Probability tells us how likely something is to happen. It’s a number between 0 (impossible) and 1 (certain). If we can count the favorable outcomes and the total outcomes, we can find the probability. Most aptitude probability questions follow a handful of patterns — once we learn those, we’re golden.
The Basics
P(E) = Number of favorable outcomes / Total number of outcomes
This is the classical definition, and it works when all outcomes are equally likely.
- P(E) = 0 means the event is impossible
- P(E) = 1 means the event is certain
- P(not E) = 1 − P(E) — this is called the complement
The complement rule is probably the most useful trick in probability. When “at least one” is hard to calculate directly, we flip it: P(at least one) = 1 − P(none).
Mutually Exclusive vs Independent Events
These are two completely different concepts, and exams love to test whether we know the difference.
Mutually Exclusive Events
Two events are mutually exclusive if they cannot happen at the same time. Think of it like: if A happens, B can’t, and vice versa.
- P(A ∩ B) = 0 (no overlap)
- P(A ∪ B) = P(A) + P(B) (just add — no double counting needed)
Example: Rolling a 3 and rolling a 5 on the same die roll — can’t happen simultaneously.
Independent Events
Two events are independent if one happening doesn’t affect the probability of the other.
- P(A ∩ B) = P(A) × P(B) (just multiply)
- P(A|B) = P(A) (knowing B happened doesn’t change A’s probability)
Example: Flipping a coin and rolling a die — the coin result doesn’t affect the die.
Key distinction: Mutually exclusive events are about overlap (or lack thereof). Independent events are about influence (or lack thereof). Mutually exclusive events with non-zero probabilities are actually DEPENDENT (if A happened, P(B) = 0, which is different from P(B) in general).
Addition Rule (OR)
When we want P(A or B), we use:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
We subtract P(A ∩ B) to avoid double-counting the overlap.
If A and B are mutually exclusive (no overlap), the formula simplifies to: P(A ∪ B) = P(A) + P(B).
Multiplication Rule (AND)
When we want P(A and B):
P(A ∩ B) = P(A) × P(B|A)
If A and B are independent: P(A ∩ B) = P(A) × P(B)
In simple language: for “AND” problems, we multiply. But we need to be careful about whether events are independent or not.
Conditional Probability
The probability of A happening given that B has already happened:
P(A|B) = P(A ∩ B) / P(B)
In simple language: we’re restricting our sample space to only the outcomes where B is true, and then asking “of those, how many also have A?”
Example: In a class, 60% study math, 40% study science, and 20% study both. What’s the probability that a student studies math given they study science?
P(Math|Science) = P(Math ∩ Science) / P(Science) = 0.20 / 0.40 = 0.50 or 50%
Bayes’ Theorem
Bayes’ theorem lets us reverse conditional probability — if we know P(B|A), we can find P(A|B).
P(A|B) = P(B|A) × P(A) / P(B)
This is incredibly powerful. Let’s see it with a practical example.
Bayes’ Theorem — A Practical Example
A factory has 3 machines: Machine A produces 50% of items, Machine B produces 30%, Machine C produces 20%. The defective rates are: A = 3%, B = 4%, C = 5%. If a randomly selected item is defective, what’s the probability it came from Machine A?
We want P(A | Defective).
P(Defective) = P(D|A)×P(A) + P(D|B)×P(B) + P(D|C)×P(C) = 0.03 × 0.50 + 0.04 × 0.30 + 0.05 × 0.20 = 0.015 + 0.012 + 0.010 = 0.037
P(A | Defective) = P(D|A) × P(A) / P(D) = 0.015 / 0.037 = 15/37 ≈ 0.405 or 40.5%
Even though Machine A has the lowest defective rate, it produces the most items, so it still contributes the most to the defective pool.
Shortcut Methods and Tricks
Trick 1: The complement is your best friend P(at least 1) = 1 − P(none). This saves enormous calculation in most “at least” problems.
Trick 2: Sequential probability For events happening one after another (like drawing cards without replacement), multiply the probabilities at each step. Just update the total and favorable counts after each step.
Trick 3: OR means add, AND means multiply This simple mnemonic covers 90% of cases. Just remember to check for independence/mutual exclusivity.
Trick 4: Odds to probability If odds in favor are a:b, then P = a/(a+b). If odds against are a:b, then P = b/(a+b).
Worked Examples
Example 1: Basic Probability
A bag contains 5 red, 3 blue, and 2 green balls. One ball is drawn at random. Find P(red) and P(not green).
Total = 5 + 3 + 2 = 10
P(red) = 5/10 = 1/2
P(not green) = 1 − P(green) = 1 − 2/10 = 4/5
Example 2: Addition Rule
A card is drawn from a standard deck of 52 cards. Find the probability that it is a King or a Heart.
P(King) = 4/52, P(Heart) = 13/52, P(King AND Heart) = 1/52 (King of Hearts)
P(King or Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13
Example 3: Independent Events
A coin is tossed and a die is rolled. What is the probability of getting a Head and a number greater than 4?
P(Head) = 1/2 P(greater than 4) = P(5 or 6) = 2/6 = 1/3
These are independent events, so: P(Head AND > 4) = 1/2 × 1/3 = 1/6
Example 4: Without Replacement
A bag has 4 red and 6 blue balls. Two balls are drawn without replacement. Find P(both red).
P(1st red) = 4/10 P(2nd red | 1st red) = 3/9 (one red ball is gone)
P(both red) = 4/10 × 3/9 = 12/90 = 2/15
Example 5: At Least One (Complement)
Three coins are tossed. Find P(at least one head).
The hard way: P(1H) + P(2H) + P(3H) = 3/8 + 3/8 + 1/8 = 7/8
The smart way: P(at least 1 head) = 1 − P(no heads) = 1 − P(all tails) = 1 − (1/2)³ = 1 − 1/8 = 7/8
Same answer, way less work!
Common Exam Variations
- Drawing balls from bags — with and without replacement
- Card problems — drawing specific suits, ranks, or combinations
- Dice problems — sum of two dice, specific outcomes
- “At least one” problems — almost always use the complement
- Conditional probability in real-world scenarios — disease testing, defective items, exam pass rates
Practice Problems
Q1: Two dice are rolled. Find the probability that the sum is 7 or 11.
Q2: A box has 3 defective and 7 non-defective items. Two items are drawn without replacement. Find the probability that both are non-defective.
Q3: The probability that A solves a problem is 2/3, and the probability that B solves it is 3/4. If both try independently, what’s the probability that the problem is solved?
Answers
A1: Total outcomes = 36. Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways. Sum 11: (5,6), (6,5) = 2 ways. These are mutually exclusive. P = (6+2)/36 = 8/36 = 2/9.
A2: P(1st non-defective) = 7/10. P(2nd non-defective | 1st non-defective) = 6/9 = 2/3. P(both) = 7/10 × 2/3 = 7/15.
A3: P(problem solved) = 1 − P(neither solves it) = 1 − P(A fails) × P(B fails) = 1 − (1/3)(1/4) = 1 − 1/12 = 11/12.