Discrete Probability Distributions – STAT 201 Study Notes

Discrete Probability Distributions

Introduction to Probability Distributions

Probability distributions are fundamental tools in statistics for describing the likelihood of various outcomes for a random variable. In the context of discrete probability distributions, the random variable takes on countable values, and each value has an associated probability.

• Discrete random variables represent counts or distinct outcomes (e.g., number of classes taken).

• A probability distribution for a discrete variable is typically presented as a table listing all possible values of the variable and their corresponding probabilities.

• For a probability distribution to be valid:

  • All probabilities must be between 0 and 1.

  • The sum of all probabilities must equal 1.

Example Table: Probability Distribution

x	P(x)
2	0.06
3	0.11
4	0.39
5	0.44

Justification: All probabilities are between 0 and 1, and their sum is 1.

Expected Value of a Probability Distribution

The expected value (or mean) of a discrete probability distribution represents the long-run average outcome if the experiment is repeated many times. It is a weighted average, where each value is weighted by its probability.

• Formula: $E(x) = ext{Expected value} = ext{mean} = ext{weighted average} = ext{sum of } x imes P(x)$ $E(x) = extstyleigsum x imes P(x)$

• Values of x are not equally likely; more probable values contribute more to the expected value.

Example Calculation

Given the table above:

$E(x) = 2(0.06) + 3(0.11) + 4(0.39) + 5(0.44) = 4.21$

Interpretation: The expected number of classes taken is 4.21. While it is not possible to take a fractional number of classes, the expected value represents the average over many students.

Application: Expected Profit Calculation

Probability distributions can be used to determine expected profit in business scenarios, such as choosing price points for event tickets.

• Suppose there are three seat types: Premium ($100), Middle ($50 or $35), Cheap ($20).

• Probabilities are assigned based on survey data for each seat type.

Example Tables: Price Point Scenarios

x (Price)	P(x)
100	0.05
50	0.20
20	0.75

x (Price)	P(x)
100	0.05
35	0.50
20	0.45

• Expected profit for first scenario: $100(0.05) + 50(0.20) + 20(0.75) = 30$

• Expected profit for second scenario: $100(0.05) + 35(0.50) + 20(0.45) = 31.5$

• Conclusion: Offering the middle level seats at a discounted rate increases expected profit.

Binomial Distribution

Definition and Conditions

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

• There must be two outcomes per trial: success or failure.

• The probability of success ($p$) must remain constant for each trial.

• The number of trials ($n$) must be fixed in advance.

• Trials must be independent; the outcome of one does not affect another.

Binomial Probability Formula

The probability of observing exactly $k$ successes in $n$ trials is given by:

$P(X = k) = inom{n}{k} p^k (1-p)^{n-k}$

• $inom{n}{k}$ is the binomial coefficient, representing the number of ways to choose $k$ successes from $n$ trials.

• $p$ is the probability of success; $1-p$ is the probability of failure.

Mean (Expected Value) of Binomial Distribution

The expected number of successes in $n$ trials is:

$E(X) = n imes p$

Factorial Review

Factorials are used in the binomial coefficient calculation.

• Definition: $n! = n imes (n-1) imes (n-2) imes ext{...} imes 1$

• $0! = 1$ and $1! = 1$

• When dividing factorials, common terms cancel out.

Calculating Binomial Probabilities

To find the probability of a range of successes (e.g., "at least k", "at most k"), use the binomial formula for each value and sum the probabilities, or use the complement rule:

• At least k: $P(X ext{ at least } k) = 1 - P(X < k)$

• At most k: $P(X ext{ at most } k) = P(X ext{ for } 0, 1, ..., k)$

• Complement rule: $P( ext{not } A) = 1 - P(A)$

Example: Lottery Tickets

• Suppose you buy 5 lottery tickets, each with a 0.10 probability of winning. Each ticket is independent.

• Probability that none win: $P(X=0) = inom{5}{0} (0.10)^0 (0.90)^5 = 0.5905$

• Probability that exactly one wins: $P(X=1) = inom{5}{1} (0.10)^1 (0.90)^4 = 0.3281$

• Probability that at least one wins: $P(X ext{ at least } 1) = 1 - P(X=0) = 1 - 0.5905 = 0.4095$

• Probability that at most one wins: $P(X=0) + P(X=1) = 0.5905 + 0.3281 = 0.9186$

• Expected number of winning tickets: $E(X) = 5 imes 0.10 = 0.50$

Summary Table: Binomial Distribution Properties

Property	Description
Number of trials ($n$)	Fixed, known in advance
Probability of success ($p$)	Constant for each trial
Independence	Each trial is independent
Outcomes	Two: success or failure
Random variable ($X$)	Number of successes in $n$ trials

Additional info: StatCrunch and other statistical software can be used to compute binomial probabilities efficiently.