Flip a fair coin once. Will it land heads? You genuinely cannot say — that single flip is a random process: the outcome is uncertain, but the long-run pattern is predictable.
Now flip it 10 times. You might get 7 heads. Flip it 1,000 times and you'll land remarkably close to 500. This is the heart of probability: the proportion of heads settles toward 0.5 as the number of trials grows. That number, 0.5, is the probability of heads.
Here's a question to chew on before we start. A weather app says "70% chance of rain tomorrow." There's only one tomorrow — you can't repeat it 1,000 times. So what does 70% even mean? Hold that thought. By the end of this lesson you'll see that probability is fundamentally a statement about long-run relative frequency, and that everything else — the rules, the Venn diagrams, the formulas — is just careful bookkeeping built on that one idea.
A random process generates outcomes that are individually uncertain but follow a predictable distribution over many repetitions. The probability of an outcome is the proportion of times it would occur in the long run — over an enormous number of repetitions.
This is the law of large numbers (LLN): as the number of trials increases, the observed relative frequency of an outcome converges to its true probability. Ten flips of a fair coin might give 70% heads; ten thousand flips will hover near 50%. The LLN does not say the counts even out (you won't necessarily get exactly 5,000 heads) — it says the proportion stabilizes.
Why this matters: "70% chance of rain" means that across all the days the model judges to be like tomorrow, it rains about 70% of the time. Probability is a long-run frequency, not a guarantee about one day.
The sample space S is the set of all possible outcomes of a random process. An event is any subset of the sample space — a collection of outcomes we care about.
Roll one fair six-sided die. The sample space is S = {1, 2, 3, 4, 5, 6}. Let event A = "roll an even number" = {2, 4, 6}. Since all six outcomes are equally likely:
P(A) = (number of outcomes in A) / (number of outcomes in S) = 3/6 = 0.5
That equally-likely shortcut only works when outcomes are equally likely — true for a fair die, false for, say, the sum of two dice.
Every probability obeys two non-negotiable rules:
A, 0 ≤ P(A) ≤ 1. A probability of 0 means impossible; a probability of 1 means certain. There is no such thing as a probability of 1.4 or −0.2.P(S) = 1. Something in the sample space must happen.The complement of event A, written Aᶜ, is the event "A does not happen." Because A and Aᶜ together cover the entire sample space and never overlap:
P(Aᶜ) = 1 − P(A)
If the probability it rains is 0.70, the probability it does not rain is 1 − 0.70 = 0.30. Simple, but it's the single most useful rule in this lesson — especially for "at least one" problems.
The union A ∪ B is the event "A or B (or both) happens." The intersection A ∩ B is "A and B both happen." The general addition rule is:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Why subtract P(A ∩ B)? Because when you add P(A) and P(B), the outcomes in the overlap get counted twice. Subtracting the intersection once removes the double-count.
[GRAPH: Venn diagram inside a rectangle labeled S (sample space). Two overlapping circles. Left circle labeled A, right circle labeled B. The left-only crescent is labeled "A only," the overlap (lens shape in the middle) is labeled "A ∩ B," the right-only crescent is labeled "B only." The region outside both circles but inside the rectangle is labeled "(A ∪ B)ᶜ — neither A nor B." Caption: "P(A ∪ B) covers both circles; the overlap A ∩ B belongs to both, so it is added once, not twice."]
Worked addition-rule example. In a class of 30 students, 18 play a sport, 12 play an instrument, and 7 do both. What's the probability a randomly chosen student plays a sport or an instrument?
Let A = plays a sport, B = plays an instrument.
P(A) = 18/30
P(B) = 12/30
P(A ∩ B) = 7/30
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= 18/30 + 12/30 − 7/30
= 23/30
≈ 0.767So about 0.767. Check it the long way: students who play a sport only = 18 − 7 = 11; instrument only = 12 − 7 = 5; both = 7. Total in the union = 11 + 5 + 7 = 23, and 23/30 ≈ 0.767. ✓
Two events are disjoint (or mutually exclusive) if they cannot happen at the same time — their intersection is empty, so P(A ∩ B) = 0. In a Venn diagram, the circles don't overlap. When events are disjoint, the general addition rule simplifies:
P(A ∪ B) = P(A) + P(B) (disjoint events only)
Rolling a single die, "roll a 2" and "roll a 5" are disjoint — one roll can't be both. So P(2 or 5) = 1/6 + 1/6 = 2/6 = 1/3. But "roll an even number" and "roll a number greater than 3" are not disjoint, because rolling a 4 or a 6 satisfies both.
Two events are independent if knowing whether one occurred does not change the probability of the other. For independent events, the multiplication rule gives the probability they both happen:
P(A ∩ B) = P(A) · P(B) (independent events only)
Worked multiplication-rule example. A free-throw shooter makes 80% of her shots, and each shot is independent of the others. What's the probability she makes two in a row?
P(make 1st) = 0.80
P(make 2nd) = 0.80
P(make both) = 0.80 × 0.80 = 0.64So 0.64. Three in a row would be 0.80³ = 0.512. The multiplication rule chains across as many independent events as you like.
These two ideas sound similar and trip up students constantly. They are opposites in an important sense.
P(A ∩ B) = 0.P(A ∩ B) = P(A)·P(B).Here's the key insight: two disjoint events with nonzero probabilities can never be independent. If A happens and A is disjoint from B, then B is now impossible — A occurring slammed B's probability down to 0. That's the opposite of "tells you nothing." Mathematically, if P(A) > 0 and P(B) > 0 but P(A ∩ B) = 0, then P(A ∩ B) = 0 ≠ P(A)·P(B) > 0, so the independence equation fails. Disjointness is a strong dependence, not independence.
A spinner lands on red with probability 0.15. What is the probability it does not land on red?
Strategy: Direct complement rule.
Solution: P(not red) = 1 − P(red) = 1 − 0.15 = 0.85.
Interpretation: Over many spins, about 85% land on something other than red.
At a coffee shop, 60% of customers order a drink, 25% order food, and 15% order both. Find the probability a randomly chosen customer orders a drink or food.
Strategy: Events overlap (15% order both), so use the general addition rule and subtract the intersection.
Solution:
P(drink ∪ food) = P(drink) + P(food) − P(drink ∩ food)
= 0.60 + 0.25 − 0.15
= 0.70Interpretation: 70% of customers order at least one of the two. Note the trap: adding 0.60 + 0.25 = 0.85 without subtracting the overlap double-counts the 15% who order both.
A two-way table classifies 200 people by whether they own a pet and whether they exercise daily.
Exercises | No Exercise | Total
Owns pet 48 | 72 | 120
No pet 32 | 48 | 80
Total 80 | 120 | 200Are "owns a pet" and "exercises daily" independent?
Strategy: Compare P(pet ∩ exercise) to P(pet)·P(exercise). If equal, independent.
Solution:
P(pet) = 120/200 = 0.60
P(exercise) = 80/200 = 0.40
P(pet)·P(exercise) = 0.60 × 0.40 = 0.24
P(pet ∩ exercise) = 48/200 = 0.24Since P(pet ∩ exercise) = 0.24 = P(pet)·P(exercise), the events are independent.
Interpretation: Owning a pet gives no information about whether someone exercises daily — the two characteristics are unrelated in this group.
A manufacturing line produces parts that are defective with probability 0.05, independently. A box contains 4 parts. What is the probability the box contains at least one defective part?
Strategy: "At least one" is the complement of "none." Finding "none" uses the multiplication rule across 4 independent parts; then subtract from 1.
Solution:
P(one part is good) = 1 − 0.05 = 0.95
P(all 4 good) = 0.95⁴ = 0.81450625
P(at least one defective) = 1 − P(all good)
= 1 − 0.81450625
= 0.18549375
≈ 0.185Interpretation: About 18.5% of boxes contain at least one defective part. Computing "at least one" directly would require adding the probabilities of exactly 1, 2, 3, and 4 defectives — the complement turns four cases into one.
Mistake 1: Treating disjoint and independent as the same thing. Students see "two separate events" and assume they're independent. Why it's wrong: disjoint events with nonzero probability are actually dependent — if one happens, the other becomes impossible. Fix: ask "Can both happen at once?" If no, they're disjoint (use P(A)+P(B)). If yes and one doesn't affect the other, they're independent (use P(A)·P(B)).
Mistake 2: Adding probabilities without subtracting the overlap. Using P(A ∪ B) = P(A) + P(B) when the events overlap double-counts the intersection and can even produce a probability above 1. Fix: default to the general addition rule P(A) + P(B) − P(A ∩ B). Only drop the intersection term after confirming the events are disjoint.
Mistake 3: Multiplying probabilities of non-independent events. P(A ∩ B) = P(A)·P(B) is valid only when A and B are independent. Applying it to dependent events (like drawing two cards without replacement) gives the wrong answer. Fix: verify independence first; if events are dependent, you need conditional probability (next lesson).
Mistake 4: Forgetting the complement on "at least one." Students try to add up every case of "1 or more." Fix: P(at least one) = 1 − P(none), and "none" is usually a clean product of independent probabilities.
Mistake 5: Reporting a probability outside [0, 1]. If your answer is 1.3 or −0.05, you made an arithmetic or rule error — stop and recheck. Every legitimate probability lives between 0 and 1.
MC 1. A fair six-sided die is rolled once. What is the probability of rolling a number greater than 4?
MC 2. If P(A) = 0.35, what is P(Aᶜ)?
MC 3. Events A and B are disjoint with P(A) = 0.3 and P(B) = 0.45. What is P(A ∪ B)?
MC 4. For events M and N, P(M) = 0.5, P(N) = 0.4, and P(M ∩ N) = 0.2. What is P(M ∪ N)?
MC 5. Two events A and B have P(A) = 0.6 and P(B) = 0.5. If they are independent, what is P(A ∩ B)?
MC 6. Which statement about two events with nonzero probability is correct?
MC 7. A basketball player makes each free throw independently with probability 0.9. What is the probability she makes three in a row?
MC 8. A component works with probability 0.8, independently of others. A system has 2 such components and works only if both work. What is the probability the system works?
MC 9 (in context). In a survey, 40% of respondents own a dog, 30% own a cat, and 12% own both. What is the probability a randomly chosen respondent owns a dog or a cat?
MC 10 (in context). A website's checkout fails with probability 0.10 on any given attempt, independently. A shopper makes 3 attempts. What is the probability at least one attempt succeeds?
Short Answer 11. Events E and F satisfy P(E) = 0.55, P(F) = 0.25, and P(E ∩ F) = 0.10. Find P(E ∪ F) and P(Eᶜ).
Short Answer 12. A quality inspector checks 5 independently produced light bulbs, each defective with probability 0.02. Find the probability that at least one bulb is defective.
Short Answer 13 (in context). At a school, 70% of students take a bus and 20% bring lunch from home. Suppose taking the bus and bringing lunch are independent. Find the probability a randomly chosen student both takes the bus and brings lunch.
Short Answer 14. Show whether events A and B are independent given P(A) = 0.5, P(B) = 0.3, and P(A ∩ B) = 0.10.
Short Answer 15 (in context). A medical test panel runs 3 independent tests on a healthy patient. Each test gives a false positive with probability 0.04. Find the probability the patient gets at least one false positive across the three tests.
The scenario. A regional airline tracks two recurring problems on its short-haul flights: a flight may depart late (event L) and a flight may have a mechanical inspection delay (event M). From a large historical database the airline estimates:
P(L) = 0.30 (a flight departs late)P(M) = 0.10 (a flight has a mechanical inspection delay)P(L ∩ M) = 0.06 (a flight is both late and has a mechanical inspection delay)(a) Find the probability that a randomly selected flight departs late or has a mechanical inspection delay (or both). (2 points)
(b) Find the probability that a randomly selected flight has neither problem. (2 points)
(c) Determine whether events L and M are independent. Justify your answer with an appropriate calculation. (3 points)
(d) Assume that delays on different flights occur independently of one another, and that each flight departs late with probability P(L) = 0.30. For 3 randomly selected flights, find the probability that at least one of the three departs late. (3 points)
(a) Using the general addition rule:
P(L ∪ M) = P(L) + P(M) − P(L ∩ M)
= 0.30 + 0.10 − 0.06
= 0.34The probability a flight departs late or has a mechanical inspection delay is 0.34.
(b) "Neither problem" is the complement of "at least one problem" (the union from part (a)):
P(neither) = 1 − P(L ∪ M) = 1 − 0.34 = 0.66The probability a flight has neither problem is 0.66.
(c) Events L and M are independent if and only if P(L ∩ M) = P(L)·P(M).
P(L)·P(M) = 0.30 × 0.10 = 0.03
P(L ∩ M) = 0.06Since 0.06 ≠ 0.03, the events are not independent. (In fact, late flights are more likely to also have a mechanical delay than independence would predict — the actual joint probability, 0.06, is double the 0.03 that independence requires.)
(d) "At least one of three departs late" is the complement of "none of the three departs late." Each flight departs on time with probability 1 − 0.30 = 0.70, and the three flights are independent:
P(none late) = (0.70)³ = 0.343
P(at least one late) = 1 − 0.343 = 0.657The probability that at least one of the three flights departs late is 0.657.
Part (a) — 2 points
P(L) + P(M) − P(L ∩ M) (subtracting the intersection).Part (b) — 2 points
1 − P(L ∪ M) (or equivalently computes it from a correct Venn breakdown).Part (c) — 3 points
P(L ∩ M) = P(L)·P(M).P(L)·P(M) = 0.03 and compares it to P(L ∩ M) = 0.06.0.06 ≠ 0.03.Part (d) — 3 points
P(none late) = 0.70³ = 0.343 using independence.Where students lose points:
P(L ∩ M) and answering 0.40 — the single most common error.1 − P(L) = 0.70 or 1 − P(M) instead of 1 − P(L ∪ M); only "neither problem" is correct.P(L ∩ M) = 0 instead of the multiplication criterion. The comparison must be P(L ∩ M) vs. P(L)·P(M).0.30³ (probability all three are late) or computing 1 − 0.30 = 0.70 and stopping. The complement of "at least one late" is "none late," which uses the on-time probability 0.70 cubed.### Multiple Choice
MC 1 — (B) 1/3. Numbers greater than 4 are {5, 6}: 2/6 = 1/3. (A) is P(exactly one specific value); (C) counts ">3"; (D) counts ">2".
MC 2 — (B) 0.65. Complement rule: 1 − 0.35 = 0.65. (A) repeats P(A); (C) is 1 − 0.30; (D) ignores that probabilities can't exceed 1.
MC 3 — (C) 0.75. Disjoint, so P(A ∪ B) = 0.3 + 0.45 = 0.75. (A) multiplies them; (B) and (D) are arithmetic slips.
MC 4 — (A) 0.70. General addition rule: 0.5 + 0.4 − 0.2 = 0.70. (B) forgets to subtract the overlap; (D) adds the intersection instead of subtracting.
MC 5 — (B) 0.30. Independent: P(A ∩ B) = 0.6 × 0.5 = 0.30. (A) uses subtraction; (C) adds and subtracts incorrectly; (D) adds the probabilities.
MC 6 — (C). Disjoint events (with nonzero probability) cannot be independent, because one occurring makes the other impossible. (A), (B), (D) all conflate the two concepts.
MC 7 — (C) 0.729. 0.9³ = 0.729. (A)/(B) is 0.3³ confusion or 3 × 0.09; (D) is a single shot.
MC 8 — (C) 0.64. Both work: 0.8 × 0.8 = 0.64. (A) is 0.8 × 0.2; (B) is 0.6²; (D) is one component.
MC 9 — (A) 0.58. General addition rule: 0.40 + 0.30 − 0.12 = 0.58. (B) forgets to subtract the overlap; (C) adds the overlap; (D) is only P(both).
MC 10 — (D) 0.999. P(at least one success) = 1 − P(all 3 fail) = 1 − 0.10³ = 1 − 0.001 = 0.999. (C) is 0.9³ (all succeed); (B) is 1 − 0.9³; (A) is 0.1³ (all fail).
### Short Answer
11. P(E ∪ F) = 0.55 + 0.25 − 0.10 = 0.70. P(Eᶜ) = 1 − 0.55 = 0.45.
12. P(at least one defective) = 1 − P(none defective) = 1 − 0.98⁵. 0.98⁵ = 0.9039207968, so 1 − 0.9039207968 ≈ 0.0961. About 0.096.
13. Independent, so P(bus ∩ lunch) = 0.70 × 0.20 = 0.14.
14. Check P(A)·P(B) = 0.5 × 0.3 = 0.15. Since P(A ∩ B) = 0.10 ≠ 0.15, the events are not independent.
15. P(at least one false positive) = 1 − P(no false positives) = 1 − 0.96³. 0.96³ = 0.884736, so 1 − 0.884736 = 0.115264 ≈ 0.115. About 0.115.
### FRQ Rubric (restated)
- (a) 2 pts: 1 pt for the general addition rule setup with the intersection subtracted; 1 pt for 0.34.
- (b) 2 pts: 1 pt for recognizing "neither" = 1 − P(L ∪ M); 1 pt for 0.66.
- (c) 3 pts: 1 pt for the criterion P(L ∩ M) = P(L)·P(M); 1 pt for computing 0.03 and comparing to 0.06; 1 pt for the conclusion not independent because 0.06 ≠ 0.03.
- (d) 3 pts: 1 pt for the complement-of-"none" approach; 1 pt for 0.70³ = 0.343; 1 pt for 0.657.
Total: 10 points.
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StatsIQ · Lesson 9 of 30 · Unit 2: Probability, Random Variables, and Probability Distributions · Phase 2: Probability
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