclusive events. Multiply when you want both of two independent events to happen. For overlapping events, use the addition rule with the intersection subtracted to avoid double counting.