The two rules that do the arithmetic of the chapter, and the two special cases that simplify them. The multiplication rule computes an AND probability as P(B) times the conditional P(A given B), and collapses to a plain product exactly when the events are independent. The addition rule computes an OR probability as the sum of the two probabilities minus their intersection, and loses the subtraction exactly when the events are mutually exclusive. Both general forms are always valid, so section 3.2's two properties turn out to be permissions to drop a term rather than separate techniques — which is why identifying them was worth the effort.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 3 — Probability Topics
Two Basic Rules of Probability
Objectives
Five outcomes. The first two are one rule and the next two are the other.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, pp. 179-182 — the section these objectives are drawn from
Warm-up
Section 3.1 found that for A = {1,2,3,4,5} and B = {4,5,6,7,8}, the union had eight members rather than ten.
Discussion prompt
Five plus five is ten, and the union had eight. Where did the other two go, and what would you have to do to the sum to fix it in general?
Hint: Ask which members were counted in both fives.
Answer:
The members 4 and 5 belong to both sets, so they were counted once in A's five and again in B's five. The sum of ten counts them twice.
Subtracting the size of the intersection once — two members — brings the total back to eight, and it does so for any pair of sets, because the intersection is exactly the part that gets double counted.
Written in probabilities rather than counts, that is the addition rule of this section. And when the two sets share nothing, the correction is zero and the plain sum is already right — which is section 3.2's mutually exclusive case appearing as a simplification rather than as a definition.
Concept
When calculating probability there are two rules to consider. The multiplication rule says that the probability of A AND B is the probability of B times the probability of A given B. The addition rule says that the probability of A OR B is the probability of A plus the probability of B minus the probability of A AND B.
the two rules — The multiplication rule: P(A AND B) equals P(B) times P(A given B), always. The addition rule: P(A OR B) equals P(A) plus P(B) minus P(A AND B), always. Independence turns the first into a plain product; mutual exclusivity removes the subtraction from the second.
\[ P(A \text{ AND } B) = P(B)P(A\mid B) \qquad P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) \]
Both general forms hold for any two events with no assumptions whatever, which is what makes them safe to reach for. The special cases are not different rules but the same rules with a term that has become zero or has been replaced: independence means P(A given B) equals P(A), and mutual exclusivity means P(A AND B) equals zero. Section 3.2's warning about the two properties now has a payoff — each one is a licence to delete something.
Figure (svg): The general multiplication and addition rules with their independent and mutually exclusive special cases shown as simplifications
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 179
Section
Section 1
Concept
If A and B are two events defined on a sample space, then the probability of A AND B is the probability of B times the probability of A given B. The rule may also be written the other way round, as the statement that the probability of A given B equals the probability of A AND B divided by the probability of B.
the multiplication rule — P(A AND B) equals P(B) times P(A given B), for any two events. It is section 3.1's definition of conditional probability rearranged, so it requires no assumptions at all — and rearranging it back recovers the conditional formula.
\[ P(A \text{ AND } B) = P(B)P(A\mid B) \qquad \text{equivalently} \qquad P(A\mid B) = \frac{P(A \text{ AND } B)}{P(B)} \]
The rule and section 3.1's conditional formula are the same equation written two ways, which is worth noticing because it means nothing new has been assumed. Which form you use depends on what the problem hands you: given a conditional and asked for a joint probability, multiply; given a joint probability and asked for a conditional, divide. The rule works for dependent events precisely because the conditional carries the dependence.
Figure (svg): A worked comparison for Carlos's two shots showing the incorrect product of 0.4225 against the correct multiplication rule answer of 0.585
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 179 — the multiplication rule and its rearrangement
Picture it
Example 3.15, where the events are dependent and the shortcut would fail.
Figure (svg): A worked comparison for Carlos's two shots showing the incorrect product of 0.4225 against the correct multiplication rule answer of 0.585
The phrase that decides everything is 'Carlos tends to shoot in streaks'. It says the second shot's probability rises after a successful first, so the conditional 0.90 replaces the unconditional 0.65, and the answer is 0.585 rather than 0.4225. That is a difference of about a quarter of the answer, produced by one clause of English — which is why section 3.1 insisted that reading the wording is the first important step.
Worked example
Example 3.15a. The conditional is given, so the rule applies directly.
\[ P(A) = 0.65, \quad P(B) = 0.65, \quad P(B\mid A) = 0.90 \]
Identify what is asked
Why: The probability that both goals are made.
\[ P(A AND B) \]
Choose the form of the rule
Why: The conditional given is P(B given A).
\[ P(B AND A) = P(B | A) P(A) \]
Substitute
Why: Point nine times point six five.
\[ (0.90) (0.65) \]
Evaluate
Why: The product.
\[ 0.585 \]
Figure (svg): The solution to Worked example both goals shown as a ladder of expressions, one row per legal move
\[ P(A \text{ AND } B) = P(B\mid A)P(A) = (0.90)(0.65) = 0.585 \]
Verify: confirm the answer sits below both individual probabilities
Why: A joint probability can never exceed either of the two probabilities separately, because both events must occur, and 0.585 is below 0.65. That check catches a rule applied upside down. Note also that the answer is above the product 0.4225: the streak makes the second shot more likely after a success, so requiring both is easier than independence would suggest, and the joint probability rises accordingly.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 180
Faded example
Helen makes free throws 75 percent of the time, and 85 percent after a made first shot.
Fill in the blanks
P(\text0.85) = P(D\mid C)P(C) = (0.6375)(0.75) = ___
Why: The conditional 0.85 replaces the unconditional 0.75 for the second shot, giving 0.6375. Assuming independence would have given 0.5625, understating the answer by about eight percentage points because the made first shot genuinely raises the chance of the second.
Worked example
The same rule read the other way, on the swim team.
\[ \text{Of } 150 \text{ members}, 75 \text{ are advanced and } 40 \text{ of those practise four times a week.} \]
Write the joint probability
Why: Forty of the whole team are both.
\[ \frac{40}{150} \]
Write the conditioning probability
Why: Seventy-five of the team are advanced.
\[ \frac{75}{150} \]
Divide
Why: The joint over the condition.
\[ \frac{\frac{40}{150}}{\frac{75}{150}} \]
Simplify
Why: The 150s cancel.
\[ \frac{40}{75} = 0.5333 \]
Figure (svg): The solution to Worked example rearranging to find a conditional shown as a ladder of expressions, one row per legal move
\[ P(\text{four}\mid\text{adv}) = \frac{40/150}{75/150} = \frac{40}{75} \approx 0.533 \]
Verify: confirm the cancellation is what the reduced sample space means
Why: The 150s cancel and leave forty over seventy-five — which is exactly counting within the advanced swimmers alone, section 3.1's reduced sample space. So the formula and the counting agree, as they must, and on a problem given in counts the counting route is faster. The formula earns its keep when the problem supplies probabilities rather than counts, which is the usual case from chapter 8 onward.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, pp. 179-181
Trap
\[ P(A) = 0.65, \quad P(B) = 0.65 \]
Compute P(A AND B) as the product of the two
Why: Both probabilities are given, so multiplying them looks like the rule.
\[ (0.65)(0.65) = 0.4225 \quad \text{(wrong: that form needs independence)} \]
Carlos shoots in streaks, so the second shot is not at 0.65 once the first has gone in — it is at 0.90.
\[ P(A \text{ AND } B) = P(B\mid A)P(A) = (0.90)(0.65) = 0.585 \]
Use the conditional the problem supplies, not the unconditional probability
Why: The general rule always takes a conditional; only independence lets it be replaced.
The giveaway is that the problem bothered to state P(B given A) at all. A conditional probability appearing in a question is almost always there because it differs from the unconditional one — if the events were independent there would be nothing to state. Section 3.2's default applies: assume dependent until shown otherwise, and here the streak clause shows the opposite of independence.
Sorting
Ask what the problem gives you and what it wants.
Sort into buckets
Sort each situation.
Item (c) is worth noticing: with a table of counts, dividing one count by another does the same job and skips the probabilities entirely. Section 3.4 builds on exactly that shortcut.
Two truths and a lie
All three concern the multiplication rule.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false in general and true only for independent events. Carlos's shots give 0.585 rather than 0.4225 because the streak makes them dependent. The general rule multiplies by a CONDITIONAL, and the plain product is the special case section 3.2 tells you how to earn.
Prediction
Commit before reasoning.
Predict first
Carlos shoots in streaks, so P(B given A) is 0.90 rather than 0.65. How does the true P(A AND B) compare with the product?
Correct: Larger than the product.
Why: The conditional 0.90 exceeds the unconditional 0.65, so multiplying by it gives a larger result: 0.585 against 0.4225. The direction follows the direction of the dependence — a positive association raises the joint probability above the product, and a negative one lowers it. Drawing two aces without replacement is the opposite case, where the conditional falls and the true joint probability is below the product.
Section
Section 2
Concept
If A and B are independent then the probability of A given B equals the probability of A. Substituting that into the multiplication rule turns it into the plain product of the two probabilities.
the independent case — When A and B are independent, P(A given B) equals P(A), so the multiplication rule P(A AND B) = P(A given B)P(B) becomes P(A AND B) = P(A)P(B). This is the third of section 3.2's conditions for independence, now seen as a consequence rather than a definition.
\[ P(A\mid B) = P(A) \;\Longrightarrow\; P(A \text{ AND } B) = P(A)P(B) \]
This closes a loop from section 3.2. There the product formula was listed as one of three equivalent conditions for independence, with no explanation of where it came from. Here it is derived: it is the general multiplication rule with the conditional replaced, which is legitimate exactly when the events are independent. So the third condition is not an extra fact to memorise but a consequence of the first.
Figure (svg): The general multiplication and addition rules with their independent and mutually exclusive special cases shown as simplifications
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 179 — the independent special case of the multiplication rule
Picture it
The general forms above, the simplified forms below.
Figure (svg): The general multiplication and addition rules with their independent and mutually exclusive special cases shown as simplifications
Reading the diagram downward is the safe direction: start from the general rule and simplify only when the condition has been established. Reading it upward — starting from the product and hoping it applies — is what produces the errors, because the simplified form gives a definite number whether or not it is entitled to.
Worked example
The same setup as Carlos, with the streak removed.
\[ \text{A shooter makes } 65\% \text{ of shots, with no streak effect.} \]
Note what independence gives
Why: The second shot's chance is unaffected.
\[ P(B | A) = P(B) = 0.65 \]
Substitute into the general rule
Why: The conditional becomes unconditional.
\[ P(A AND B) = P(A) P(B) \]
Compute
Why: Point six five squared.
\[ 0.4225 \]
Compare with the streaky case
Why: 0.4225 against 0.585.
Figure (svg): The solution to Worked example independent shots shown as a ladder of expressions, one row per legal move
\[ P(A \text{ AND } B) = (0.65)(0.65) = 0.4225 \]
Verify: confirm the general rule gives the same answer
Why: Applying the general rule with P(B given A) equal to 0.65 gives 0.65 times 0.65, which is the same 0.4225. The general rule is never wrong; the product is a shortcut through it. That is why the safe habit is to write the general form first and substitute the simplification, rather than reaching for the product and hoping independence holds.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, pp. 179-180
Faded example
On the swim team, P(novice) is 0.1867 and P(practises four times) is 0.5333.
Fill in the blanks
P(\text0.0996)P(\text0.0667) = (0.1867)(0.5333) \approx ___, \text___ ___
Why: The product is 0.0996 and the true joint probability is 0.0667, so the two differ and the events are dependent. The joint being lower says novices are under-represented among those who train four times a week, which is the substantive finding rather than just a failed test.
Worked example
Example 3.16e. The product form used as a test rather than as a formula.
\[ \text{Are being a novice and practising four times a week independent?} \]
Compute the joint probability
Why: Ten novices practise four times a week.
\[ \frac{10}{150} = 0.0667 \]
Compute each probability
Why: Twenty-eight novices; eighty practise four times.
\[ 0.1867\text{ and } 0.5333 \]
Multiply them
Why: The product the test compares against.
\[ 0.0996 \]
Compare
Why: The joint against the product.
\[ 0.0667\text{ is not } 0.0996 \]
Figure (svg): The solution to Worked example testing independence with the rule shown as a ladder of expressions, one row per legal move
\[ P(\text{nov AND four}) = 0.0667 \ne 0.0996 = P(\text{nov})P(\text{four}) \]
Verify: confirm what the direction of the gap says
Why: The joint probability is below the product, so novices practise four times a week LESS often than independence would predict — which makes sense, since serious training is what makes a swimmer advanced. The test does more than return a yes or no: the direction of the discrepancy describes the association. This is exactly what section 12's correlation coefficient will measure for numerical variables and what section 11.3's chi-square test will test formally.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 181
Trap
\[ P(A \text{ AND } B) = P(A)P(B) \text{, by the multiplication rule} \]
Treat the product as the multiplication rule itself
Why: It is the version most often seen, so it looks like the rule.
\[ \text{used on dependent events, this is simply false} \]
On the swim team it would give 0.0996 where the truth is 0.0667, an error of half as much again.
\[ P(A \text{ AND } B) = P(B)P(A\mid B) \text{, always; the product only when independent} \]
Write the general rule first, then simplify if entitled
Why: The product is a conclusion, not a starting point.
The same equation does two quite different jobs and it is worth keeping them apart. Used as a FORMULA on events already known to be independent, it computes a joint probability. Used as a TEST on events whose relationship is unknown, comparing the product with the true joint probability decides whether they are independent. Confusing the two amounts to assuming what was to be shown.
Discrimination
Ask whether independence has been established.
Sort into buckets
Sort each situation.
Two truths and a lie
All three concern the independent case.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. A joint probability above the product simply means the events are positively associated — Carlos's 0.585 against 0.4225 is exactly that, and nothing is miscalculated. The product equals the joint probability only under independence, and it can be either larger or smaller otherwise.
Prediction
Commit before reasoning.
Predict first
The true joint probability is below the product of the two probabilities. What does that say about the events?
Correct: Negatively associated.
Why: The joint probability falling short of the product means the two events co-occur less often than independence would predict, so knowing one occurred lowers the chance of the other. The swim team shows it: novices train four times a week less often than the overall rates would suggest. Mutual exclusivity is the extreme case where the joint probability falls all the way to zero, but any shortfall indicates negative association.
Section
Section 3
Concept
If A and B are defined on a sample space, then the probability of A OR B is the probability of A plus the probability of B minus the probability of A AND B.
the addition rule — P(A OR B) equals P(A) plus P(B) minus P(A AND B), for any two events. The subtraction corrects for the outcomes in both events, which the plain sum counts twice.
\[ P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) \]
Section 3.1 met this as a fact about counting: A with five members and B with five members had a union of eight, because the two shared members were counted twice in the ten. Dividing every count by the size of the sample space turns that identity into the addition rule, so nothing new is being asserted — the rule is the counting fact expressed in probabilities.
Figure (svg): Two overlapping circles with the intersection marked as counted twice, and an arrow to the addition rule showing the intersection subtracted once
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 179 — the addition rule
Picture it
Adding the two circles covers the overlap twice.
Figure (svg): Two overlapping circles with the intersection marked as counted twice, and an arrow to the addition rule showing the intersection subtracted once
The picture makes the correction inevitable rather than arbitrary. Shading A covers the lens once and shading B covers it again, so the sum of the two probabilities has counted those outcomes twice while the union contains them once. Subtracting the intersection once removes the surplus exactly, whatever the size of the overlap.
Worked example
Example 3.15b, using the joint probability computed earlier.
\[ P(A) = 0.65, \quad P(B) = 0.65, \quad P(A \text{ AND } B) = 0.585 \]
Write the rule
Why: Sum minus intersection.
\[ P(A) + P(B) - P(A AND B) \]
Substitute
Why: The three known values.
\[ 0.65 + 0.65 - 0.585 \]
Add the first two
Why: The uncorrected sum.
\[ 1.30 \]
Subtract the intersection
Why: Removing the double count.
\[ 0.715 \]
Figure (svg): The solution to Worked example either goal shown as a ladder of expressions, one row per legal move
\[ P(A \text{ OR } B) = 0.65 + 0.65 - 0.585 = 0.715 \]
Verify: confirm why the correction was essential here
Why: The uncorrected sum is 1.30, which is impossible — no probability can exceed one. That alone proves the subtraction is not optional, and it is the fastest check available on any addition-rule problem: if the sum of the two probabilities exceeds one, the events certainly overlap and the intersection must be removed. The corrected 0.715 is comfortably below one and above each individual probability, as a union must be.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 180
Faded example
P(A) = 0.4, P(B) = 0.5 and P(A AND B) = 0.15.
Fill in the blanks
P(A \text0.15 B) = 0.4 + 0.5 - 0.75 = ___
Why: Nine tenths minus 0.15 gives 0.75. The answer must lie between the larger individual probability, 0.5, and the sum, 0.9 — and 0.75 does, which is a quick sanity check on any union.
Worked example
The swim team, where the intersection is a cell rather than a given.
\[ \text{P(advanced OR practises four times a week)} \]
Find each probability
Why: Seventy-five advanced; eighty practise four times.
\[ \frac{75}{150}\text{ and } \frac{80}{150} \]
Find the intersection
Why: Forty are both.
\[ \frac{40}{150} \]
Apply the rule
Why: Sum minus intersection.
\[ \frac{75 + 80 - 40}{150} \]
Evaluate
Why: One hundred and fifteen of one hundred and fifty.
\[ \frac{115}{150} = 0.7667 \]
Figure (svg): The solution to Worked example a union from a table shown as a ladder of expressions, one row per legal move
\[ P = \frac{75 + 80 - 40}{150} = \frac{115}{150} \approx 0.767 \]
Verify: confirm by counting the members directly
Why: The members satisfying at least one condition are the 75 advanced, plus the 30 intermediates and 10 novices who train four times a week — a total of 115, which matches. Counting directly is the check, and it works because with a table the union can be assembled from disjoint pieces. Note that the 40 advanced swimmers who also train four times a week were counted once in the 75 and would have been counted again in the 80, which is precisely what the subtraction removes.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 181
Error analysis
P(A) = 0.65, P(B) = 0.65 and P(A AND B) = 0.585. Four students compute P(A OR B).
Annotate
On: \( \begin{aligned} &(1)\; 0.65 + 0.65 = 1.30 \\ &(2)\; 0.65 + 0.65 - 0.4225 = 0.8775 \\ &(3)\; 0.65 \times 0.65 = 0.4225 \\ &(4)\; 0.65 + 0.65 - 0.585 = 0.715 \end{aligned} \)
Error (2) is the subtle one and it is a compound of two lessons: it uses the right rule with the wrong ingredient, because it silently assumed the independence that section 3.2's default forbids. The intersection to subtract is always the true P(A AND B), computed by the multiplication rule with a conditional if necessary.
Estimation
P(A) = 0.6 and P(B) = 0.5, and nothing is known about their overlap.
Predict first
What range must P(A OR B) lie in?
Correct: Between 0.6 and 1.
Why: The union is at least as large as the larger of the two, which is 0.6, since A alone already gives that much. And it cannot exceed 1. The upper end 1.1 is impossible for a probability, and the sum exceeding one proves the events must overlap by at least 0.1. Bounding a union this way is often enough to check an answer without knowing the intersection at all.
Two truths and a lie
All three concern the addition rule.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Carlos's two probabilities sum to 1.30 and both are correct — what the excess proves is that the events must overlap, by at least 0.30. A sum above one is evidence about the intersection, not evidence of an error.
Prediction
Commit before reasoning.
Predict first
You know P(A) and P(B) but not P(A AND B). Can you compute P(A OR B)?
Correct: No: the intersection must be found or bounded.
Why: The two individual probabilities do not determine the overlap, which can be anything from zero up to the smaller of the two. That is why every addition-rule problem needs a third ingredient, and why so much of this chapter is about finding it — by a contingency table in section 3.4, by a tree in section 3.5, or by the multiplication rule here.
Section
Section 4
Concept
If A and B are mutually exclusive then the probability of A AND B is zero. Substituting that into the addition rule removes the correction, leaving the probability of A OR B as the plain sum of the two probabilities.
the mutually exclusive case — When A and B cannot occur together, P(A AND B) is zero, so P(A OR B) = P(A) + P(B) - 0 = P(A) + P(B). The correction is not omitted; it is present and equal to zero.
\[ P(A \text{ AND } B) = 0 \;\Longrightarrow\; P(A \text{ OR } B) = P(A) + P(B) \]
The book's Example 3.14 is the cleanest case. Klaus can afford only one vacation, so choosing New Zealand and choosing Alaska are mutually exclusive; with probabilities 0.6 and 0.35 the chance he chooses one of them is simply 0.95, and the remaining 0.05 is the probability he goes nowhere. Note what makes the events exclusive here: not the geography but the budget, which is a fact about the situation rather than about the destinations.
Figure (svg): Two overlapping circles with the intersection marked as counted twice, and an arrow to the addition rule showing the intersection subtracted once
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, pp. 179-180 — the mutually exclusive case, with Klaus's vacation
Picture it
When the circles do not meet, the sum is already correct.
Figure (svg): Two overlapping circles with the intersection marked as counted twice, and an arrow to the addition rule showing the intersection subtracted once
It is worth thinking of the correction as always present and sometimes zero, rather than as sometimes absent. That framing keeps the general rule in view and makes the special case a substitution rather than a different formula — which matters because the commonest error in this section is applying the simplified form to events that do overlap.
Worked example
Example 3.14. Mutual exclusivity comes from the budget.
\[ P(A) = 0.6 \text{ (New Zealand)}, \quad P(B) = 0.35 \text{ (Alaska)} \]
Establish mutual exclusivity
Why: He can afford only one, so he cannot choose both.
\[ P(A AND B) = 0 \]
Apply the addition rule
Why: Sum minus a zero intersection.
\[ 0.6 + 0.35 - 0 \]
Evaluate
Why: The sum.
\[ 0.95 \]
Interpret the remainder
Why: One minus the union.
Figure (svg): The solution to Worked example Klaus's vacation shown as a ladder of expressions, one row per legal move
\[ P(A \text{ OR } B) = 0.6 + 0.35 = 0.95, \quad P(\text{neither}) = 0.05 \]
Verify: confirm the leftover probability is meaningful
Why: The two options and 'neither' between them exhaust the possibilities, so their probabilities must total one — and 0.6 plus 0.35 plus 0.05 does. The 0.05 is not a rounding artefact but a real possibility the problem allows: Klaus might take no vacation at all. Had the two probabilities summed to more than one, mutual exclusivity would have been impossible and the setup contradictory, which is a useful consistency check on any problem of this shape.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 180
Sorting
Ask whether a single outcome could satisfy both events.
Sort into buckets
Sort each pair.
Item (e) is exclusive because of the budget rather than because of anything about the destinations. Mutual exclusivity often comes from a constraint in the situation rather than from the events being logically incompatible, and reading the problem for such a constraint is part of setting it up.
Worked example
Example 3.16d. Two categories that cannot overlap by construction.
\[ \text{P(advanced AND intermediate)} \]
Ask whether a member can be both
Why: The levels are exclusive categories.
State the intersection
Why: Nothing satisfies both.
\[ P = 0 \]
Conclude about exclusivity
Why: The intersection is empty.
Compute their union
Why: The subtraction vanishes.
\[ \frac{75 + 47}{150} = 0.8133 \]
Figure (svg): The solution to Worked example mutually exclusive on the swim team shown as a ladder of expressions, one row per legal move
\[ P = \frac{75}{150} + \frac{47}{150} = \frac{122}{150} \approx 0.813 \]
Verify: confirm against the third category
Why: The remaining members are the 28 novices, and 122 plus 28 is 150, so the union of advanced and intermediate is exactly the complement of novice. That gives a second route to the same answer: one minus 28 over 150 is 122 over 150. The three levels partition the team — mutually exclusive and between them exhaustive — which is the structure that makes both routes work.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 181
Trap
\[ P(\text{advanced}) = 0.5, \quad P(\text{practises four times}) = 0.533 \]
Add them to get the probability of either
Why: Both are given, so the sum looks like the union.
\[ 0.5 + 0.533 = 1.033 \quad \text{(impossible)} \]
Forty members are both advanced and training four times a week, so they were counted twice.
\[ P = 0.5 + 0.533 - \frac{40}{150} = 0.767 \]
Check whether the two events can co-occur before dropping the subtraction
Why: Only genuine mutual exclusivity licenses the plain sum.
A sum exceeding one is the loudest possible signal, but the error is just as real when the sum stays below one and no alarm sounds — that is the dangerous case. The reliable habit is to ask whether a single member could satisfy both descriptions. A swimmer can be advanced and train four times a week, so the events overlap; a swimmer cannot be advanced and intermediate, so those do not.
Faded example
Klaus, with probabilities 0.6 and 0.35 for two vacations he cannot both take.
Fill in the blanks
P(A \text0 B) = 0.6 + 0.35 - 0.95 = ___
Why: The intersection is zero because Klaus can afford only one vacation, so the correction vanishes and the union is the plain sum of 0.95. The remaining 0.05 is the probability that he takes no vacation at all, which the problem leaves open.
Two truths and a lie
All three concern the mutually exclusive case.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. A sum below one is perfectly consistent with overlap — the swim team's advanced and four-times events have probabilities summing to 1.03, but two events at 0.3 and 0.4 could overlap heavily and still sum to 0.7. A small sum proves nothing about the intersection, which is why it must be found rather than inferred.
Prediction
Commit before reasoning.
Predict first
Two events have probabilities summing to 1.2. What follows?
Correct: They must overlap by at least 0.2.
Why: The union cannot exceed one, and the addition rule says the union is 1.2 minus the intersection, so the intersection must be at least 0.2. That is a genuine deduction about the overlap from the two individual probabilities alone, and it is the reason mutual exclusivity is impossible whenever the probabilities sum past one. Nothing is wrong with either probability.
Section
Section 5
Concept
Each rule needs one ingredient beyond the two individual probabilities. The multiplication rule needs a conditional probability, and the addition rule needs the intersection. Identifying which rule the question calls for, and then locating that extra ingredient, is the whole of the setup.
choosing a rule — An AND question calls for the multiplication rule and requires a conditional probability. An OR question calls for the addition rule and requires the intersection. The special cases of section 3.2 supply that ingredient for free: independence gives the conditional, and mutual exclusivity gives the intersection as zero.
\[ \text{AND} \;\to\; \text{multiply, needs } P(A\mid B) \qquad \text{OR} \;\to\; \text{add, needs } P(A \text{ AND } B) \]
The two rules interlock, and a full problem usually needs both. Carlos's union required his intersection, which required the multiplication rule with the conditional 0.90; so the addition rule's extra ingredient was supplied by the other rule. That is the normal pattern, and it is why the two are presented together as the two basic rules rather than separately.
Figure (svg): Two columns showing which rule to use for an AND question and for an OR question
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, pp. 179-182 — the two rules applied together
Picture it
What each rule computes, and what each one needs.
Figure (svg): Two columns showing which rule to use for an AND question and for an OR question
The fourth row is the practical content. Neither rule works from the two individual probabilities alone, so every problem of this kind has a third quantity to be found, and the difficulty of a problem is almost always the difficulty of finding it. Sections 3.4 and 3.5 are two displays whose whole purpose is to make that quantity easy to read off.
Worked example
Example 3.16 worked end to end on the swim team.
\[ 150 \text{ members}: 75 \text{ advanced}, 47 \text{ intermediate}, \text{ rest novice}; 40, 30, 10 \text{ practise four times} \]
Find the novice count
Why: One hundred fifty minus seventy-five minus forty-seven.
\[ 28\text{ novices} \]
P(novice)
Why: Twenty-eight of one hundred fifty.
\[ 0.1867 \]
P(practises four times)
Why: Forty plus thirty plus ten.
\[ \frac{80}{150} = 0.5333 \]
P(advanced AND four times)
Why: The cell where both hold.
\[ \frac{40}{150} = 0.2667 \]
Figure (svg): The solution to Worked example a full problem, both rules shown as a ladder of expressions, one row per legal move
\[ P(\text{nov}) = \frac{28}{150}, \quad P(\text{four}) = \frac{80}{150}, \quad P(\text{adv AND four}) = \frac{40}{150} \]
Verify: confirm the level counts partition the team
Why: Seventy-five advanced plus forty-seven intermediate plus twenty-eight novice is exactly 150, so every member is in one level and none in two. That check matters because part d asks whether advanced and intermediate are mutually exclusive, and the partition is why they are. It also confirms the novice count, which the problem does not state directly and which every probability involving novices depends on.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, pp. 180-181
Sorting
Read for AND or OR, however it is phrased.
Sort into buckets
Sort each question.
Item (b) is the one whose wording hides the operation: 'at least one' is the reliable English for a union, exactly as section 3.1 said. Anything phrased as at least one, either, or any of should route to the addition rule.
Worked example
Carlos again, showing how the two rules chain.
\[ \text{Find } P(A \text{ OR } B) \text{ given } P(A), P(B) \text{ and } P(B\mid A) \]
Identify the rule for the question
Why: It asks for OR.
List what the rule needs
Why: Both probabilities and the intersection.
Get the intersection from the other rule
Why: Multiplication, using the given conditional.
\[ (0.90) (0.65) = 0.585 \]
Now apply the addition rule
Why: Sum minus intersection.
\[ 0.65 + 0.65 - 0.585 = 0.715 \]
Figure (svg): The solution to Worked example the ingredient comes from the other rule shown as a ladder of expressions, one row per legal move
\[ P(A \text{ OR } B) = 0.65 + 0.65 - (0.90)(0.65) = 0.715 \]
Verify: confirm the order of operations was forced
Why: The addition rule could not be applied first, because its third term was unknown. The multiplication rule could be applied immediately, because everything it needs was given. So the order is determined by what the problem supplies rather than chosen — and the general strategy follows: write down the rule the question calls for, see which term is missing, and go and get that term. Almost every multi-step probability problem has this shape.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, p. 180
Trap
\[ \text{the problem gives } P(A) \text{ and } P(B); \text{ so multiply them} \]
Pick the operation that uses the numbers on the page
Why: Two numbers are given, so an operation combining two numbers seems called for.
\[ \text{answers whatever question that operation happens to answer} \]
The question decides the rule. If it asked for OR, a product answers something else entirely and will be far too small.
\[ \text{read the question} \;\to\; \text{choose the rule} \;\to\; \text{find the missing term} \]
Let the wording choose the rule, then go looking for what it needs
Why: The given numbers are rarely the complete set of ingredients.
Almost every problem in this chapter withholds one ingredient deliberately, and finding it is the exercise. Reaching instead for whatever operation fits the given numbers produces a confident answer to an unasked question, which is harder to detect than an arithmetic slip because nothing about the number looks wrong.
Ranking
You are given P(A), P(B) and P(B given A), and asked for P(A OR B).
Put in order
Why: Read, choose the rule, find the missing term, substitute. The order is forced by what is available: the multiplication rule can run immediately and the addition rule cannot, so the detour through the other rule comes in the middle rather than at the start.
Two truths and a lie
All three concern choosing between the rules.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and it is the point of the whole idea. Each rule needs a third ingredient — a conditional for multiplication, the intersection for addition — and the two individual probabilities never determine it. That third quantity is what sections 3.4 and 3.5 exist to make findable.
Explain it
A classmate has memorised 'multiply for AND, add for OR' and is getting answers wrong.
Discussion prompt
In three sentences or fewer, tell them what the slogan leaves out.
Hint: Ask what each rule needs besides the two probabilities.
Answer:
Tell them the slogan is right about which rule but silent about the ingredients: multiplying needs a CONDITIONAL, not the other event's plain probability, and adding needs the INTERSECTION subtracted.
Those two extra terms are exactly what the slogan drops, and dropping them is only legal when the events are independent or mutually exclusive respectively.
The safe version is 'multiply by a conditional for AND, add and subtract the overlap for OR', and simplify only after checking which of section 3.2's properties actually holds.
Comparison
Fill the blanks. Every row of the right column is a term becoming zero or unconditional.
Comparison matrix
| Rule | General form, always valid | Special case |
|---|---|---|
| Multiplication (AND) | P(A AND B) = P(B) P(A given B) | independent: P(A AND B) = P(A)P(B) |
| Addition (OR) | P(A OR B) = P(A) + P(B) - P(A AND B) | mutually exclusive: P(A OR B) = P(A) + P(B) |
| Extra ingredient needed | multiplication needs a conditional | addition needs the intersection |
| What the special case costs | nothing, if it has been established | a wrong answer, if it has not |
The two general forms are worth writing down first every time. They never fail, and each special case is a substitution into them rather than a separate formula to remember — which is one fewer thing to get wrong under pressure.
Pattern
Six steps. The first two decide everything and the arithmetic is trivial once they are settled.
Step four is where section 3.2 pays off, and its defaults apply: assume dependent and not mutually exclusive unless you can show otherwise, because both shortcuts give definite wrong answers when unearned.
OpenStax Introductory Business Statistics 2e, §3.3 Two Basic Rules of Probability §3.3 Two Basic Rules of Probability
Check
The general multiplication rule.
Check your understanding
P(A) = 0.5 and P(B given A) = 0.4. What is P(A AND B)?
Answer: A
Why: The multiplication rule gives P(A) times P(B given A), which is 0.5 times 0.4, or 0.20. No independence assumption is needed because the conditional was supplied.
Check
The addition rule with an overlap.
Check your understanding
P(A) = 0.7, P(B) = 0.6 and P(A AND B) = 0.45. What is P(A OR B)?
Answer: A
Why: Adding 0.7 and 0.6 gives 1.3, and subtracting the intersection of 0.45 leaves 0.85 — a legal probability, larger than either event alone.
Check
Spot the unearned shortcut.
Check your understanding
Carlos makes 65 percent of shots but shoots in streaks, with P(second given first) = 0.90. A student computes P(both) as 0.65 times 0.65. What is wrong?
Answer: A
Why: The plain product is the independent special case. Here the conditional 0.90 differs from the unconditional 0.65, so the general rule must be used, giving 0.585 rather than 0.4225.
Real world
An insurer prices a policy covering two risks. Flooding has probability 0.02 in a given year and a burst pipe has probability 0.03. The pricing model computes the chance of at least one claim as 0.05 and the chance of both as 0.0006.
Discussion prompt
Which rule has been used for each figure, what has been assumed, and in what direction would a hard winter change the answers?
Hint: Ask whether the two risks could share a cause.
Answer:
The 0.05 is the addition rule with the subtraction dropped, which assumes the two risks are mutually exclusive — that a house cannot suffer both in the same year. That is plainly false, so the figure is slightly too high, though only by the size of the intersection.
The 0.0006 is the multiplication rule with the product form, which assumes independence. That is the more serious assumption, because a hard freeze causes burst pipes and the subsequent thaw causes flooding, so the two risks share a cause.
A hard winter pushes the two figures in opposite directions. The true joint probability rises well above 0.0006, since P(flood given burst pipe) is far higher than 0.02. And because the intersection grows, the true probability of at least one claim falls slightly below 0.05.
\[ P(\text{both}) = P(\text{pipe}) \cdot P(\text{flood} \mid \text{pipe}) \;\gg\; (0.03)(0.02) \]
The direction matters commercially. An insurer under-pricing the joint event is exposed precisely in the years when many policies claim at once, which is when the capital is needed — the same common-cause problem section 3.2 raised for redundant power supplies. This is why catastrophe models are built around dependence between risks rather than around each risk separately, and why the general forms of both rules are the ones a serious model uses.
Commit first
Answer, then rate your confidence honestly.
Predict first
What does the addition rule subtract, and why?
Correct: The intersection, because the plain sum counts it twice.
\[ P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B): \quad \text{the last term is the double count} \]
Why: Adding P(A) and P(B) includes the shared outcomes in both terms, so they appear twice in a union that contains them once. Subtracting P(A AND B) removes the surplus exactly. The product is the intersection only for independent events, so option three is right only by accident; and the correction is always present, merely equal to zero when the events are mutually exclusive.
Explain it
They have added two probabilities and got 1.3.
Discussion prompt
In three sentences or fewer, tell them what the impossible answer proves and how to fix it.
Hint: Ask what a probability above one would mean.
Answer:
Point out that no probability can exceed one, so the answer is not merely wrong but impossible — and that impossibility is itself informative.
The plain sum only works for events that cannot both happen; here the excess of 0.3 above one proves the two events must overlap by at least that much.
The fix is the general addition rule: add the two probabilities and then subtract the probability that both occur, which brings the answer back below one.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the multiplication rule, use the conditional the problem gives rather than the unconditional probability. For the shortcut, check independence explicitly and default to dependent. For the addition rule, remember the intersection is a third quantity that must be found, often by the other rule. For choosing, translate 'both' as AND and 'at least one' as OR before doing anything else. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write both general rules in full and box them. Under each, write its special case and the condition that licenses it, with an arrow from the general form to the simplified one. Beside the addition rule, draw two overlapping circles and mark the lens as counted twice, with a note saying what the subtraction removes. In the middle, work Carlos completely: given P(A) = 0.65, P(B) = 0.65 and P(B given A) = 0.90, compute P(A AND B) and then P(A OR B), showing every substitution, and beside it write the wrong answers you would get by assuming independence for each. At the bottom left, build the swim team table: 150 members, 75 advanced, 47 intermediate, the rest novice, with 40, 30 and 10 practising four times a week. Fill in every cell and both totals, then compute P(novice), P(practises four times), P(advanced AND practises four times), and P(advanced OR practises four times). At the bottom right, test whether novice and practising four times are independent by comparing the product with the true joint probability, and write one sentence on what the direction of the gap means.
Check the swim team table by adding the three level totals: they must come to exactly 150, which confirms the novice count you had to derive. And check your union of 115 against the direct count of 75 advanced plus the 30 intermediates and 10 novices who train four times — if the two disagree, the intersection was subtracted wrongly.
Recap
Two rules, two special cases, and the habit of finding the missing ingredient.
| If you see | Then |
|---|---|
| A question asking for both | Multiplication rule, with a conditional |
| A question asking for at least one | Addition rule, minus the intersection |
| A conditional stated in the problem | The events are probably dependent; use it |
| Two probabilities summing above one | The events must overlap; never add plainly |
| Categories that cannot both apply | Mutually exclusive: the correction is zero |
| A joint probability below the product | Negative association between the events |
| A missing term in the rule you need | Get it from the other rule, or from a table |
Section 3.4 supplies the display that makes the missing ingredient easy to find. A contingency table lays out the counts for every combination of two variables, so a joint probability is one cell, a marginal is a row or column total, and a conditional is a cell divided by its row or column — which turns most of this chapter's arithmetic into reading.
OpenStax Introductory Statistics 2e, §3.3 Two Basic Rules of Probability §3.3, pp. 179-182 — everything on these slides traces back here
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