Probability
Describe chance with sample spaces, event rules, conditional probability, independence, and expected value.
Reasoning about uncertainty
A random experiment has an uncertain individual result, even when its possible outcomes are known. Probability describes how likely an event is under a specified model.
- Construct sample spaces and events.
- Distinguish equally likely models from empirical estimates.
- Use complements, unions, conditional probability, and independence.
- Compute and interpret expected values in decision problems.
Experiments, sample spaces, and events
Before calculating a chance, describe exactly what can happen in one trial. Does the order matter, and what counts as a single outcome? Two sensible-looking lists can answer different questions if those choices differ. We will build the sample space first so that every later count refers to the same experiment.
A sample point is one outcome. The sample space \(S\) contains every possible outcome. An event \(E\) is a subset of \(S\).
| Experiment | Sample space |
|---|---|
| Toss one coin | \(\{H,T\}\) |
| Roll one die | \(\{1,2,3,4,5,6\}\) |
| Toss two coins in order | \(\{HH,HT,TH,TT\}\) |
Order matters when the experiment distinguishes a first and second result. A tree diagram records a branch for each possible next result.
Three coin tosses
Show worked solution
Two stock movements
Show worked solution
There are nine possible ordered outcomes, but their probabilities need not be equal. Outcome counting alone does not justify assigning a probability of five ninths.
Classical probability and counting
Counting favorable outcomes works only after we decide which elementary outcomes are equally likely. For example, different totals from a pair of dice need not have the same chance. Count the underlying outcomes before treating a list of event labels as an equally likely sample space.
If a finite sample space has equally likely outcomes, probability is the favorable count divided by the total count.
\[P(E)=\frac{|E|}{|S|}\]A fair die
Show worked solution
The favorable sets are {2, 4, 6} and {3, 6}, respectively.
Choose a pair
Show worked solution
There are \(\binom52=10\) equally likely unordered pairs. Only one pair contains no man.
\[P(\text{at least one man})=1-\frac1{10}=\frac9{10}\]For exactly one woman, choose one of two women and one of three men: \(2\cdot3=6\) pairs.
\[P(\text{exactly one woman})=\frac6{10}=\frac35\]Empirical probability
Sometimes we do not have a model that makes all outcomes equally likely. Instead, we use observed repetitions to estimate how often an event occurs. The relative frequency summarizes the available evidence; it is an estimate that may change with more observations.
An empirical probability uses observed relative frequency. It estimates a chance from data rather than assuming equally likely outcomes.
\[\widehat P(E)=\frac{\text{observations in }E}{\text{total observations}}\]| Payment method | Transactions |
|---|---|
| Credit card | \(320\) |
| Debit card | \(120\) |
| Cash | \(45\) |
| Mobile payment | \(15\) |
| Total | \(500\) |
Payment behavior
Show worked solution
This is an estimate based on the observed transactions. It does not guarantee that exactly this fraction will use credit next week.
| Audit duration in days | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) |
|---|---|---|---|---|---|
| Frequency | \(12\) | \(28\) | \(45\) | \(25\) | \(10\) |
An audit finishes early
Show worked solution
The total frequency is \(120\). Include the \(5\)-day and \(6\)-day categories.
\[\widehat P(X\le6)=\frac{12+28}{120}=\frac13\]Complements and mutually exclusive events
Sometimes it is easier to describe how an event fails than to count all the ways it succeeds. That is where the complement helps. Separately, ask whether two events can occur together: being mutually exclusive concerns overlap, while being complements also requires covering every possible outcome.
- \(0\le P(E)\le1\); impossible events have probability zero and certain events have probability one.
- \(E^c=\{x\in S:x\notin E\}\) is the complement of \(E\).
- \(P(E^c)=1-P(E)\)
- Mutually exclusive events cannot occur together: \(A\cap B=\varnothing\).
A complement
Show worked solution
Mutually exclusive and independent mean different things. Two mutually exclusive events with positive probabilities cannot be independent: one occurring rules out the other.
The addition rule: include the overlap once
When an event can happen in either of two ways, it is tempting simply to add their probabilities. Pause and ask whether one outcome can belong to both events. Drawing the overlap first explains exactly why a correction may be needed.
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]The union means “A or B or both.” Adding the two probabilities counts the overlap twice, so subtract it once.
Two payment services
Show worked solution
| Credit only | Both | GCash only | Neither |
|---|---|---|---|
| \(300\) | \(200\) | \(400\) | \(100\) |
Do not simply add probabilities unless you know the events are mutually exclusive.
Conditional probability changes the reference group
What “given” means
For an event of positive probability, conditional probability is the probability of its intersection with the event of interest divided by the probability of the conditioning event.
The condition becomes the new reference group. Count the desired outcomes inside that group and divide by the probability or count of the group, which must be nonzero.
New information can change which outcomes we are considering. Think of the condition as narrowing the group before we count successes. The denominator must describe that smaller group, not the original experiment.
The probability of \(A\) given \(B\) is calculated within the group where \(B\) has already occurred.
\[P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0\]| Owns a washing machine | Metro | Region | Total |
|---|---|---|---|
| Yes | \(51\) | \(33\) | \(84\) |
| No | \(25\) | \(22\) | \(47\) |
| Total | \(76\) | \(55\) | \(131\) |
Read the condition carefully
Show worked solution
The first denominator is all regional respondents. The second is all owners. Reversing the condition changes the question.
Independence and the multiplication rule
Conditional probability showed us how new information can change a chance. Independence is the special situation in which that information does not change it. Ask whether knowing the first result affects the second before multiplying the original probabilities. Drawing without replacement is a useful reminder that the experiment can change after the first outcome.
Events are independent if \(P(A\cap B)=P(A)P(B)\).
\[P(A\cap B)=P(B)P(A\mid B)\]When \(P(B)>0\), independence means that learning \(B\) occurred does not change the probability of \(A\).
The conditional multiplication formula requires \(P(B)>0\). The independence product formula also applies to events of probability zero.
Independent tosses
Show worked solution
Check independence from a table
Show worked solution
These proportions are not equal, so exact independence does not hold in this dataset. This arithmetic comparison alone does not establish causation or statistical significance.
Predict: disjoint or independent?
A bag has 3 green and 2 white tokens. Draw one token uniformly. Let G mean green and W mean white. Predict how the two events are related.
Follow every route to the event
A partition of a sample space is a collection of disjoint events whose union is the whole space. If the positive-probability events \(B_1,\ldots,B_k\) form a partition, the law of total probability is \(P(A)=\sum_iP(B_i)P(A\mid B_i)\). We multiply along each route and add disjoint routes. This connects sample-space trees with conditional probability; it does not require independence.
Select a box, then a token
A fair coin selects Box A or Box B. Box A contains \(9\) red and \(1\) blue token; Box B contains \(2\) red and \(8\) blue tokens. Draw one token uniformly from the selected box. Find the probability of red. Given that the result is red, find the probability Box A was selected.
Show worked solution
The disjoint red routes have probabilities \((1/2)(9/10)=9/20\) and \((1/2)(2/10)=2/20\). Thus \(P(R)=11/20\). Restricting to red outcomes gives \(P(A\mid R)=(9/20)/(11/20)=9/11\). The original box choice is fair, but the observed color changes our information about that choice.
Averaging two conditional probabilities equally works here only because the box-selection probabilities are equal. With unequal selection chances, use those chances as weights; do not pool tokens unless that matches the experiment.
Challenge: pairwise does not mean jointly independent
Toss two independent fair coins. Let A mean the first is heads, B the second is heads, and C that the two results match. Show that each pair of these events is independent, although all three together are not.
Show worked solution
The four outcomes are equally likely. Each event has probability \(1/2\). Every pairwise intersection consists only of HH, so each has probability \(1/4=(1/2)(1/2)\). But the triple intersection is also HH, of probability \(1/4\ne1/8=P(A)P(B)P(C)\). Testing pairs alone cannot justify multiplying three probabilities.
Expected value: a weighted average
What the expectation predicts
For a finite discrete random variable, expected value is the sum of each possible value multiplied by its probability.
It is a probability-weighted average, useful for comparing repeated payoffs or long-run costs. It need not be a possible outcome of one trial and does not guarantee what happens on the next trial. For a financial example, distinguish gross revenue from net gain after costs.
Knowing which outcomes are possible does not yet tell us the average result over repeated trials. Expected value combines each outcome with its probability, giving more influence to more likely outcomes. We will identify the payoff or net gain for each case before taking that weighted average.
A discrete random variable assigns a real number to each outcome and has a finite or countably infinite set of possible values.
For a finite set of values, multiply each distinct value by its probability, then add. The probabilities of all distinct values must sum to one.
For infinitely many possible values, a finite expected value requires the sum of absolute values weighted by their probabilities to converge.\[E(X)=\sum_i x_iP(X=x_i)\]
A return model
Show worked solution
The weighted average is not a guaranteed realized return. In this model, none of the individual outcomes equals the expected value.
Campaign revenue and cost
Show worked solution
Expected net revenue is positive, but a loss is still possible.
Compare separate bidding choices
Show worked solution
These are three separate decisions, not three outcomes of one probability distribution. The first has the greatest expected revenue under these assumptions; costs and other constraints are not included.
Practice: select the right model
1 · Two stages
Show worked solution
Each of the \(2\) coin outcomes can accompany each of \(6\) die outcomes, giving \(2\cdot6=12\).
2 · An event
Show worked solution
Use the complement: no heads means TTT.
\[P(\text{at least one head})=1-\frac18=\frac78\]3 · A frequency table
Show worked solution
4 · Only one service
Show worked solution
5 · A conditional denominator
Show worked solution
Restrict the denominator to Metro respondents.
6 · Dependence without replacement
Show worked solution
The first ace changes the remaining deck; do not multiply four fifty-seconds by itself.
7 · Expected payoff
Show worked solution
The model gives an average loss of five pesos per play over many repetitions, not a five-peso loss on every play.
8 · Explain the assumption
Show worked solution
Counting possible outcomes does not show equal likelihood. A probability model or empirical evidence is needed before assigning probabilities.
Additional practice: mixed exercises
Sample spaces and empirical probabilities
Give exact fractions where possible.
- The president, secretary, and auditor arrive in random order. List all possible arrival orders.
- A fair coin is tossed and a fair six-sided die is rolled. List the sample space.
- Two cards are drawn in order without replacement from a standard \(52\)-card deck. Describe the sample space and its size.
- Audit durations of \(5,6,7,8,9\) days have frequencies \(12,28,45,25,10\). Find the probability of exactly \(7\) days and of fewer than \(9\) days.
- Of \(50\) managers, \(34\) support a budgeting system. There are \(6\) Finance managers. Among \(11\) Operations managers, \(2\) have no opinion. Among \(33\) Marketing managers, \(5\) oppose it. Find the probabilities of support, Finance membership, no opinion given Operations, and opposition given Marketing.
- Among \(1000\) respondents, \(500\) use credit cards, \(600\) use GCASH, and \(200\) use both. Find the probabilities of either, GCASH only, and neither.
Show worked solution
- Label them \(P,S,A\). The sample space is \(\{PSA,PAS,SPA,SAP,APS,ASP\}\), with \(3!=6\) orders.
- \(S=\{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}\). There are \(2\cdot6=12\) outcomes.
- All ordered pairs \((c_1,c_2)\) of distinct cards; \(52\cdot51=2652\) outcomes. A card cannot be repeated.
- Total \(120\). Exactly \(7\): \(45/120=3/8\). Fewer than \(9\): \((12+28+45+25)/120=11/12\).
- Respectively \(34/50=17/25\), \(6/50=3/25\), \(2/11\), and \(5/33\). Conditional denominators use only the named department.
- Either: \((500+600-200)/1000=0.9\). GCASH only: \((600-200)/1000=0.4\). Neither: \(1-0.9=0.1\).
Complete the sample space
Distinguish possible outcomes from equally likely outcomes.
- Each of two stocks, JFC then ALI, can increase \((I)\), decrease \((D)\), or remain unchanged \((U)\). List the sample space and the events where at least one, and exactly one, increases.
- A box contains chicks with disease \(A\), \(B\), or \(C\). Record the disease of a first and second chick drawn without replacement. Assuming at least two chicks of each type are present, list a sample space. How many outcomes have the same disease? How many have different diseases?
- Among \(131\) surveyed households, \(51\) of the \(76\) Metro Manila households and \(33\) of the \(55\) Region 1 households own a washing machine. Find the probability that a household owns one, and the probability that it is from Region 1.
Show worked solution
- \(S=\{II,ID,IU,DI,DD,DU,UI,UD,UU\}\). At least one: \(\{II,ID,IU,DI,UI\}\). Exactly one: \(\{ID,IU,DI,UI\}\). These outcomes need not have equal probabilities.
- \(S=\{AA,AB,AC,BA,BB,BC,CA,CB,CC\}\). Same disease: \(AA,BB,CC\), giving \(3\) outcomes. Different: \(AB,AC,BA,BC,CA,CB\), giving \(6\). Disease counts are needed to calculate probabilities; these nine types are not necessarily equally likely.
- There are \(51+33=84\) owners, so \(P(\text{owns})=84/131\). The regional probability is \(55/131\). Both use the full sample as denominator because neither question is conditional.
Which rule should I use?
| Question wording | Useful idea |
|---|---|
| Not / at least one | Complement |
| A or B | Addition rule; check overlap |
| A given B | Conditional probability; restrict the group |
| A and B | Multiplication rule; check independence |
| Long-run average payoff | Expected value |
Audit assumptions and decisions
A correct probability calculation begins with a model. In these problems, first identify what is equally likely, what information is given, and whether one event changes the chance of another. Only then choose a formula.
Interpretation · A flag is not a certainty
A school file-checking system reviews \(1000\) records. Exactly \(20\) contain an error. It flags \(18\) of those and also flags \(98\) error-free records. One flagged record is selected uniformly. What is the probability it contains an error? Compare this with the fraction of erroneous records that are flagged.
Hint
The two questions condition on different groups: flagged records and erroneous records.
Show worked solution
In contrast, \(P(\text{flag}\mid\text{error})=18/20=0.9\). Catching most errors does not mean most flagged records have errors: the larger error-free group contributes many flags. The denominators express the different reference groups.
Reasoning · Equal average, different uncertainty
At a classroom reward station, option A awards exactly \(2\) tokens. Option B awards \(10\) tokens with probability \(1/5\) and zero otherwise. Compare expected tokens and the probability of receiving no tokens. Does equal expectation make the outcomes interchangeable for every student?
Hint
Expected value is a probability-weighted average; also inspect the possible outcomes.
Show worked solution
Option A has zero probability of no reward, whereas B has probability \(4/5\). Equal expected values describe equal long-run averages under repeated use of the models; they do not give the same distribution or guarantee an average result on a single trial. A student who needs at least one token certainly obtains it with A, but only with probability \(1/5\) with B.
Advanced / Honors · Independence and exclusion
Prove that mutually exclusive events \(A,B\) can be independent only if at least one has probability zero. Prove the converse as well.
Hint
Use the product definition of independence, which remains meaningful when a conditioning event has probability zero.
Show worked solution
Mutual exclusion gives \(A\cap B=\varnothing\), hence \(P(A\cap B)=0\). If the events are independent, \(P(A)P(B)=0\); because these are nonnegative real numbers, at least one is zero. Conversely, if the events are mutually exclusive and one has probability zero, then \(P(A\cap B)=0=P(A)P(B)\), so they are independent. Do not divide by a zero probability to use a conditional formula.
