Probability is counting with a denominator. If you can count favorable outcomes and total outcomes, you can find probability. This module covers the foundations: classical probability, mutually exclusive events, independent events, conditional probability, and expected value β the building blocks for all probabilistic reasoning.
If all outcomes are equally likely:
\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{|A|}{|S|} \]
where S is the sample space (set of all possible outcomes).
Example: rolling a die, P(even) = 3/6 = 1/2.
\(Always between 0 and 1: 0 \leq P(A) \leq 1. P(S) = 1 (something must happen).\)
Two events that can't happen at the same time.
If A and B are mutually exclusive:
\[ P(A \text{ or } B) = P(A) + P(B) \]
Example: drawing a card, P(ace or king) = 4/52 + 4/52 = 8/52 = 2/13.
P(A and B) = 0 for mutually exclusive events.
When events can overlap (not disjoint):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
This is probability's version of inclusion-exclusion.
Example: P(heart or face card) = P(heart) + P(face) - P(heart face card) = 13/52 + 12/52 - 3/52 = 22/52.
\[ P(\text{not } A) = 1 - P(A) \]
Also written P(A^{c}) = 1 - P(A).
Same idea as complementary counting, just divided by total. "At least one" \((\to\) complement is\) "none."
Two events are independent if the occurrence of one doesn't affect the probability of the other.
If A and B are independent:
\[ P(A \text{ and } B) = P(A) \times P(B) \]
\(Example: flipping two coins, P(both heads) = 1/2 \times 1/2 = 1/4.\)
\(Independence \neq mutual exclusivity.\) In fact, if two events are both mutually exclusive AND have positive probability, they can't be independent.
Probability of A given that B has occurred:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Read: "probability of A given B." We restrict our sample space to only outcomes where B happens.
\(Rearranged: P(A \cap B) = P(B) \times P(A|B)\) β the general multiplication rule.
If A and B are independent, then P(A|B) = P(A) (knowing B doesn't change P(A)).
With replacement: \(put the item back, probabilities stay the same \to independent trials.\)
Without replacement: \(don't put it back, probabilities change \to dependent trials, use conditional probability.\)
\(Example: drawing 2 cards from a deck. With replacement: P(both red) = 26/52 \times 26/52 = 1/4. Without replacement: P(both red) = 26/52 \times 25/51 = 25/102.\)
The long-run average value of a random variable.
\[ E[X] = \sum x_i \cdot P(X = x_i) \]
Multiply each outcome by its probability, add them up.
\(Example: roll a fair die. E[X] = 1 \times (1/6) + 2 \times (1/6) + ... + 6 \times (1/6) = 21/6 = 3.5\)
Linearity of expectation: E[X + Y] = E[X] + E[Y], always, even if X and Y are dependent. This is incredibly powerful.
When outcomes are continuous (points on a line, in a plane), probability = favorable area / total area.
Same classical probability idea, just with measure instead of counting.
Example: "a point chosen randomly in a square..." β use geometric probability.
For discrete uniform probability, P(A) = |A| / |S| is just counting divided by counting.
All your combinatorics tools apply to both numerator and denominator.
Strategy: count favorable, count total, divide. Make sure you count both the same way (both ordered or both unordered).
Method 1: Counting approach (combinations).
Step 1: Total ways to draw 2 marbles from 8.
\((\binom{8}{2}\) = 28\)
Step 2: Favorable ways (both red from 3 reds).
\((\binom{3}{2}\) = 3\)
Step 3: Probability = favorable / total.
\[ P(\text{both red}) = \frac{\binom{3}{2}}{\binom{8}{2}} = \frac{3}{28} \]
Method 2: Sequential conditional probability.
P(first red) = 3/8
P(second red | first red) = 2/7 (one red removed, 7 left total)
\(P(both red) = 3/8 \times 2/7 = 6/56 = 3/28\) β
Answer: 3/28
"At least one" \((\to\) use complement: P(at least one 6) = 1 - P(no sixes)\)
Step 1: \(Total outcomes: 6 \times 6 = 36\)
Step 2: \(Outcomes with no sixes: 5 \times 5 = 25 (each die has 5 non-6 faces)\)
Step 3: P(no sixes) = 25/36
Step 4: Complement.
\[ P(\text{at least one 6}) = 1 - \frac{25}{36} = \frac{11}{36} \]
Direct check (inclusion-exclusion):
P(first die 6 OR second die 6) = P(first 6) + P(second 6) - P(both 6)
= 1/6 + 1/6 - 1/36 = 6/36 + 6/36 - 1/36 = 11/36 β
Answer: 11/36
\(Expected value = sum of (outcome \times probability).\)
Step 1: Expected winnings from the die roll.
Each outcome 1,2,3,4,5,6 has probability 1/6.
\[ E[\text{winnings}] = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} \]
= (1 + 2 + 3 + 4 + 5 + 6) / 6 = 21/6 = 3.5
Step 2: Subtract the cost to find net gain.
Net gain = Winnings - Cost = 3.5 - 5.0 = -1.5
Expected profit is -$1.50 (you lose money on average).
Interpretation: Over many games, you'd expect to lose an average of $1.50 per game. This is a losing game for the player.
Answer: -1.5