Counting and Probability Reference
Probability and Binomial Models
Probability models measure likelihood, while binomial models organize repeated yes-or-no trials.
Solved Examples
Jump directly to a worked example by probability rule or model.
Fact Table
| Idea | Formula | Watch For |
|---|---|---|
| Basic probability | $P(A)=\frac{\text{favorable outcomes}}{\text{total outcomes}}$ | All outcomes must be equally likely for simple counting. |
| Union | $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ | Subtract overlap once. |
| Conditional probability | $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ | The condition becomes the new sample space. |
| Binomial model | $P(X=k)=\binom nk p^k(1-p)^{n-k}$ | Use only fixed, independent trials with the same success probability. |
Content Formulas
Complement
$$P(A^c)=1-P(A)$$
Independent Events
$$P(A\cap B)=P(A)P(B)$$
Binomial Coefficient
$$\binom nk=\frac{n!}{k!(n-k)!}$$
Classic Examples
Find a Basic Probability
A fair six-sided die is rolled. Find $P(\text{even})$.
Solution Steps
- List the equally likely outcomes.
- Count the favorable even outcomes.
- Write and simplify the probability.
$$\{1,2,3,4,5,6\}$$$$\text{even outcomes}=\{2,4,6\}$$$$\boxed{P(\text{even})=\frac36=\frac12}$$
Use the Addition Rule
One card is drawn from a standard deck. Find $P(\text{heart or face card})$.
Solution Steps
- Count hearts and face cards.
- Subtract the face cards that are already hearts.
- Divide the favorable count by $52$.
$$13\text{ hearts}+12\text{ face cards}$$$$13+12-3=22\text{ favorable cards}$$$$\boxed{P(\text{heart or face})=\frac{22}{52}=\frac{11}{26}}$$
Find a Conditional Probability
A card is known to be a face card. Find $P(\text{king}\mid\text{face card})$.
Solution Steps
- Use only face cards as the conditioned sample space.
- Count kings among those face cards.
- Write the conditional probability.
$$\text{face cards}=12$$$$\text{kings among face cards}=4$$$$\boxed{P(\text{king}\mid\text{face card})=\frac4{12}=\frac13}$$
Find an Exact Binomial Probability
A fair coin is tossed four times. Find the probability of exactly two heads.
Solution Steps
- Use $n=4$, $k=2$, and $p=\frac12$.
- Choose the two toss positions that are heads.
- Multiply by the probability of each four-toss arrangement.
$$P(X=2)=\binom42\left(\frac12\right)^2\left(\frac12\right)^2$$$$\binom42=6$$$$\boxed{P(X=2)=6\left(\frac12\right)^4=\frac38}$$