Probability Questions Placement
Probability Questions for Placement Exams
Last Updated: March 2026
Introduction to Probability
Probability is a fundamental topic in quantitative aptitude that measures the likelihood of an event occurring. It plays a crucial role in placement exams for companies like TCS, Infosys, Wipro, Cognizant, Amazon, Microsoft, and Google, as well as in sarkari exams such as SSC CGL, Bank PO, Railway, and UPSC CSAT.
Why Probability is Important:
- Tests logical reasoning and analytical thinking
- Applications in data science, machine learning, and risk analysis
- Frequently asked in technical and non-technical rounds
- Foundation for advanced topics in statistics and mathematics
Important Formulas and Concepts
Basic Probability Formula
$$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$
Range of Probability
$$0 \leq P(E) \leq 1$$
Types of Events
| Event Type | Description | Formula |
|---|---|---|
| Sure Event | Always occurs | P(E) = 1 |
| Impossible Event | Never occurs | P(E) = 0 |
| Complementary Event | Event not occurring | P(E') = 1 - P(E) |
| Mutually Exclusive | Cannot occur together | P(A ∩ B) = 0 |
| Independent Events | One doesn't affect other | P(A ∩ B) = P(A) × P(B) |
Key Formulas
Addition Rule:
- P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
- For mutually exclusive: P(A ∪ B) = P(A) + P(B)
Multiplication Rule:
- P(A ∩ B) = P(A) × P(B|A)
- For independent events: P(A ∩ B) = P(A) × P(B)
Conditional Probability: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
Bayes' Theorem: $$P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}$$
Combinatorial Formulas
- Permutations: $P(n,r) = \frac{n!}{(n-r)!}$
- Combinations: $C(n,r) = \frac{n!}{r!(n-r)!}$
Shortcut Tips
- At least one = 1 - P(none)
- Neither A nor B = P(A' ∩ B') = 1 - P(A ∪ B)
- Either A or B but not both = P(A) + P(B) - 2P(A ∩ B)
- For dice: Sum probabilities follow a symmetric pattern (2→1/36, 3→2/36...7→6/36...12→1/36)
Practice Questions
Level 1: Easy
Question 1: Basic Coin Toss What is the probability of getting at least one head when two coins are tossed?
<details> <summary>Solution</summary>Total outcomes = {HH, HT, TH, TT} = 4 Favorable outcomes (at least one head) = {HH, HT, TH} = 3
$$P(\text{at least one head}) = \frac{3}{4}$$
Shortcut: P(at least one head) = 1 - P(no heads) = 1 - 1/4 = 3/4
</details>Question 2: Single Die Roll What is the probability of getting a number greater than 4 when a fair die is rolled?
<details> <summary>Solution</summary>Total outcomes = {1, 2, 3, 4, 5, 6} = 6 Favorable outcomes = {5, 6} = 2
$$P(\text{> 4}) = \frac{2}{6} = \frac{1}{3}$$
</details>Question 3: Card Probability What is the probability of drawing a king from a standard deck of 52 cards?
<details> <summary>Solution</summary>Total cards = 52 Number of kings = 4
$$P(\text{King}) = \frac{4}{52} = \frac{1}{13}$$
</details>Question 4: Complementary Probability If the probability of passing an exam is 0.7, what is the probability of failing?
<details> <summary>Solution</summary>$$P(\text{fail}) = 1 - P(\text{pass}) = 1 - 0.7 = 0.3$$
</details>Question 5: Marble Selection A bag contains 5 red, 3 blue, and 2 green marbles. What is the probability of drawing a red marble?
<details> <summary>Solution</summary>Total marbles = 5 + 3 + 2 = 10 Red marbles = 5
$$P(\text{Red}) = \frac{5}{10} = \frac{1}{2}$$
</details>Question 6: Simple AND Probability Two coins are tossed. What is the probability of getting heads on both?
<details> <summary>Solution</summary>P(Head on first coin) = 1/2 P(Head on second coin) = 1/2
Since independent: $$P(\text{HH}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$
</details>Question 7: Odd/Even Die What is the probability of getting an odd number when rolling a fair die?
<details> <summary>Solution</summary>Odd numbers = {1, 3, 5} = 3 Total outcomes = 6
$$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$$
</details>Question 8: Face Cards What is the probability of drawing a face card (Jack, Queen, King) from a deck?
<details> <summary>Solution</summary>Face cards per suit = 3 (J, Q, K) Total face cards = 3 × 4 = 12
$$P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$$
</details>Question 9: Ball Selection Without Replacement A box has 4 white and 6 black balls. What is the probability of drawing a white ball first?
<details> <summary>Solution</summary>Total balls = 10 White balls = 4
$$P(\text{white}) = \frac{4}{10} = \frac{2}{5}$$
</details>Question 10: Sum of Two Dice What is the probability of getting a sum of 7 when two dice are rolled?
<details> <summary>Solution</summary>Total outcomes = 6 × 6 = 36 Favorable combinations for sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways
$$P(\text{sum = 7}) = \frac{6}{36} = \frac{1}{6}$$
</details>Level 2: Medium
Question 11: Conditional Probability - Cards Two cards are drawn from a deck without replacement. What is the probability that both are aces?
<details> <summary>Solution</summary>P(First card is ace) = 4/52 = 1/13 P(Second card is ace | First was ace) = 3/51 = 1/17
$$P(\text{both aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$$
</details>Question 12: At Least One Three coins are tossed. What is the probability of getting at least one tail?
<details> <summary>Solution</summary>Shortcut Method: P(at least one tail) = 1 - P(no tail) = 1 - P(all heads)
P(all heads) = (1/2)³ = 1/8
$$P(\text{at least one tail}) = 1 - \frac{1}{8} = \frac{7}{8}$$
</details>Question 13: Committee Selection A committee of 3 is to be formed from 5 men and 4 women. What is the probability that the committee has at least one woman?
<details> <summary>Solution</summary>Total ways to select 3 from 9 = C(9,3) = 84 Ways with no women (all men) = C(5,3) = 10
P(at least one woman) = 1 - P(no women) $$= 1 - \frac{10}{84} = 1 - \frac{5}{42} = \frac{37}{42}$$
</details>Question 14: Mutually Exclusive Events The probability that A solves a problem is 1/2 and B solves it is 1/3. What is the probability that at least one of them solves the problem?
<details> <summary>Solution</summary>P(A solves) = 1/2, P(B solves) = 1/3 Assuming independence:
P(at least one solves) = 1 - P(none solves) = 1 - (1/2) × (2/3) = 1 - 1/3 = 2/3
</details>Question 15: Lottery Problem In a lottery with 100 tickets, 5 are prizes. If a person buys 4 tickets, what is the probability of winning at least one prize?
<details> <summary>Solution</summary>Total ways to choose 4 tickets = C(100,4) Ways to choose 4 non-prize tickets = C(95,4)
P(at least one prize) = 1 - P(no prizes) $$= 1 - \frac{C(95,4)}{C(100,4)}$$
$$= 1 - \frac{95 \times 94 \times 93 \times 92}{100 \times 99 \times 98 \times 97}$$
$$= 1 - \frac{77311680}{94109400} ≈ 1 - 0.821 = 0.179$$
</details>Question 16: Letter Arrangement What is the probability that in a random arrangement of the letters of the word "PROBABILITY", the two B's are together?
<details> <summary>Solution</summary>Total letters = 11 (P,R,O,B,A,B,I,L,I,T,Y) Repeated: B appears twice, I appears twice
Total arrangements = 11!/(2!×2!)
Treat two B's as one unit: 10 units to arrange Arrangements with B's together = 10!/2!
$$P(\text{B's together}) = \frac{10!/2!}{11!/(2!×2!)} = \frac{10! × 2! × 2!}{2! × 11!} = \frac{2}{11}$$
</details>Question 17: Defective Items A box contains 10 items, 3 of which are defective. If 4 items are selected at random, what is the probability that exactly 2 are defective?
<details> <summary>Solution</summary>Ways to choose 2 defective from 3 = C(3,2) = 3 Ways to choose 2 good from 7 = C(7,2) = 21 Total favorable = 3 × 21 = 63
Total ways to choose 4 from 10 = C(10,4) = 210
$$P(\text{exactly 2 defective}) = \frac{63}{210} = \frac{3}{10}$$
</details>Question 18: Birthday Problem (Simplified) What is the probability that in a group of 3 people, at least two have the same birth month?
<details> <summary>Solution</summary>P(at least two same) = 1 - P(all different months)
P(all different) = (12/12) × (11/12) × (10/12) = 110/144 = 55/72
$$P(\text{at least two same}) = 1 - \frac{55}{72} = \frac{17}{72}$$
</details>Question 19: Spinner Probability A spinner has numbers 1 to 8. What is the probability of getting an even number or a number greater than 5?
<details> <summary>Solution</summary>Even numbers: {2, 4, 6, 8} Numbers > 5: {6, 7, 8} Union: {2, 4, 6, 7, 8} = 5 outcomes
$$P(\text{even or > 5}) = \frac{5}{8}$$
Using formula: P(even) = 4/8, P(>5) = 3/8, P(even and >5) = 2/8 P(even or >5) = 4/8 + 3/8 - 2/8 = 5/8
</details>Question 20: Sequential Drawing An urn contains 5 red and 3 black balls. Two balls are drawn one after another without replacement. What is the probability that the first is red and the second is black?
<details> <summary>Solution</summary>P(First red) = 5/8 P(Second black | First red) = 3/7
$$P(\text{Red then Black}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$$
</details>Level 3: Hard
Question 21: Complex Card Problem Three cards are drawn from a deck. What is the probability that they are all of different suits?
<details> <summary>Solution</summary>Total ways to draw 3 cards = C(52,3) = 22100
For different suits: Choose 3 suits from 4 = C(4,3) = 4 From each chosen suit, pick 1 card: 13 × 13 × 13 = 2197
Favorable outcomes = 4 × 2197 = 8788
$$P(\text{different suits}) = \frac{8788}{22100} = \frac{2197}{5525} = \frac{169}{425}$$
</details>Question 22: Bayes' Theorem Application A factory has three machines A, B, and C producing 30%, 45%, and 25% of total output respectively. Their defective rates are 2%, 3%, and 4% respectively. If an item is selected at random and found to be defective, what is the probability it was produced by machine B?
<details> <summary>Solution</summary>Using Bayes' Theorem:
P(A) = 0.30, P(B) = 0.45, P(C) = 0.25 P(D|A) = 0.02, P(D|B) = 0.03, P(D|C) = 0.04
P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) = 0.02(0.30) + 0.03(0.45) + 0.04(0.25) = 0.006 + 0.0135 + 0.010 = 0.0295
$$P(B|D) = \frac{P(D|B) \times P(B)}{P(D)} = \frac{0.03 \times 0.45}{0.0295} = \frac{0.0135}{0.0295} = \frac{27}{59}$$
</details>Question 23: Circular Arrangement 8 people sit around a circular table. What is the probability that two specific people sit together?
<details> <summary>Solution</summary>Total arrangements in circle = (8-1)! = 7!
Treat the two specific people as one unit: 7 units Arrangements in circle = (7-1)! = 6! The two people can arrange in 2 ways
Favorable = 2 × 6!
$$P(\text{together}) = \frac{2 \times 6!}{7!} = \frac{2 \times 720}{5040} = \frac{2}{7}$$
</details>Question 24: Dice Sum Conditions Three dice are thrown. What is the probability that the sum is at least 16?
<details> <summary>Solution</summary>Total outcomes = 6³ = 216
Sum ≥ 16 means sum = 16, 17, or 18
Sum = 18: (6,6,6) → 1 way Sum = 17: (6,6,5) and permutations → 3 ways Sum = 16: (6,6,4), (6,5,5) and permutations → 3 + 3 = 6 ways
Total favorable = 1 + 3 + 6 = 10
$$P(\text{sum ≥ 16}) = \frac{10}{216} = \frac{5}{108}$$
</details>Question 25: Geometric Probability Two numbers x and y are chosen at random from the interval [0,1]. What is the probability that x + y < 1 and xy < 3/16?
<details> <summary>Solution</summary>Sample space is unit square with area = 1
For x + y < 1: below the line x + y = 1 (triangle with area 1/2)
For xy < 3/16 in region x + y < 1: The curve xy = 3/16 intersects x + y = 1 at points where: x(1-x) = 3/16 → 16x - 16x² = 3 → 16x² - 16x + 3 = 0 x = (16 ± √(256-192))/32 = (16 ± 8)/32 = 3/4 or 1/4
Area below xy = 3/16 in the triangle = ∫[1/4 to 3/4] (1-x - 3/(16x)) dx = [x - x²/2 - (3/16)ln|x|] from 1/4 to 3/4
= (3/4 - 9/32 - (3/16)ln(3/4)) - (1/4 - 1/32 - (3/16)ln(1/4)) = (15/32 + (3/16)ln(4/3)) - (7/32 + (3/16)ln(4)) = 8/32 + (3/16)(ln(4/3) - ln(4)) = 1/4 + (3/16)ln(1/3) = 1/4 - (3/16)ln(3)
</details>Question 26: Complex Selection From a group of 6 boys and 4 girls, a committee of 5 is to be formed. What is the probability that it contains more boys than girls?
<details> <summary>Solution</summary>Total ways = C(10,5) = 252
More boys than girls means:
- 3 boys, 2 girls: C(6,3) × C(4,2) = 20 × 6 = 120
- 4 boys, 1 girl: C(6,4) × C(4,1) = 15 × 4 = 60
- 5 boys, 0 girls: C(6,5) × C(4,0) = 6 × 1 = 6
Favorable = 120 + 60 + 6 = 186
$$P(\text{more boys}) = \frac{186}{252} = \frac{31}{42}$$
</details>Question 27: Game Theory Probability A and B alternately throw a pair of dice. A wins if he throws a sum of 6 before B throws a sum of 7. If A starts, what is his probability of winning?
<details> <summary>Solution</summary>P(A throws 6) = 5/36, P(B throws 7) = 6/36 = 1/6 P(A doesn't throw 6) = 31/36, P(B doesn't throw 7) = 5/6
Let P = probability that A wins
A can win on first throw, or if both miss and then A wins: P = 5/36 + (31/36)(5/6)P
P(1 - 155/216) = 5/36 P(61/216) = 5/36 P = (5/36) × (216/61) = 30/61
</details>Question 28: Restricted Permutations The letters of the word 'MISSISSIPPI' are arranged at random. What is the probability that all four S's are separated?
<details> <summary>Solution</summary>Total letters: 11 (M-1, I-4, S-4, P-2) Total arrangements = 11!/(4!×4!×2!)
For S's to be separated: First arrange non-S letters: M, I, I, I, I, P, P Number of arrangements = 7!/(4!×2!) = 105
This creates 8 gaps (including ends): M_I_I_I_I_P_P Choose 4 gaps from 8 for S's: C(8,4) = 70
Favorable arrangements = 105 × 70 = 7350
$$P(\text{S's separated}) = \frac{7350 \times 4! \times 4! \times 2!}{11!}$$
= (7350 × 24 × 24 × 2) / 39916800 = 8467200 / 39916800 = 7/33
</details>Question 29: Multiple Condition Probability Four cards are drawn from a deck. What is the probability that there is at least one card from each suit?
<details> <summary>Solution</summary>Total ways = C(52,4) = 270725
For at least one from each suit with 4 cards: exactly 2 cards from one suit and 1 from each of the other three.
Choose which suit has 2 cards: 4 ways Choose 2 cards from that suit: C(13,2) = 78 Choose 1 from each remaining suit: 13 × 13 × 13 = 2197
Favorable = 4 × 78 × 2197 = 685464
$$P = \frac{685464}{270725} = \frac{685464}{270725}$$
Simplifying: = 685464/270725 = 685464 ÷ 270725 ≈ 2.53 (error in calculation)
Correct approach: Actually, let's recalculate: C(52,4) = 270725
4 × C(13,2) × 13³ = 4 × 78 × 2197 = 685464
But this exceeds total outcomes! Error: We need exactly one suit with 2 cards.
Correct: 4 × C(13,2) × 13 × 13 × 13 / C(52,4) = 685464/270725
This gives > 1, so error. Let me recheck C(52,4): 52×51×50×49/(4×3×2×1) = 6497400/24 = 270725 ✓
The issue is in the problem - with only 4 cards and 4 suits, getting "at least one from each" means exactly one from each suit (not 2 from one suit).
Correct calculation for exactly one from each suit: 13⁴ = 28561
$$P = \frac{28561}{270725} = \frac{2197}{20825}$$
</details>Question 30: Advanced Conditional Probability In a multiple-choice test with 5 questions, each with 4 options (one correct), a student either knows the answer or guesses. The probability that the student knows an answer is 2/3. If the student gets a question correct, what is the probability that they actually knew the answer?
<details> <summary>Solution</summary>For a single question (using Bayes' theorem):
P(Knows) = 2/3, P(Guesses) = 1/3 P(Correct | Knows) = 1 P(Correct | Guesses) = 1/4
P(Correct) = P(Correct|Knows)P(Knows) + P(Correct|Guesses)P(Guesses) = 1 × (2/3) + (1/4) × (1/3) = 2/3 + 1/12 = 8/12 + 1/12 = 9/12 = 3/4
$$P(\text{Knows} | \text{Correct}) = \frac{P(\text{Correct} | \text{Knows}) \times P(\text{Knows})}{P(\text{Correct})}$$
$$= \frac{1 \times (2/3)}{3/4} = \frac{2/3}{3/4} = \frac{2}{3} \times \frac{4}{3} = \frac{8}{9}$$
</details>Companies & Exams Asking Probability
Top Companies
- TCS, Infosys, Wipro, Cognizant - Basic to medium level
- Amazon, Microsoft, Google - Medium to hard, application-based
- Goldman Sachs, JP Morgan - Hard, focus on conditional probability
- Mu Sigma, Fractal - Data science oriented problems
Government Exams
- SSC CGL/CHSL - Moderate difficulty, combinatorics-based
- Bank PO (SBI, IBPS) - Card and ball problems, moderate
- Railway Exams - Basic probability concepts
- UPSC CSAT - Logical probability questions
- State PSCs - Mixed difficulty
Preparation Tips
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Master the Basics: Ensure you understand fundamental concepts like sample space, events, and probability axioms before moving to complex problems.
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Practice Combinatorics: Most probability problems require counting techniques. Master permutations and combinations thoroughly.
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Learn Shortcut Methods: Use complementary probability (1 - P(not E)) for "at least" problems to save time.
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Visualize with Diagrams: For conditional probability, draw tree diagrams or Venn diagrams to understand the problem better.
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Categorize Problem Types: Group problems by type (cards, dice, balls, arrangements) to identify patterns quickly.
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Time Management: In exams, don't spend more than 2-3 minutes on a probability question. If stuck, mark and move on.
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Review Mistakes: Maintain an error log to track commonly missed concepts and review them regularly.
Frequently Asked Questions (FAQs)
Q1: Is probability difficult to learn for placement exams?
Probability builds on basic counting principles. With systematic practice of 2-3 weeks, most students can handle medium-level questions confidently. Focus on understanding rather than memorizing formulas.
Q2: How many probability questions typically appear in placement exams?
Most companies include 2-4 probability questions in their aptitude tests. Product-based companies and consulting firms tend to ask more challenging problems compared to service-based companies.
Q3: What is the most common mistake in probability problems?
The most common error is counting the sample space incorrectly or missing some favorable outcomes. Always double-check whether order matters (permutations) or not (combinations).
Q4: Should I use Venn diagrams for solving probability problems?
Venn diagrams are excellent for problems involving union, intersection, and conditional probability. They help visualize relationships between events and reduce calculation errors.
Q5: How is probability different in technical vs HR interviews?
Technical interviews focus on calculation-based problems (cards, dice, expected value). HR interviews may use probability puzzles to assess logical thinking and problem-solving approach under pressure.
Best of luck with your placement preparation! Practice consistently and you'll master probability in no time.