Dice, coins, and playing cards — these three objects make up about 80% of all probability questions in aptitude exams. The reason they’re so popular is that they have clean, well-defined sample spaces. Once we know the sample spaces inside out, these problems become almost mechanical. Let’s master each one.
Coins — The Simplest Probability
Sample Space
- 1 coin: {H, T} → 2 outcomes
- 2 coins: {HH, HT, TH, TT} → 4 outcomes
- 3 coins: 8 outcomes
- n coins: 2ⁿ outcomes
The number of ways to get exactly r heads in n tosses = nCr. This is because we’re choosing which r of the n tosses show heads.
Common Coin Probabilities (3 coins)
| Outcome | Ways | Probability |
|---|---|---|
| 0 heads (TTT) | 3C0 = 1 | 1/8 |
| Exactly 1 head | 3C1 = 3 | 3/8 |
| Exactly 2 heads | 3C2 = 3 | 3/8 |
| 3 heads (HHH) | 3C3 = 1 | 1/8 |
| At least 1 head | 7 | 7/8 |
| At most 1 head | 4 | 4/8 = 1/2 |
The power of complement: P(at least 1 head in n tosses) = 1 − P(all tails) = 1 − (1/2)ⁿ. No counting needed!
Dice — The King of Probability Questions
Sample Space
- 1 die: {1, 2, 3, 4, 5, 6} → 6 outcomes
- 2 dice: 6 × 6 = 36 outcomes
- n dice: 6ⁿ outcomes
Single Die Probabilities
| Event | Probability |
|---|---|
| Any specific number | 1/6 |
| Even number (2, 4, 6) | 3/6 = 1/2 |
| Odd number (1, 3, 5) | 3/6 = 1/2 |
| Prime number (2, 3, 5) | 3/6 = 1/2 |
| Greater than 4 (5, 6) | 2/6 = 1/3 |
| Less than 3 (1, 2) | 2/6 = 1/3 |
Two Dice — Sum Probabilities
This is the most tested topic. There are 36 total outcomes for two dice. Here’s how many ways each sum can occur:
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
| P | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
Pattern to memorize: The ways go 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 — a symmetric pyramid peaking at sum = 7. Sum = 7 is the most likely outcome with 6 ways.
The combinations for sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
Useful Two-Dice Facts
- P(sum = 7) = 6/36 = 1/6 (the highest for any specific sum)
- P(doubles) = 6/36 = 1/6 → (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
- P(sum is even) = 18/36 = 1/2
- P(sum is odd) = 18/36 = 1/2
- P(at least one 6) = 1 − P(no 6) = 1 − (5/6)² = 1 − 25/36 = 11/36
Playing Cards — The Most Complex Sample Space
A standard deck has 52 cards organized as follows:
Single Card Drawn
| Event | Favorable | Probability |
|---|---|---|
| A specific card (e.g., Ace of Spades) | 1 | 1/52 |
| A specific suit (e.g., Hearts) | 13 | 13/52 = 1/4 |
| A specific rank (e.g., any King) | 4 | 4/52 = 1/13 |
| A face card (J, Q, K) | 12 | 12/52 = 3/13 |
| A black card | 26 | 26/52 = 1/2 |
| Not a heart | 39 | 39/52 = 3/4 |
Two Cards Drawn
When drawing 2 cards from 52 without replacement:
Total ways = 52C2 = 1326
Both Kings: 4C2 / 52C2 = 6/1326 = 1/221
Both from same suit: 4 × 13C2 / 52C2 = 4 × 78 / 1326 = 312/1326 = 4/17
One King and one Queen: (4C1 × 4C1) / 52C2 = 16/1326 = 8/663
”At Least” and “At Most” Problems
These are the trickiest question types, but the complement rule makes them easy.
”At Least One” = 1 − P(None)
Example: Two cards drawn from 52. P(at least one ace)?
P(at least one ace) = 1 − P(no ace) = 1 − 48C2/52C2 = 1 − 1128/1326 = 198/1326 = 33/221
”At Most One” = P(None) + P(Exactly One)
Example: Three coins tossed. P(at most one head)?
P(at most 1 head) = P(0 heads) + P(1 head) = 1/8 + 3/8 = 4/8 = 1/2
Shortcut Methods and Tricks
Trick 1: Dice sum = 7 shortcut For any single die value x on one die, the other die needs (7-x). Since x can be 1 through 6, there are always 6 ways. This is why P(sum=7) = 1/6.
Trick 2: Card counting by elimination Instead of counting favorable, sometimes it’s faster to count unfavorable and subtract. “Not a face card” = 52 − 12 = 40 cards.
Trick 3: “At least” always means complement Whenever we see “at least,” our first instinct should be: 1 − P(complement). It’s faster in almost every case.
Trick 4: With/Without replacement matters hugely With replacement: probabilities stay the same each draw. Without replacement: we must update the denominator AND numerator after each draw.
Trick 5: Drawing simultaneously vs one by one Drawing 2 cards simultaneously = drawing 2 cards one by one without replacement. The math is the same. Use combinations for simultaneous, sequential multiplication for one-by-one.
Worked Examples
Example 1: Coin Problem
4 coins are tossed. Find P(exactly 2 heads).
Total outcomes = 2⁴ = 16 Favorable (exactly 2 heads) = 4C2 = 6
P = 6/16 = 3/8
Example 2: Two Dice Problem
Two dice are rolled. Find P(sum > 9).
Sum = 10: (4,6), (5,5), (6,4) → 3 ways Sum = 11: (5,6), (6,5) → 2 ways Sum = 12: (6,6) → 1 way
Total favorable = 6
P = 6/36 = 1/6
Example 3: Card Problem
Two cards are drawn from a deck. Find P(both are red).
Red cards = 26
P = 26C2 / 52C2 = (26 × 25/2) / (52 × 51/2) = 325/1326 = 25/102
Step by step (sequential): P(1st red) × P(2nd red | 1st red) = 26/52 × 25/51 = 25/102. Same answer!
Example 4: At Least One Problem
A die is rolled 3 times. Find P(at least one 6).
P(no 6 in a single roll) = 5/6
P(no 6 in 3 rolls) = (5/6)³ = 125/216
P(at least one 6) = 1 − 125/216 = 91/216
Example 5: Mixed Problem
From a well-shuffled deck, 3 cards are drawn. What is the probability that all 3 are face cards?
Face cards = 12
P = 12C3 / 52C3 = 220 / 22100 = 1/100.45…
Let’s compute exactly: 12C3 = (12 × 11 × 10)/6 = 220. 52C3 = (52 × 51 × 50)/6 = 22100.
P = 220/22100 = 22/2210 = 11/1105
The Birthday Problem (Fun Bonus)
This is a famous probability puzzle: In a group of 23 people, there’s a greater than 50% chance that two people share a birthday. Sounds wild, right?
Here’s the logic: P(all different birthdays) = 365/365 × 364/365 × 363/365 × … × 343/365
For 23 people, this product works out to about 0.493, so P(at least one shared birthday) = 1 − 0.493 ≈ 0.507 or 50.7%.
At 50 people, the probability jumps to about 97%. Our intuition is terrible at this — which is exactly why probability is so important to study!
Common Exam Variations
- Two dice sum problems — P(sum = k), P(sum > k), P(sum is prime)
- Card drawing — with specific suit/rank constraints, with or without replacement
- Coin sequences — P(exactly r heads in n tosses), P(at least r heads)
- Bag problems — balls of different colors, drawing with/without replacement
- Conditional dice — “given that the sum is even, find P(both are same)“
Practice Problems
Q1: Two dice are rolled. Find the probability that the product of the numbers is 12.
Q2: 5 coins are tossed simultaneously. Find P(at least 4 heads).
Q3: From a deck of 52 cards, two cards are drawn without replacement. Find the probability that one is a spade and the other is a heart.
Answers
A1: Pairs giving product 12: (2,6), (3,4), (4,3), (6,2) → 4 favorable outcomes. P = 4/36 = 1/9.
A2: P(at least 4 heads) = P(4 heads) + P(5 heads) = 5C4/2⁵ + 5C5/2⁵ = 5/32 + 1/32 = 6/32 = 3/16.
A3: Ways to pick 1 spade and 1 heart = 13C1 × 13C1 = 169. Total ways = 52C2 = 1326. P = 169/1326 = 13/102.