Calculate conditional probability using Bayes’ Theorem
Solve counting problems
Calculate the average value of a random event
Bayes’ Theorem
The Main Idea
Bayes’ Theorem is a fundamental principle in probability theory and statistics that describes how to update the probabilities of hypotheses when given evidence. It serves as a method to revise existing predictions or theories (prior probabilities) in light of new data. The theorem connects the conditional and marginal probabilities of two random events, providing a mathematical framework for updating beliefs based on new evidence. In its simplest form, Bayes’ Theorem calculates the probability of an event based on prior knowledge of conditions that might be related to the event.
A certain disease has an incidence rate of [latex]2\%[/latex]. If the false negative rate is [latex]10\%[/latex] and the false positive rate is [latex]1\%[/latex], compute the probability that a person who tests positive actually has the disease.
Imagine [latex]10,000[/latex] people who are tested. Of these [latex]10,000[/latex], [latex]200[/latex] will have the disease; [latex]10\%[/latex] of them, or [latex]20[/latex], will test negative and the remaining [latex]180[/latex] will test positive. Of the [latex]9800[/latex] who do not have the disease, [latex]98[/latex] will test positive. So of the [latex]278[/latex] total people who test positive, [latex]180[/latex] will have the disease. Thus
The basic counting rule, also known as the Fundamental Counting Principle, is a key concept in probability that allows for calculating the total number of possible outcomes for a series of events. It states that if there are [latex]n[/latex] ways for one event to occur and [latex]m[/latex] ways for another independent event to occur, then there are [latex]n[/latex] times [latex]m[/latex] ways for both to occur. This principle applies when the outcome of one event does not influence the outcome of another. For example, if you have [latex]3[/latex] shirts and [latex]4[/latex] pants, then you have [latex]3[/latex] times [latex]4[/latex], or [latex]12[/latex], different outfits to choose from.
There are [latex]21[/latex] novels and [latex]18[/latex] volumes of poetry on a reading list for a college English course. How many different ways can a student select one novel and one volume of poetry to read during the quarter?
There are [latex]21[/latex] choices from the first category and [latex]18[/latex] for the second, so there are [latex]21 \cdot 18 = 378[/latex] possibilities.
Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta) and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?
There are [latex]3[/latex] choices for an appetizer, [latex]5[/latex] for the main course and [latex]2[/latex] for dessert, so there are [latex]3 \cdot 5 \cdot 2 = 30[/latex] possibilities.
Permutations
The Main Idea
Factorials are mathematical operations applied to non-negative integers. The factorial of a number [latex]n[/latex], denoted as [latex]n![/latex], is the product of all positive integers less than or equal to [latex]n[/latex]. For example, the factorial of [latex]4[/latex] (denoted as [latex]4![/latex]) is [latex]4\times3\times2\times1[/latex], which equals [latex]24[/latex]. Notably, the factorial of [latex]0[/latex] is defined as [latex]1[/latex], reflecting the convention that there is exactly one way to arrange zero items.
Permutations are a key concept in combinatorics and deal with the arrangement of items in a specific order. The principle of permutations is used when order matters – that is, when the arrangement of items makes a difference. For instance, in a race, it matters who finishes first, second, or third. The number of permutations of [latex]n[/latex] objects taken [latex]r[/latex] at a time is denoted by [latex]P(n, r)[/latex] and is given by:
[latex]P(n, r) = \frac{n!}{(n-r)!}[/latex],
where [latex]![/latex] denotes a factorial, the product of an integer and all the integers below it, down to [latex]1[/latex]. For example, the permutations of three different items ([latex]A[/latex], [latex]B[/latex], and [latex]C[/latex]) are [latex]ABC, ACB, BAC, BCA, CAB, \text{ and } CBA[/latex], giving a total of [latex]6[/latex] permutations.
How many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a [latex]16[/latex]-member board of directors of a non-profit organization?
We want to choose [latex]4[/latex] people out of [latex]16[/latex] without replacement and where the order of selection is important.
The Main Idea Combinations are a key concept in combinatorics and statistics, used to determine the number of possible selections of items from a larger set, where the order of selection does not matter. For a given set of [latex]n[/latex] items, you can select [latex]r[/latex] items in [latex]C(n, r)[/latex] ways, where [latex]C[/latex] represents the combination function, also denoted as [latex]nCr[/latex] or “[latex]n[/latex] choose [latex]r[/latex]“. This is calculated as [latex]C(n, r) = \frac{n!}{r!(n-r)!}[/latex], where [latex]![/latex] denotes factorial. For instance, if you have [latex]5[/latex] books and you want to choose [latex]2[/latex] to take on a trip, there are [latex]C(5, 2) = 10[/latex] different pairs you could choose.
A charity benefit is attended by [latex]25[/latex] people at which three [latex]$50[/latex] gift certificates are given away as door prizes. Assuming no person receives more than one prize, how many different ways can the gift certificates be awarded?
Using the Basic Counting Rule, there are [latex]25[/latex] choices for the first person, [latex]24[/latex] remaining choices for the second person, and [latex]23[/latex] for the third, so there are [latex]25 · 24 · 23 = 13,800[/latex] ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the [latex]13,800[/latex] possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the [latex]13,800[/latex] possible outcomes. But either way Abe, Bea and Cindy each get [latex]$50[/latex], so it doesn’t really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are [latex]6[/latex]: ABC ACB BAC BCA CAB CBA. How can we be sure that we have counted them all? We are really just choosing [latex]3[/latex] people out of [latex]3[/latex], so there are [latex]3 · 2 · 1 = 6[/latex] ways to do this; we didn’t really need to list them all. We can just use permutations! So, out of the [latex]13,800[/latex] ways to select [latex]3[/latex] people out of [latex]25[/latex], six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted six times. Thus the [latex]13,800[/latex] figure is six times too big. The number of distinct three-person groups will be [latex]\frac{13,800}{6} = 2300[/latex].
A group of four students is to be chosen from a [latex]35[/latex]-member class to represent the class on the student council. How many ways can this be done?
Since we are choosing [latex]4[/latex] people out of [latex]35[/latex] without replacement where the order of selection is not important there are [latex]{}_{35}{{C}_{4}}=\frac{35\cdot34\cdot33\cdot32}{4\cdot3\cdot2\cdot1} = 52\mbox{,}360[/latex] combinations. View the following for more explanation of the combinations examples.
The expected value, also known as the expectation or the mean, is a fundamental concept in probability and statistics. It’s essentially a weighted average of possible outcomes of a random variable, where each outcome is weighted by its probability. In other words, the expected value gives the long-term average result of repeated trials of the same experiment. It is calculated by summing the products of each outcome and its associated probability. Importantly, the expected value may not always correspond to an outcome that is actually possible; rather, it gives a measure of the center of the distribution of the random variable.
Expected value notation: [latex]E(x)=\sum_{i}^{n}P(x_i)x_i[/latex]
In the casino game roulette, a wheel with [latex]38[/latex] spaces ([latex]18[/latex] red, [latex]18[/latex] black, and [latex]2[/latex] green) is spun. In one possible bet, the player bets [latex]$1[/latex] on a single number. If that number is spun on the wheel, then they receive [latex]$36[/latex] (their original [latex]$1 + $35[/latex]). Otherwise, they lose their [latex]$1[/latex]. On average, how much money should a player expect to win or lose if they play this game repeatedly?
Suppose you bet [latex]$1[/latex] on each of the [latex]38[/latex] spaces on the wheel, for a total of [latex]$38[/latex] bet. When the winning number is spun, you are paid [latex]$36[/latex] on that number. While you won on that one number, overall you’ve lost [latex]$2[/latex]. On a per-space basis, you have “won” [latex]\frac{-$2}{$38} ≈ -$0.053[/latex]. In other words, on average you lose [latex]5.3[/latex] cents per space you bet on. We call this average gain or loss the expected value of playing roulette. Notice that no one ever loses exactly [latex]5.3[/latex] cents: most people (in fact, about [latex]37[/latex] out of every [latex]38[/latex]) lose [latex]$1[/latex] and a very few people (about [latex]1[/latex] person out of every [latex]38[/latex]) gain [latex]$35[/latex] (the [latex]$36[/latex] they win minus the [latex]$1[/latex] they spent to play the game). There is another way to compute the expected value without imagining what would happen if we play every possible space. There are [latex]38[/latex] possible outcomes when the wheel spins, so the probability of winning is [latex]\frac{1}{38}[/latex]. The complement, the probability of losing, is [latex]\frac{37}{38}[/latex]. Summarizing these along with the values, we get this table:
Outcome
Probability of outcome
[latex]$35[/latex]
[latex]\frac{1}{38}[/latex]
[latex]-$1[/latex]
[latex]\frac{37}{38}[/latex]
Notice that if we multiply each outcome by its corresponding probability we get [latex]\$35\cdot \frac{1}{38}=0.9211[/latex] and [latex]-\$1\cdot \frac{37}{38}=-0.9737[/latex], and if we add these numbers we get:
[latex]0.9211 + (-0.9737) ≈ -0.053[/latex], which is the expected value we computed above.