🎲 Probability Basics

The fundamental concepts of probability that form the foundation for all probability problems.

🎯 Key Ideas

Basic Probability

The probability of an event $A$ is the ratio of favorable outcomes to total outcomes: $$P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$$

Conditional Probability

The probability of event $A$ given that event $B$ has occurred: $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$

Independence

Events $A$ and $B$ are independent if: $$P(A \cap B) = P(A) \cdot P(B)$$

Law of Total Probability

For mutually exclusive and exhaustive events $B_1, B_2, \ldots, B_n$: $$P(A) = \sum_{i=1}^{n} P(A \mid B_i) \cdot P(B_i)$$

💡 Micro-Examples

Basic Probability

  • Problem: What’s the probability of rolling a 6 on a fair die?
  • Solution: $P(6) = \frac{1}{6}$ (1 favorable outcome out of 6 total)

Conditional Probability

  • Problem: Given that a card is red, what’s the probability it’s a heart?
  • Solution: $P(\text{heart} \mid \text{red}) = \frac{P(\text{heart} \cap \text{red})}{P(\text{red})} = \frac{13/52}{26/52} = \frac{1}{2}$

Independence

  • Problem: Are consecutive coin flips independent?
  • Solution: Yes! $P(\text{heads on flip 2} \mid \text{heads on flip 1}) = P(\text{heads on flip 2}) = \frac{1}{2}$

Law of Total Probability

  • Problem: What’s the probability of rain given that 60% of days are cloudy and it rains on 30% of cloudy days?
  • Solution: $P(\text{rain}) = P(\text{rain} \mid \text{cloudy}) \cdot P(\text{cloudy}) + P(\text{rain} \mid \text{not cloudy}) \cdot P(\text{not cloudy}) = 0.3 \cdot 0.6 + 0.1 \cdot 0.4 = 0.22$

⚠️ Common Traps & Fixes

Trap: Confusing independence with disjointness

  • Wrong: “If events are independent, they can’t happen together”
  • Right: Independence means $P(A \cap B) = P(A)P(B)$, not $P(A \cap B) = 0$

Trap: Forgetting to check if events are mutually exclusive

  • Wrong: “What’s the probability of rolling a 1 or 2?” = $P(1) + P(2) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$
  • Right: This is actually correct! Rolling 1 and rolling 2 are mutually exclusive.

Trap: Misapplying conditional probability

  • Wrong: “What’s the probability of drawing a heart given that it’s red?” = $\frac{13}{52} = \frac{1}{4}$
  • Right: $P(\text{heart} \mid \text{red}) = \frac{P(\text{heart} \cap \text{red})}{P(\text{red})} = \frac{13/52}{26/52} = \frac{1}{2}$

🏆 AMC-Style Worked Example

Problem: A bag contains 3 red balls and 2 blue balls. You draw 2 balls without replacement. What’s the probability that both balls are red?

Solution:

  • Method 1: Direct counting

    • Total ways to draw 2 balls: $\binom{5}{2} = 10$
    • Ways to draw 2 red balls: $\binom{3}{2} = 3$
    • Probability: $\frac{3}{10} = 0.3$

  • Method 2: Sequential probability

    • First ball red: $P(\text{first red}) = \frac{3}{5}$
    • Second ball red given first red: $P(\text{second red} \mid \text{first red}) = \frac{2}{4} = \frac{1}{2}$
    • Both red: $P(\text{both red}) = \frac{3}{5} \cdot \frac{1}{2} = \frac{3}{10}$

Key insight: Both methods give the same answer! Choose the one that’s easier for the specific problem.

🔗 Related Topics

Next: Expected Value → Distributions