Probability

A2.K.1Permutations & combinations01/04

Permutations and combinations.

Step01of 04

Both COUNT arrangements of items, but they answer different questions.

Permutation	→order MATTERS — different orders count separately
Combination	→order does NOT matter — only the chosen set counts

"Choosing a president, VP, secretary" → permutation (the roles distinguish the order). "Choosing a 3-person committee" → combination (no roles, just the set of three).

A2.K.1 — Permutations and combinations.

Step 01 of 04

Both COUNT arrangements of items, but they answer different questions.

Permutation	→order MATTERS — different orders count separately
Combination	→order does NOT matter — only the chosen set counts

"Choosing a president, VP, secretary" → permutation (the roles distinguish the order). "Choosing a 3-person committee" → combination (no roles, just the set of three).

Step 02 of 04

Formulas.

P (n, k) = \frac{n !}{( n - k )!}

C (n, k) = (k n) = \frac{n !}{k ! ( n - k )!}

$n!$ ("n factorial") = $n \cdot (n - 1) \cdot \dots \cdot 1$ . By convention $0! = 1$ .

The combination formula has an extra $k!$ in the denominator — that "kills" the orderings within the chosen set, since order doesn't matter.

Step 03 of 04

Worked examples.

How many ways to arrange the letters in $M A T H$ ? $P (4, 4) = 4! = 24$ .

How many ways to pick 3 desserts from a menu of 8? Order doesn't matter: $C (8, 3) = \frac{8 !}{3 ! \cdot 5 !} = \frac{8 \cdot 7 \cdot 6}{6} = 56$ .

How many ways to seat 5 people in a row from a group of 10? Order matters: $P (10, 5) = \frac{10 !}{5 !} = 30240$ .

Step 04 of 04

Quick decision tree.

"Arrange / order / lineup / first-second-third"	→permutation
"Choose / select / pick / committee / group"	→combination
"How many subsets of size k"	→combination
Tied to specific positions	→permutation

Key insight

Order matters → permutation,

P (n, k) = n! / (n - k)!

. Order doesn't → combination,

C (n, k) = n! / (k! (n - k)!)

. The

k!

denominator in the combination formula erases internal orderings.

A2.K.2 — Conditional probability.

Step 01 of 04

Conditional probability measures the chance of event $A$ GIVEN that event $B$ already happened.

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

Reads "P of A given B." The numerator is the joint probability of both events; the denominator is the probability of the conditioning event. Knowing $B$ shrinks the universe of possibilities.

Step 02 of 04

Worked example with a deck of cards. Draw one card. What's the probability it's a king, given that it's a face card?

$P (face card) = 12/52$ (J, Q, K of each suit). $P (king AND face card) = 4/52$ (just the four kings — all kings are face cards).

P (king ∣ face) = \frac{4/52}{12/52} = \frac{4}{12} = \frac{1}{3}

Step 03 of 04

Tree diagrams are the cleanest tool for multi-stage problems.

Example: A bag has 3 red and 2 blue marbles. Draw 2 without replacement. What's $P (2nd is red ∣ 1st is red)$ ?

After drawing one red, 4 marbles remain (2 red, 2 blue):

P (2nd red ∣ 1st red) = 2/4 = 1/2

Step 04 of 04

Multiplication rule. Rearrange the conditional formula to compute joint probability.

P (A \cap B) = P (A) \cdot P (B ∣ A)

From the marble example: $P (both red)$ = $(3/5) (2/4) = 6/20 = 3/10$ .

Key insight

P (A ∣ B) = P (A \cap B) / P (B)

— divide joint by the "given" event's probability. Multiplication rule reverses this:

P (A \cap B) = P (A) \cdot P (B ∣ A)

. Tree diagrams handle sequential events cleanly.

A2.K.3 — Independent and dependent events.

Step 01 of 04

Independent events: the outcome of one does NOT affect the probability of the other. Dependent: it does.

Coin flips, separate dice rolls	→independent
Drawing two cards WITH replacement	→independent (deck resets)
Drawing two cards WITHOUT replacement	→dependent (deck shrinks)
Weather today vs. weather tomorrow	→dependent (correlated)

Step 02 of 04

Test for independence.

A and B independent ⟺ P (A \cap B) = P (A) \cdot P (B)

Equivalently, $P (A ∣ B) = P (A)$ — knowing $B$ doesn't change the probability of $A$ .

Step 03 of 04

Worked example — independent. Roll two dice. What's the probability of getting a 6 on both?

P (both 6) = (1/6) (1/6) = 1/36

Multiply because independent. The first die's roll has no effect on the second.

Dependent — drawing without replacement. Bag with 5 red, 5 blue. Probability of drawing red, then blue?

P (R) \cdot P (B ∣ R) = \frac{5}{10} \cdot \frac{5}{9} = \frac{25}{90} = \frac{5}{18}

Step 04 of 04

"Or" calculations — addition rule.

P (A \cup B) = P (A) + P (B) - P (A \cap B)

Subtracting the intersection avoids double-counting. For mutually exclusive events (can't both happen), $P (A \cap B) = 0$ and the formula simplifies to $P (A) + P (B)$ .

Key insight

Independent → multiply unconditional probabilities. Dependent → multiply, but use conditional probability for later events. "Or" → add the two probabilities and subtract the overlap. Mutually exclusive ⇒ overlap is zero.

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