Quiz 2

Combinatorics and Basic Probability

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# Combinatorics and Basic Probability ## 10.1 Counting Principles **Product rule:** If task A has $m$ outcomes and task B has $n$, then the sequence (A then B) has $m \times n$ outcomes. **Sum rule:** If tasks A and B are mutually exclusive, then A or B has $m + n$ outcomes.

Combinatorics and Basic Probability

10.1 Counting Principles

Product rule: If task A has mm outcomes and task B has nn, then the sequence (A then B) has m×nm \times n outcomes. Sum rule: If tasks A and B are mutually exclusive, then A or B has m+nm + n outcomes.

10.2 Permutations and Combinations

Permutation (order matters): P(n,k)=n!(n−k)!P(n, k) = \frac{n!}{(n-k)!} Combination (order doesn't matter): (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}

10.3 Basic Probability

Probability: P(A)=∣A∣∣Ω∣P(A) = \frac{|A|}{|\Omega|} (for equally likely outcomes) Conditional probability: P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)} Bayes' theorem: P(A∣B)=P(B∣A)P(A)P(B)P(A|B) = \frac{P(B|A)P(A)}{P(B)}

✅ Practice Questions

Q1: How many ways to choose a committee of 3 from 10 people?
Solution
(103)=10!3!7!=120\binom{10}{3} = \frac{10!}{3!7!} = 120 ways. Q2: What's the probability of getting exactly 2 heads in 3 coin flips? Solution
Total outcomes: 23=82^3 = 8. Favorable: choose 2 of 3 flips to be heads = (32)=3\binom{3}{2} = 3. Probability = 3/83/8. Join Discord PreviousUncountable SetsNextGraph Theory Introduction
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