Math 131: Discrete Distributions & Combinatorics

• Math 131, class 11 from Thursday, February 20.
• The course is divided into three areas: describing and summarizing data, probability, and statistical inference.
• There will be a midterm after the first two areas and a final covering everything at the end.
• Midterm #1 target date is January 30.
• Midterm #2 target date is March 11.
• The final target date is April 22.
• Overview of probability topics includes foundations, independent events/conditional probabilities, Bayes Theorem, recap, discrete random variables, continuous random variables, and normal distributions.
• Combinatorics involves figuring out ways to count things or combinations.

Plan for Today

• Chit-chat Drills, discussion, discrete distributions, and expected value/mean are planned.
• The discrete distributions include: Bernoulli, Categorical, Binomial, Geometric, and Poisson.

Drills

• Consider drawing a random card from a standard deck.
• Where A = draw a queen, B = draw a red card, C = draw a heart and D = draw an ace
• Describe the following: ANC, (AND) UC, -BUD
• Consider the following sets of real numbers:
• A = [-10, 10], B = [0, 1], C = (1, 2], D = [0, 1)
• Describe the following: BUC, CND, AN-(CUD)
• Tadej Pogocar is an excellent cyclist in all conditions but he is considered to be better in the cold/wet than in the heat
• He has an 80% chance of winning in the rain, a 60% chance of winning if it is cloudy and a 30% chance of winning if it is sunny.
• There is 40% chance of rain with other options being equally likely, so what is the probability that Pogacar wins?
• There is a map provided with nodes that list percentage wins, and asks the probability that you'll end up in a specific location.

Discussion

• Canadians, on average, are spending $173 per date according to a survey of some 2,500 adults.
• Questions to consider: Median or mean? Per person or total?

Rule of Succession

• This is a method for estimating probability based on previous observations.
• For instance, if Michael has worn his green hat to class every session so far, estimating the likelihood he wears it to the next class
• Pseudocounts involves the idea to increase the count of total observations by 2 and the count of "successful" observations by 1.
• Probability is estimated as (10+1)/(10+2) = 11/12

Discrete Probability Distributions

• This is chapter 4 in the textbook.
• There is a link provided to assets.openstax.org
• The Hypergeometric distribution will not be covered.

Discrete Probability

• A discrete random variable is a random process where the outcomes are discrete.
• Capital letters from the end of the alphabet denote discrete random variables, such as X and Y. Things categorized as discrete include:
• Coin flips, dice rolls, counts, and rankings. Things not categorized as discrete include:
• Heights, weights, times, and intervals.
• Sometimes, it is easier to pretend that a discrete variable is continuous.

Probability Mass Function

• Probability mass function is a function that tells us the probability that an outcome of a discrete random variable is exactly equal to one of the sample space values.
• P(X=x)
• A pmf would tell us the probability of rolling a 4 on a die (P(X=4)), but it would not tell us the probability of rolling an even number.

Bernoulli Distribution

• This a random variable that can take the value 1 or 0.
• P(X=1) = p
• P(X=0) = q = 1-p
• A common example is biased coin flips but it can be applied to anything with a True/False, Success/Failure or Yes/No answer.

Categorical Distribution

• It is like the Bernoulli Distribution except it allows for more than 2 values (X₁,..., Xn).
• P(X=xᵢ) = pᵢ for i<n
• P(X=xn) = 1 - p₁ - p₂ - … - pn-1

Binomial Distribution

• If there is a Bernoulli Trial with probability of success p and n independent trials, then the total number of successes follows a binomial distribution.
• Write as X ~ Bin(n, p)
• How many heads in 10 coin flips can be shown as X ~ Bin(10, 0.5).
• The pmf for a binomial distribution is: n! P(X = k) = pk (1-p)n-k k!(n − k)!
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• Where "!" is the factorial function; 5! = 54321 and 0! = 1

Example - Binomial Distribution

• Calculate the probability of getting 5 heads from 10 coin flips: is 0.246

Geometric Distribution

• It is useful to show the amount of the Bernoulli Trials to run before the first success.
• P(X = k) = (1-p)k-1p
• For example: how many tails before you get a head or how many people do you need to ask out before you get a date?

Craps

• The "shooter" rolls two dice.
• If the total is 7 or 11, they win.
• If the total is 2, 3, or 12, they lose.
• If the total is anything else, that becomes their "point."
• Once a point has been set, the shooter continues to roll:
• If they roll 7, they lose.
• If they match their point, they win.

Geometric Distribution

• Used to calculate the probability that the game lasts more than 6 rolls after the point has been set, assuming e point is 4.
• In the game above to calculate:
• P(4 or 7) = P(4) + P(7) = P({1,3}) + P({3,1}) + P({2,2}) + P({6,1}) + ... + P({3,3}) = 9/36
• P(more than 6 rolls) = 1 - P(¬(more than 6 rolls)) = 1 - P(5 or fewer rolls) = 1 - P(X=1) - P(X=2) - P(X=3) - P(X=4) - P(X=5) = 1 - (27/36)°(9/36) - (27/36)1(9/36) - ... - (27/36)5(9/36) = 0.176

Poisson Distribution

• This is the count of events that happen in a time interval, assuming events happen at a fixed rate and arrival times are independent.
• For instance, consider: the number of customers per hour, goals scored per half/quarter/game, or the number of insurance.
• The equations is: P(X = k) = λke-λ k!
• |
• λ (lambda) is a parameter known as the rate
• e = 2.718281828459...
• The average number of goals in a World Cup soccer match is approximately 2.5.

Poisson Distribution Examples

• To work out the likelihood that a match has zero goals, the calculations is: P(X = 0) = .082
• To work out the likelihood that a match has six goals. P(X = 6) = .028

Expected Value

• It is a weighted sum of the possible outcomes where the weights are the probabilities.
• Normally written E[X]
• If X is a dice roll, then: E[X] = (1/6)*1 + (1/6)*2 + (1/6)*3 + (1/6)*4 + (1/6)*5 + (1/6)*6 = 21/6 = 3.5
• If repeated rolling of a dice, the mean of the scores observed would get closer and closer to the expected value as the number of rolls increased.
• This is the law of large numbers

Expected Value Equations

• X ~ Bernoulli(p) ⇒ E[X] = p
• X ~ Bin(n, p) ⇒ E[X] = np
• X ~ Geo(p) ⇒ E[X] = 1/p
• X ~ Poisson(λ) ⇒ E[X] = λ