Probability, Counting, and Discrete Distributions: Study Notes

Probability Rules and Conditional Probability

General and Special Multiplication Rules

Understanding how to calculate the probability of two events both occurring is fundamental in probability theory. The multiplication rules help us determine these probabilities, especially when events are dependent or independent.

• General Multiplication Rule: For any two events E and F, the probability that both occur is given by: $P(E \text{ and } F) = P(E)P(F|E)$

• Special Multiplication Rule (for Independent Events): If E and F are independent, then: $P(E \text{ and } F) = P(E)P(F)$

• Conditional Probability: The probability of event E occurring given that F has occurred is: $P(E|F) = \frac{P(E \text{ and } F)}{P(F)}$

Example: Suppose P(L) = 0.25, P(M) = 0.35, and P(M|L) = 0.20. Then:

• $P(L \text{ and } M) = P(L)P(M|L) = 0.25 \times 0.20 = 0.05$

• $P(L|M) = \frac{P(L \text{ and } M)}{P(M)} = \frac{0.05}{0.35} \approx 0.143$

• $P(L \text{ or } M) = P(L) + P(M) - P(L \text{ and } M) = 0.25 + 0.35 - 0.05 = 0.55$

Independence of Events

Two events are independent if the occurrence of one does not affect the probability of the other.

• If L and M are independent, then $P(L \text{ and } M) = P(L)P(M)$

• $P(L|M) = P(L)$ and $P(M|L) = P(M)$

Example: If P(L) = 0.20 and P(M) = 0.30, and L, M are independent:

• $P(L \text{ and } M) = 0.20 \times 0.30 = 0.06$

• $P(L \text{ or } M) = 0.20 + 0.30 - 0.06 = 0.44$

• $P(L|M) = 0.20$

Conditional Probability in Context

• Example: 20% of nursing students are both tired and cranky, and 30% are tired. Probability a tired student is also cranky: $P(\text{Cranky}|\text{Tired}) = \frac{P(\text{Tired and Cranky})}{P(\text{Tired})} = \frac{0.20}{0.30} \approx 0.67$

• Example: 40% of men have false teeth, and 30% of those also wear a toupee. Probability a man has false teeth and wears a toupee: $P(\text{False Teeth and Toupee}) = P(\text{False Teeth}) \times P(\text{Toupee}|\text{False Teeth}) = 0.40 \times 0.30 = 0.12$

Probability with Replacement and Without Replacement

• With Replacement: Each selection is independent; probabilities remain the same for each draw.

• Without Replacement: Each selection affects the next; probabilities change after each draw.

Example: An urn has 5 red and 10 blue marbles (15 total).

• Without Replacement: Probability both marbles are red: $P(\text{both red}) = \frac{5}{15} \times \frac{4}{14} = \frac{20}{210} \approx 0.095$

• With Replacement: Probability all three marbles are red: $P(\text{all red}) = \left(\frac{5}{15}\right)^3 = \frac{1}{27} \approx 0.037$

Probability of "At Least One" Events

• To find the probability of at least one event occurring, use the complement rule: $P(\text{at least one}) = 1 - P(\text{none})$

• Example: Probability of at least one tail in 4 flips of a biased coin (P(H) = 0.6): $P(\text{all heads}) = 0.6^4 = 0.1296$ $P(\text{at least one tail}) = 1 - 0.1296 = 0.8704$

Probability with Cards

• Example: Probability of drawing 2 Aces from a deck without replacement: $P(\text{both Aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} \approx 0.0045$

Probability from Two-Way Tables

Two-way tables summarize data for two categorical variables. Probabilities can be calculated using row and column totals.

Subject	Blue	Green	Yellow	Row Sum
Math	10	40	100	150
English	40	50	50	140
Science	30	100	80	210
Column Sum	80	190	230	500

• $P(\text{Math}) = \frac{150}{500} = 0.30$

• $P(\text{Green}) = \frac{190}{500} = 0.38$

• $P(\text{Math}|\text{Green}) = \frac{40}{190} \approx 0.211$

• $P(\text{Green}|\text{Math}) = \frac{40}{150} \approx 0.267$

• $P(\text{Math and Green}) = \frac{40}{500} = 0.08$

• $P(\text{Math or Green}) = P(\text{Math}) + P(\text{Green}) - P(\text{Math and Green}) = 0.30 + 0.38 - 0.08 = 0.60$

Counting Principles: Permutations and Combinations

Permutations

Permutations count the number of ways to arrange objects where order matters.

• Number of permutations of n objects taken r at a time: $P(n, r) = \frac{n!}{(n - r)!}$

• Example: Number of ways to arrange 5 brothers out of 15 for clean-up duty: Since order does not matter, use combinations (see below).

• Example: Number of 5-letter words from RSTUVWXYZ (no repeats): $P(8, 5) = \frac{8!}{(8-5)!} = \frac{40320}{6} = 6720$

Combinations

Combinations count the number of ways to select objects where order does not matter.

• Number of combinations of n objects taken r at a time: $C(n, r) = \frac{n!}{r!(n - r)!}$

• Example: Number of ways to select 5 brothers from 15 for clean-up duty: $C(15, 5) = \frac{15!}{5!10!} = 3003$

Discrete Probability Distributions

Mean and Standard Deviation of a Discrete Random Variable

A discrete probability distribution lists each possible value a random variable can take, along with its probability.

• Mean (Expected Value): $\mu = E[X] = \sum x P(X = x)$

• Variance: $\sigma^2 = \sum x^2 P(X = x) - \mu^2$

• Standard Deviation: $\sigma = \sqrt{\sigma^2}$

x	P(X = x)
1	0.1
2	0.2
3	0.2
4	0.2
5	0.3

• $P(X = 2) = 0.2$

• $P(X > 2) = 0.2 + 0.2 + 0.3 = 0.7$

• $P(X \geq 3) = 0.2 + 0.2 + 0.3 = 0.7$

• $P(X < 4) = 0.1 + 0.2 + 0.2 = 0.5$

• $\mu = 1 \times 0.1 + 2 \times 0.2 + 3 \times 0.2 + 4 \times 0.2 + 5 \times 0.3 = 0.1 + 0.4 + 0.6 + 0.8 + 1.5 = 3.4$

• $\sigma^2 = (1^2 \times 0.1 + 2^2 \times 0.2 + 3^2 \times 0.2 + 4^2 \times 0.2 + 5^2 \times 0.3) - (3.4)^2$ $= (0.1 + 0.8 + 1.8 + 3.2 + 7.5) - 11.56 = 13.4 - 11.56 = 1.84$

• $\sigma = \sqrt{1.84} \approx 1.36$

Binomial Distribution

Definition and Properties

The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success.

• Probability Mass Function: $P(X = k) = C(n, k) p^k (1-p)^{n-k}$

• Mean: $\mu = np$

• Variance: $\sigma^2 = np(1-p)$

Example: X ~ Binomial(n = 12, p = 0.4)

• $P(X = 5) = C(12, 5) (0.4)^5 (0.6)^7$

• $P(X < 5) = \sum_{k=0}^{4} P(X = k)$

• $P(X \geq 5) = 1 - P(X < 5)$

Application: In a sample of 12 people, probability exactly 5 suffer from anxiety (p = 0.4):

• $P(X = 5) = C(12, 5) (0.4)^5 (0.6)^7$

• For less than or at least 5, sum the probabilities accordingly.

Standard Normal Distribution and Z-Scores

Z-Score Calculation

The z-score measures how many standard deviations a value x is from the mean μ.

• $z = \frac{x - \mu}{\sigma}$

• Example: If $x = 35, \mu = 10, \sigma = 5$: $z = \frac{35 - 10}{5} = 5$

• Finding x from z: $x = \mu + z\sigma$ Example: $z = 2.25, \mu = 12, \sigma = 5$ $x = 12 + 2.25 \times 5 = 23.25$

Areas Under the Standard Normal Curve

• Use z-tables to find areas (probabilities) under the curve.

• To the left of z: $P(Z < z)$

• To the right of z: $P(Z > z) = 1 - P(Z < z)$

• Between two z-values: $P(a < Z < b) = P(Z < b) - P(Z < a)$

• Example:

  • $P(Z < -1.14)$: Look up -1.14 in the z-table.

  • $P(Z > -1.14) = 1 - P(Z < -1.14)$

  • $P(-1.1 < Z < 2.41) = P(Z < 2.41) - P(Z < -1.1)$

Assumption of Independence in Large Populations

• When sampling from a large population, individual selections can be considered independent if the sample size is small relative to the population.

Additional info: For all probability calculations, ensure that the sum of probabilities in a discrete distribution equals 1. For binomial and normal calculations, use statistical tables or technology as appropriate.