Probability Questions FOR Placement

Meta Description: Master Probability questions for placement with step-by-step solutions, essential formulas, shortcuts, and PYQs from top IT companies.

Introduction

If you're preparing for campus drives or off-campus recruitment drives, mastering Probability questions for placement is non-negotiable. This topic forms the backbone of quantitative aptitude sections, testing logical reasoning, counting fundamentals, and numerical accuracy. Nearly every major recruiter, including TCS, Infosys, Wipro, Cognizant, and Accenture, includes at least 2-4 probability questions in their aptitude rounds. The difficulty spectrum is wide: beginners face straightforward dice and card problems, while advanced rounds test conditional probability, independent events, and multi-step combinatorial logic. This guide is structured to take you from foundational concepts to placement-ready proficiency, with 23 fully solved problems covering every exam pattern you’ll encounter in 2026.

Key Formulas & Concepts

Probability measures the likelihood of an event occurring. Below are the core formulas you must memorize:

1. Basic Probability:
   P(E) = n(E) / n(S)
   Where n(E) = number of favorable outcomes, n(S) = total possible outcomes in sample space.

2. Complementary Rule:
   P(not E) = 1 - P(E)
   Highly useful when calculating "at least one" or "none" scenarios.

3. Addition Rule (Union):
   P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
   Subtracts intersection to avoid double-counting overlapping events.

4. Conditional Probability:
   P(A | B) = P(A ∩ B) / P(B), where P(B) > 0
   Probability of A occurring given that B has already occurred.

5. Multiplication Rule:
   P(A ∩ B) = P(A) × P(B | A)
   Sequential or dependent events. For independent events: P(A ∩ B) = P(A) × P(B)

6. Independent Events:
   Occurrence of one event doesn't affect the other.
   Test: If P(A ∩ B) = P(A) × P(B), events A and B are independent.

7. Mutually Exclusive Events:
   Events cannot occur simultaneously: P(A ∩ B) = 0.

Contextual Basics:

• Dice: Single die → n(S) = 6. Two dice → n(S) = 36.
• Cards: Standard deck → 52 cards (4 suits × 13 ranks). 12 face cards, 4 aces.
• Balls/Bags: Use combinations ⁿCᵣ = n! / [r!(n-r)!] for "without replacement" draws.

Solved Examples (Basic Level)

Q1. A fair coin is tossed twice. What is the probability of getting exactly one head?
Sample space S = {HH, HT, TH, TT} → n(S) = 4
Favorable (exactly 1 head) = {HT, TH} → n(E) = 2
P(E) = 2/4 = 1/2

Q2. A single die is rolled. Find the probability of obtaining a number greater than 4.
S = {1,2,3,4,5,6}
Favorable: {5,6} → 2 outcomes
P = 2/6 = 1/3

Q3. A card is drawn at random from a standard deck. What is the probability of drawing a King or a Queen?
P(K) = 4/52, P(Q) = 4/52. Kings and Queens don’t overlap.
P(King ∪ Queen) = 4/52 + 4/52 = 8/52 = 2/13

Q4. A bag contains 5 red and 3 blue balls. One ball is drawn at random. Probability it is red?
Total balls = 8. Red = 5.
P(Red) = 5/8

Q5. Two dice are thrown. What is the probability of getting a sum of 11?
Total outcomes = 36
Sum = 11 → {(5,6), (6,5)} = 2 outcomes
P = 2/36 = 1/18

Practice Questions (Medium Level)

Q1. From a bag containing 6 white and 4 black balls, two balls are drawn without replacement. Find the probability both are white.
P(1st White) = 6/10 = 3/5
After drawing 1 white: 5 white, 9 total remain.
P(2nd White | 1st White) = 5/9
P(Both) = (3/5) × (5/9) = 1/3

Q2. Find the probability of drawing an Ace or a heart from a standard deck.
P(Ace) = 4/52, P(Heart) = 13/52, P(Ace of Hearts) = 1/52
P(A ∪ H) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13

Q3. If P(A) = 0.6 and P(B) = 0.5, and A, B are independent events, find P(A ∩ B).
For independent: P(A ∩ B) = P(A) × P(B) = 0.6 × 0.5 = 0.30

Q4. A bag has 3 green, 4 yellow, and 5 red marbles. One is drawn. Given it’s not green, what’s the probability it’s yellow? (Conditional)
"Not green" → 4Y + 5R = 9 possible. Yellow favorable = 4.
P(Y | not G) = 4/9 = 0.444

Q5. Two coins are tossed. Given that at least one is heads, find the probability both are heads.
Sample space for "at least one head" = {HH, HT, TH} → 3 outcomes
Favorable "both heads" = {HH} → 1 outcome
P = 1/3

Q6. A die is rolled and a card is drawn from a deck. Find probability of rolling an even number AND drawing a spade.
P(Even die) = 3/6 = 1/2. P(Spade) = 13/52 = 1/4.
Events independent → (1/2) × (1/4) = 1/8

Q7. Find the probability of getting a total of 9 when two fair dice are rolled.
Pairs: (3,6), (4,5), (5,4), (6,3) → 4 outcomes.
Total outcomes = 36 → P = 4/36 = 1/9

Q8. From a deck of 52 cards, what is the probability that the drawn card is a face card or a number greater than 8 but less than 12?
Face cards = 12. Numbers 9, 10 → 4 each × 4 suits = 8 cards. Overlap? None (9,10 are not face).
Total favorable = 12 + 8 = 20 → P = 20/52 = 5/13

Tricky Questions (Advanced Level)

Q1. A bag initially contains 5 white and 5 black balls. A ball is drawn, noted, and replaced with 2 more of the same color. Then a second ball is drawn. Find the probability the second ball is white.
Case 1: First ball white (P=1/2) → Bag becomes 7W, 5B. P(2nd W|Case1) = 7/12
Case 2: First ball black (P=1/2) → Bag becomes 5W, 7B. P(2nd W|Case2) = 5/12
P(2nd W) = (1/2 × 7/12) + (1/2 × 5/12) = (7+5)/24 = 12/24 = 1/2
(Note: Polyá Urn symmetry often preserves marginal probability)

Q2. In an office, 60% use Excel, 40% use PowerPoint, and 25% use both. If a randomly selected employee uses PowerPoint, what’s the probability they also use Excel?
P(Uses Excel | Uses PPT) = P(E ∩ PPT) / P(PPT)
= 0.25 / 0.40 = 0.625 (62.5%)

Q3. Three fair dice are rolled. What is the probability that at least two show the same number?
Use complement: P(All different)
1st die: any. 2nd die: 5/6 (different from 1st). 3rd die: 4/6 (different from 1st & 2nd).
P(All different) = 1 × 5/6 × 4/6 = 20/36 = 5/9
P(At least 2 same) = 1 - 5/9 = 4/9

Q4. A box contains 4 defective and 6 good items. Two items are drawn without replacement. Given that both are from the same type, find the probability they are both defective.
P(Both D) = 4/10 × 3/9 = 12/90
P(Both G) = 6/10 × 5/9 = 30/90
P(Same Type) = 12/90 + 30/90 = 42/90
P(Both D | Same) = (12/90) / (42/90) = 12/42 = 2/7

Q5. Two students, A and B, solve a problem independently. Probabilities are 1/3 and 2/5 respectively. If the problem is solved, what’s the probability it was solved by both?
P(At least 1 solves) = P(A∪B) = 1/3 + 2/5 - (1/3×2/5) = (5+6-2)/15 = 9/15 = 3/5
P(Both solve) = P(A∩B) = 2/15
P(Both | Solved) = (2/15) / (9/15) = 2/9

Common Mistakes to Avoid

• ❌ Treating dependent events as independent: When drawing without replacement, probabilities change. Always adjust the denominator and numerator for the next step.
• ❌ Adding probabilities for overlapping events directly: Using P(A) + P(B) without subtracting P(A ∩ B) leads to overcounting.
• ❌ Confusing mutually exclusive with independent: Mutually exclusive means P(A∩B)=0. Independent means P(A∩B)=P(A)×P(B). They can’t coexist unless one has probability 0.
• ❌ Miscounting sample space for dice/cards: Forgetting (1,6) and (6,1) are distinct outcomes in dice doubles, or missing suit-rank combinations in cards.
• ❌ Mishandling "at least one" vs "exactly one": "At least one" is almost always faster using 1 - P(none). Using direct calculation increases error risk.
• ❌ Ignoring order vs combination context: Using permutations when order doesn't matter (or vice versa) distorts favorable counts. Stick to ⁿCᵣ for bag draws.

Shortcut Tricks

1. Complement Rule for "At Least One"
   Instead of calculating 1, 2, or 3 successes separately, use P(at least 1) = 1 - P(none).
   Example: 3 coins tossed. P(at least 1 tail) = 1 - P(HHH) = 1 - 1/8 = 7/8.

2. Symmetry in Independent Coin/Bag Problems
   When drawing without replacement from identical groups (e.g., 5W, 5B), the probability of drawing a specific color on the nth draw (without knowing prior draws) equals the initial proportion.
   Example: In Q1 above, P(2nd White) = 1/2 directly by symmetry.

3. Conditional Shortcut: P(A∩B) = P(A) × P(B|A)
   If you know intersection and one marginal, instantly find conditional: P(B|A) = P(A∩B) / P(A). Skip rewriting sample space.

Previous Year Questions from Top Companies

1. TCS NQT (Pattern Adapted):
Two fair dice are thrown simultaneously. What is the probability that the product of numbers on their faces is even?
Product is even unless both numbers are odd.
P(Both Odd) = (3/6) × (3/6) = 1/4
P(Even Product) = 1 - 1/4 = 3/4

2. Infosys:
A bag has 6 green, 5 yellow, and 9 blue balls. Two balls are drawn randomly without replacement. What is the probability both are of different colors?
Easier via complement: 1 - P(Same color)
Total = 20.
P(GG) = (6C2)/(20C2) = 15/190
P(YY) = 10/190, P(BB) = 36/190
P(Same) = (15+10+36)/190 = 61/190
P(Different) = 1 - 61/190 = 129/190

3. Wipro Elite:
A card is picked from a shuffled deck. If it’s red, what’s the probability it’s a face card?
Red cards = 26. Red face cards = 6 (J,Q,K of hearts & diamonds)
P = 6/26 = 3/13

4. Cognizant GenC:
If P(A)=0.4, P(B)=0.3, and P(A∪B)=0.6, find P(A|B).
P(A∩B) = P(A)+P(B)-P(A∪B) = 0.4+0.3-0.6 = 0.1
P(A|B) = P(A∩B)/P(B) = 0.1/0.3 = 1/3

5. Accenture:
A number from 1 to 30 is chosen at random. What is the probability it’s prime OR divisible by 4?
Primes ≤30: {2,3,5,7,11,13,17,19,23,29} → 10
Multiples of 4: {4,8,12,16,20,24,28} → 7
Overlap (prime & mult of 4): None.
P = (10+7)/30 = 17/30 = 17/30

Quick Revision

• 🔹 Always define Sample Space n(S) before calculating probabilities.
• 🔹 Use 1 - P(None) for all "at least one" problems.
• 🔹 Independent: Multiply. Overlapping: Add & Subtract intersection. Conditional: Divide intersection by given.
• 🔹 Without replacement → decrease denominator & adjust numerator. With replacement → keep same.
• 🔹 Dice doubles count only once in combinations, but as 2 outcomes in ordered pairs.
• 🔹 Verify if events are truly independent; real-world placement questions often hide dependencies.
• 🔹 Memorize card deck structure: 52 total, 26 red/black, 12 faces, 4 suits × 13.
• 🔹 Practice 15-20 mixed problems weekly to build pattern recognition for 2026 campus tests.