Core Concepts
Concepts & Key Formulas
1. Fundamental Counting Principle
If one event can occur in m ways and a second event can occur in n ways, the total number of ways both events can occur is m × n.
Total ways = n₁ × n₂ × n₃ × ⋯
MULTIPLY independent choices
Order of events does NOT matter for counting total
Quick Example
A shop offers 3 sizes and 4 colors. How many combinations?
Answer: 3 × 4 = 12 combinations
2. Permutation (Order Matters)
The number of ways to arrange r items selected from n distinct items, where order matters.
P(n, r) = n! / (n − r)!
Examples:
P(5, 2) = 5! / 3! = 5 × 4 = 20
P(4, 4) = 4! = 24
ORDER matters → Permutation
n! = n×(n-1)×⋯×2×1
P(n,n) = n!
Quick Example
How many ways can 4 runners finish 1st, 2nd, 3rd in a race of 6?
Answer: P(6,3) = 6!/(6-3)! = 6×5×4 = 120
3. Combination (Order Does NOT Matter)
The number of ways to choose r items from n distinct items where order does not matter.
C(n, r) = n! / [r! × (n − r)!] also written as ⁿCᵣ or C(n,r)
Examples:
C(5, 2) = 5! / (2! × 3!) = 10
C(6, 3) = 6! / (3! × 3!) = 20
ORDER doesn't matter → Combination
C(n,r) = C(n, n-r) [symmetry]
C(n,0)=C(n,n)=1
Quick Example
How many ways to choose a committee of 3 from 7 people?
Answer: C(7,3) = 7!/(3!×4!) = 35
4. Basic Probability
The probability of an event is the ratio of favorable outcomes to total possible outcomes. Probability is always between 0 and 1.
P(A) = (Number of favorable outcomes) / (Total outcomes)
P(A) + P(A') = 1 ← Complement rule
P(A') = 1 − P(A)
0 ≤ P(A) ≤ 1
P(impossible)=0
P(certain)=1
Quick Example
Rolling a standard die: P(even number)?
Answer: 3 even numbers (2,4,6) out of 6 → P = 3/6 = 1/2
5. Addition Rule (OR)
Used when we want the probability that event A OR event B occurs.
General: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Mutually Exclusive (A∩B = ∅):
P(A ∪ B) = P(A) + P(B)
OR → Add (subtract overlap)
Mutually exclusive: cannot occur together
Quick Example
A card drawn from 52-card deck: P(Heart OR Face card)? Hearts=13, Face=12, Heart Face=3
Answer: 13/52 + 12/52 − 3/52 = 22/52 = 11/26
6. Multiplication Rule (AND)
Used when we want the probability that event A AND event B both occur.
General: P(A ∩ B) = P(A) × P(B|A)
Independent: P(A ∩ B) = P(A) × P(B)
P(B|A) = conditional probability of B given A
AND → Multiply
Independent: one event doesn't affect the other
Quick Example
Flipping a fair coin twice: P(Heads AND Heads)?
Answer: 1/2 × 1/2 = 1/4
7. Conditional Probability
The probability of event B occurring given that A has already occurred.
P(B|A) = P(A ∩ B) / P(A) [P(A) ≠ 0]
Key check: If P(B|A) = P(B), then A and B are INDEPENDENT
B given A: restrict sample space to A
Read "|" as "given that"
Quick Example
P(A)=0.4, P(A∩B)=0.2. Find P(B|A).
Answer: P(B|A) = 0.2 / 0.4 = 0.5
8. Binomial Probability
Used when an experiment has exactly two outcomes (success/failure), is repeated n times, and each trial is independent with constant probability p of success.
P(X = k) = C(n, k) × pᵏ × (1−p)^(n−k)
where: n = number of trials
k = number of successes
p = probability of success per trial
n independent trials
2 outcomes only
C(n,k) accounts for arrangements
Quick Example
Fair coin flipped 3 times: P(exactly 2 Heads)?
Answer: C(3,2)×(1/2)²×(1/2)¹ = 3×1/4×1/2 = 3/8
Exam Questions
Practice Exam
20 Multiple Choice Questions
01
Fundamental Counting
Easy
A restaurant offers 4 appetizers, 5 main courses, and 3 desserts. If a meal consists of one item from each category, how many different meals are possible?
02
Permutation
Easy
In how many ways can 5 different books be arranged on a shelf?
03
Combination
Easy
A committee of 3 people is chosen from a group of 8 people. How many different committees are possible?
04
Basic Probability
Easy
A bag contains 4 red, 3 blue, and 5 green marbles. One marble is drawn at random. What is the probability it is NOT green?
05
Permutation
Medium
How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if no digit may be repeated?
06
Addition Rule
Medium
In a class, P(student plays soccer) = 0.4, P(student plays basketball) = 0.3, and P(student plays both) = 0.1. What is the probability a randomly chosen student plays soccer OR basketball?
07
Combination
Medium
A team of 4 players is selected from 6 men and 4 women. If the team must include exactly 2 women, how many different teams are possible?
08
Multiplication Rule
Medium
Two cards are drawn without replacement from a standard 52-card deck. What is the probability that both cards are Aces?
09
Conditional Probability
Medium
A die is rolled. Given that the result is greater than 3, what is the probability that the result is even?
10
Complement Rule
Medium
A fair coin is flipped 4 times. What is the probability of getting at least one Head?
11
Binomial Probability
Medium
A fair coin is flipped 5 times. What is the probability of getting exactly 3 Heads?
12
Permutation with Restriction
Hard
How many ways can 6 people be seated in a row if 2 specific people must sit together?
13
Probability with Combination
Medium
A hand of 5 cards is dealt from a standard 52-card deck. What is the probability that all 5 cards are Hearts?
14
Conditional Probability
Hard
In a group of 100 students: 60 study Math, 50 study Science, and 30 study both. A student is chosen at random. Given that the student studies Math, what is the probability they also study Science?
15
Combination — Symmetry
Medium
Which of the following is equal to C(10, 3)?
16
Binomial Probability
Hard
A biased coin has probability 2/3 of landing Heads. It is flipped 4 times. What is the probability of getting exactly 2 Heads?
17
Counting — Distinct Arrangements
Hard
How many distinct arrangements are there of all the letters in the word MISSISSIPPI? (M=1, I=4, S=4, P=2)
18
Probability — Independence
Hard
Events A and B are independent. P(A) = 0.3 and P(B) = 0.5. What is P(A ∪ B)?
19
Permutation — Circular
Hard
In how many ways can 6 people be seated at a circular table? (Rotations of the same arrangement are considered identical.)
20
Combined Concepts
Hard
From a standard 52-card deck, 2 cards are drawn without replacement. What is the probability of drawing a King first and then a Queen second?
Final Score
0
out of 20
Complete Answer Key & Explanations
Q01
Answer: C — 60
Fundamental Counting
4 appetizers × 5 mains × 3 desserts = 60
By the Fundamental Counting Principle, multiply the number of independent choices: 4 × 5 × 3 = 60 distinct meals.
Q02
Answer: D — 120
Permutation
P(5,5) = 5! = 5 × 4 × 3 × 2 × 1 = 120
Arranging all 5 books is a permutation of 5 items taken 5 at a time. Since order matters (different arrangements), 5! = 120.
Q03
Answer: B — 56
Combination
C(8,3) = 8! / (3! × 5!) = (8×7×6)/(3×2×1) = 336/6 = 56
A committee has no order — choosing person A,B,C is the same as B,C,A. So we use combinations: C(8,3) = 56.
Q04
Answer: B — 7/12
Complement Rule
Total marbles = 4+3+5 = 12
P(green) = 5/12
P(NOT green) = 1 − 5/12 = 7/12
Using the complement rule: P(not green) = 1 − P(green) = 1 − 5/12 = 7/12. Alternatively: (4 red + 3 blue) / 12 = 7/12.
Q05
Answer: C — 60
Permutation
P(5,3) = 5! / (5−3)! = 5! / 2! = (5×4×3) = 60
We select and arrange 3 digits from 5, with no repetition and order mattering (hundreds, tens, units are distinct positions). P(5,3) = 5×4×3 = 60.
Q06
Answer: C — 0.6
Addition Rule
P(S ∪ B) = P(S) + P(B) − P(S ∩ B)
= 0.4 + 0.3 − 0.1 = 0.6
The general addition rule subtracts the overlap (both sports) to avoid double-counting: 0.4 + 0.3 − 0.1 = 0.6.
Q07
Answer: B — 90
Combination with Restriction
Choose 2 women from 4: C(4,2) = 6
Choose 2 men from 6: C(6,2) = 15
Total = 6 × 15 = 90
We must pick exactly 2 women (from 4) AND exactly 2 men (from 6). These are independent choices, so multiply: C(4,2) × C(6,2) = 6 × 15 = 90.
Q08
Answer: B — 1/221
Multiplication Rule (Dependent)
P(1st Ace) = 4/52
P(2nd Ace | 1st Ace) = 3/51
P(both Aces) = (4/52) × (3/51) = 12/2652 = 1/221
Without replacement, drawing the first Ace reduces both total cards and Aces remaining. Multiply dependent probabilities: (4/52)×(3/51) = 12/2652 = 1/221.
Q09
Answer: C — 2/3
Conditional Probability
Given > 3: sample space = {4, 5, 6} (3 outcomes)
Even numbers in {4,5,6}: {4, 6} → 2 outcomes
P(even | >3) = 2/3
Given the result is greater than 3, the restricted sample space is {4,5,6}. Of these, even numbers are {4,6}. So P = 2/3.
Q10
Answer: D — 15/16
Complement Rule
P(no Heads in 4 flips) = (1/2)⁴ = 1/16
P(at least one Head) = 1 − 1/16 = 15/16
It is much easier to use the complement. The only way to NOT get at least one Head is to get all Tails: (1/2)⁴ = 1/16. So P(at least one Head) = 1 − 1/16 = 15/16.
Q11
Answer: B — 5/16
Binomial Probability
P(X=3) = C(5,3) × (1/2)³ × (1/2)²
= 10 × (1/8) × (1/4)
= 10/32 = 5/16
Using the binomial formula with n=5, k=3, p=1/2: C(5,3)×(1/2)⁵ = 10/32 = 5/16. C(5,3)=10.
Q12
Answer: C — 240
Permutation with Restriction
Treat the 2 people as 1 unit → 5 units total
Arrange 5 units: 5! = 120 ways
Arrange the 2 together: 2! = 2 ways
Total = 120 × 2 = 240
Glue the 2 specific people into one "super-person." Now arrange 5 items in a row: 5!=120 ways. Then the 2 within their unit can swap: ×2. Total = 240.
Q13
Answer: B — C(13,5) / C(52,5)
Probability with Combination
Favorable: choose 5 Hearts from 13 → C(13,5)
Total: choose any 5 cards from 52 → C(52,5)
P = C(13,5) / C(52,5)
When drawing without replacement and order doesn't matter, we use combinations. Favorable outcomes = C(13,5); total outcomes = C(52,5). Order among 5 cards is irrelevant, so no extra factorial needed.
Q14
Answer: C — 1/2
Conditional Probability
P(Math) = 60/100 = 3/5
P(Math ∩ Science) = 30/100 = 3/10
P(Science | Math) = (3/10) / (3/5) = (3/10) × (5/3) = 1/2
By the conditional probability formula: P(Science|Math) = P(Math ∩ Science)/P(Math) = (30/100)/(60/100) = 30/60 = 1/2.
Q15
Answer: B — C(10, 7)
Combination Symmetry
Symmetry property: C(n, r) = C(n, n−r)
C(10, 3) = C(10, 10−3) = C(10, 7)
Check: C(10,3) = 120 = C(10,7) ✓
The symmetry property states C(n,r) = C(n, n−r). So C(10,3) = C(10,7) because choosing 3 items to INCLUDE is equivalent to choosing 7 items to EXCLUDE.
Q16
Answer: C — 8/27
Binomial Probability
n=4, k=2, p=2/3, q=1/3
P(X=2) = C(4,2) × (2/3)² × (1/3)²
= 6 × (4/9) × (1/9)
= 6 × 4/81
= 24/81 = 8/27
Binomial: C(4,2)×(2/3)²×(1/3)² = 6×4/9×1/9 = 24/81 = 8/27. Note: (1−2/3) = 1/3 is the probability of Tails.
Q17
Answer: A/B — 34,650
Permutations of Repeated Elements
MISSISSIPPI: 11 letters
M=1, I=4, S=4, P=2
Arrangements = 11! / (1! × 4! × 4! × 2!)
= 39,916,800 / (1×24×24×2)
= 39,916,800 / 1152
= 34,650
When items repeat, divide by the factorial of each repeated count to avoid counting identical arrangements. 11!/(1!·4!·4!·2!) = 34,650.
Q18
Answer: C — 0.65
Independence + Addition Rule
Independent → P(A∩B) = P(A)×P(B) = 0.3×0.5 = 0.15
P(A∪B) = P(A) + P(B) − P(A∩B)
= 0.3 + 0.5 − 0.15 = 0.65
Since A and B are independent, P(A∩B) = 0.15. Then apply the general addition rule: 0.3 + 0.5 − 0.15 = 0.65.
Q19
Answer: C — 120
Circular Permutation
Circular permutations of n people = (n−1)!
= (6−1)! = 5! = 120
In a circle, one person's position is fixed (we only count relative arrangements). So the formula is (n−1)! = 5! = 120. Linear arrangements (720) over-count by a factor of 6 (rotations).
Q20
Answer: C — 4/663
Sequential Probability (Dependent)
P(King 1st) = 4/52
P(Queen 2nd | King 1st) = 4/51
P(King then Queen) = (4/52) × (4/51)
= 16/2652 = 4/663
Order matters here (King THEN Queen). After drawing a King, there are 51 cards left, all 4 Queens still present. P = (4/52)×(4/51) = 16/2652 = 4/663.
Answer Key & Explanations
Q1 — Fundamental Counting
✓ C: 60
4 × 5 × 3 = 60 meals
Q2 — Permutation
✓ D: 120
5! = 120 arrangements
Q3 — Combination
✓ B: 56
C(8,3) = 56 committees
Q4 — Complement
✓ B: 7/12
1 − 5/12 = 7/12
Q5 — Permutation
✓ C: 60
P(5,3) = 5×4×3 = 60
Q6 — Addition Rule
✓ C: 0.6
0.4+0.3−0.1 = 0.6
Q7 — Combination+Restriction
✓ B: 90
C(4,2)×C(6,2)=6×15=90
Q8 — Multiplication Rule
✓ B: 1/221
(4/52)×(3/51)=1/221
Q9 — Conditional Probability
✓ C: 2/3
Restrict to {4,5,6}: 2 even → 2/3
Q10 — Complement
✓ D: 15/16
1−(1/2)⁴=1−1/16=15/16
Q11 — Binomial
✓ B: 5/16
C(5,3)×(1/2)⁵=10/32=5/16
Q12 — Permutation+Restriction
✓ C: 240
5!×2!=120×2=240
Q13 — Probability+Combination
✓ B: C(13,5)/C(52,5)
Favorable/Total using combinations
Q14 — Conditional Probability
✓ C: 1/2
P(S|M)=30/60=1/2
Q15 — Combination Symmetry
✓ B: C(10,7)
C(n,r)=C(n,n−r): C(10,3)=C(10,7)
Q16 — Binomial
✓ C: 8/27
C(4,2)×(2/3)²×(1/3)²=24/81=8/27
Q17 — Repeated Permutation
✓ A: 34,650
11!/(1!4!4!2!)=34,650
Q18 — Independence + Addition
✓ C: 0.65
0.3+0.5−(0.3×0.5)=0.65
Q19 — Circular Permutation
✓ C: 120
(6−1)!=5!=120
Q20 — Sequential Probability
✓ C: 4/663
(4/52)×(4/51)=16/2652=4/663