Unit 4, lesson 17

Probability Rules and Events

How a probability model assigns probabilities to events in a sample space, how the complement and addition rules work, what unions, intersections and disjoint events are, and how probability relates to relative frequency.

  • Video 17
  • Events, Conditioning and Counting

A process model says that each part from a production line is defective with probability 0.030.03. One day an inspector checks 200 parts and finds 9 defective, an observed rate of 0.0450.045. Is the model wrong, and is a run of sound parts now due? Answering needs a precise language for chance. This lesson sets up sample spaces, events and the rules for complements and unions, and it separates a probability in a model from the relative frequency seen in one finite run.

Key terms

Sample space and events

Definition The sample space Ω\Omega of a chance process is the set of all its possible outcomes, where an outcome is one complete result. An event is a set of outcomes to which the model assigns a probability, and it occurs when the outcome that happens belongs to it. In a finite sample space every set of outcomes is usually taken to be an event. A probability model gives each event AA a number P(A)P(A) with 0≤P(A)≤10 \le P(A) \le 1 and P(Ω)=1P(\Omega) = 1, and for any finite or infinite sequence of disjoint events the probability of their union is the sum of their probabilities. For a finite sample space the model is fixed by giving each outcome a probability, with these outcome probabilities nonnegative and adding to 11. Then P(A)P(A) is the sum of the probabilities of the outcomes in AA. If Ω\Omega has nn outcomes, all equally likely, and AA contains kk of them,

P(A)=kn.P(A) = \frac{k}{n}.

In plain words List everything that can happen, then name the collections of outcomes you want to ask about. Counting outcomes is a valid way to find probabilities only when the model makes the outcomes equally likely. Having a list of labels is not enough to justify that.

Example For one roll of a die, Ω={1,2,3,4,5,6}\Omega = \{1, 2, 3, 4, 5, 6\} and the event “even” is A={2,4,6}A = \{2, 4, 6\}. If the die is fair, each outcome has probability 1/61/6 and P(A)=3/6=1/2P(A) = 3/6 = 1/2. Now take a weighted die model that gives the six outcomes the weights 1,1,2,2,3,31, 1, 2, 2, 3, 3 out of a total of 1212. The event “at least five” is {5,6}\{5, 6\}, and its probability is (3+3)/12=1/2(3 + 3)/12 = 1/2, while counting two of the six labels would give the wrong value 1/31/3.

Complement rule

Definition The complement AcA^c of an event AA is the set of outcomes in Ω\Omega that are not in AA. The events AA and AcA^c are disjoint and together make up Ω\Omega, so P(A)+P(Ac)=P(Ω)=1P(A) + P(A^c) = P(\Omega) = 1. This gives the complement rule

P(Ac)=1−P(A).P(A^c) = 1 - P(A).

In plain words Either an event happens or it does not, so the two probabilities share the total of 11. When an event is awkward to count directly, its complement is often simpler. The standard case is “at least one”, whose complement is “none”.

Example Roll two fair dice and treat the 3636 ordered pairs of results as equally likely. Let SS be the event of at least one six. Its complement “no six” contains the 5×5=255 \times 5 = 25 pairs that use only the faces 11 to 55, so

P(S)=1−2536=1136.P(S) = 1 - \frac{25}{36} = \frac{11}{36}.

Counting the pairs with a six directly gives 6+6−1=116 + 6 - 1 = 11, the same answer, but it needs care not to count the pair (6,6)(6, 6) twice.

Union and intersection

Definition The union A∪BA \cup B is the event containing every outcome that lies in AA, in BB or in both. It occurs when at least one of the two events occurs. The intersection A∩BA \cap B is the event containing the outcomes that lie in both, and it occurs when both events occur. In probability the word “or” means the union, which includes the outcomes where both events happen, unless “exactly one” is said. Since A∩BA \cap B lies inside AA and AA lies inside A∪BA \cup B, every model has P(A∩B)≤P(A)≤P(A∪B)P(A \cap B) \le P(A) \le P(A \cup B).

In plain words Union is the inclusive “or” and intersection is “and”. Joining events with “or” never removes outcomes, and joining them with “and” never adds any.

Example For one fair die let A={2,4,6}A = \{2, 4, 6\} be “even” and B={4,5,6}B = \{4, 5, 6\} be “at least four”. Then A∪B={2,4,5,6}A \cup B = \{2, 4, 5, 6\} and A∩B={4,6}A \cap B = \{4, 6\}, so P(A∪B)=4/6=2/3P(A \cup B) = 4/6 = 2/3 and P(A∩B)=2/6=1/3P(A \cap B) = 2/6 = 1/3. The event “exactly one of the two” is {2,5}\{2, 5\}, which leaves out the overlap.

Disjoint events

Definition Events AA and BB are disjoint, also called mutually exclusive, if they share no outcomes, that is, if A∩B=∅A \cap B = \varnothing. Events A1,…,AkA_1, \dots, A_k are disjoint if every two of them are disjoint. They are exhaustive if their union is the whole sample space, so that at least one of them occurs. Events that are both disjoint and exhaustive split Ω\Omega into pieces of which exactly one occurs, and their probabilities add to 11.

In plain words Disjoint events cannot happen together, and exhaustive events leave nothing out. The two properties are separate, and a collection of events can have either one without the other. Disjointness is also different from independence, which the next lesson defines. When two disjoint events both have positive probability, learning that one occurred rules out the other, which is strong information rather than a lack of connection.

Example For one die roll, “odd” {1,3,5}\{1, 3, 5\} and “even” {2,4,6}\{2, 4, 6\} are disjoint and exhaustive, and under a fair model their probabilities 1/21/2 and 1/21/2 add to 11. The events {1}\{1\} and {2}\{2\} are disjoint but not exhaustive. The events “at most four” {1,2,3,4}\{1, 2, 3, 4\} and “at least three” {3,4,5,6}\{3, 4, 5, 6\} are exhaustive but not disjoint, since both contain 33 and 44. Under a fair model their probabilities add to 4/6+4/6=4/34/6 + 4/6 = 4/3, more than 11, because the overlap is counted twice.

Addition rule

Definition For any two events AA and BB in a probability model,

P(A∪B)=P(A)+P(B)−P(A∩B).\begin{aligned} P(A \cup B) &= P(A) + P(B) \\[0.4em] &\quad - P(A \cap B). \end{aligned}

The reason is that A∪BA \cup B splits into the disjoint pieces AA and B∩AcB \cap A^c, while BB splits into the disjoint pieces B∩AB \cap A and B∩AcB \cap A^c. It is the two-event case of the inclusion-exclusion principle. When AA and BB are disjoint, the last term is 00 and P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B). More generally, if A1,…,AkA_1, \dots, A_k are disjoint, then P(A1∪⋯∪Ak)=P(A1)+⋯+P(Ak)P(A_1 \cup \dots \cup A_k) = P(A_1) + \dots + P(A_k).

In plain words Adding P(A)P(A) and P(B)P(B) counts the overlap twice, once inside each event. Subtracting P(A∩B)P(A \cap B) once leaves every outcome of the union counted exactly once. Without that correction the sum can even exceed 11.

Example A constructed model has 2020 equally likely outcomes. Event AA contains 88 of them, event BB contains 77, and 33 belong to both. Then

P(A∪B)=8+7−320=1220=0.6,\begin{aligned} P(A \cup B) &= \frac{8 + 7 - 3}{20} \\[0.4em] &= \frac{12}{20} = 0.6, \end{aligned}

and by the complement rule the probability that neither event occurs is 1−0.6=0.41 - 0.6 = 0.4. In a diagram the regions hold 55 outcomes in AA only, 33 in both, 44 in BB only and 88 in neither, and 5+3+4+8=205 + 3 + 4 + 8 = 20. If instead AA and BB had 88 and 77 outcomes with none in common, the overlap term would be 00 and P(A∪B)P(A \cup B) would be 15/20=0.7515/20 = 0.75.

Long-run relative frequency

Definition If a chance process is repeated nn times and an event AA occurs in mm of them, the relative frequency of AA in those trials is m/nm/n. The long-run interpretation of probability applies when the trials are modeled as independent repetitions under the same conditions, each giving AA the probability P(A)P(A). In that model, for every margin ε>0\varepsilon > 0, the probability that m/nm/n differs from P(A)P(A) by more than ε\varepsilon tends to 00 as nn grows, which is the law of large numbers. It does not say that the relative frequency moves steadily toward P(A)P(A), and it does not say that a finite run must balance out.

In plain words A probability belongs to the model and a relative frequency belongs to the data. Over many independent repetitions the two tend to be close, but a finite run varies, and an accurate count of what happened is still not the model’s probability. Short runs often show streaks and imbalances, and in a model of independent trials later outcomes do not compensate for earlier ones.

Example Suppose 1818 of 3030 rolls of a die come up even, a relative frequency of 18/30=0.618/30 = 0.6. That does not change the fair model’s probability 1/21/2. Under the fair model with independent rolls, 1818 or more even results in 3030 rolls has probability about 0.180.18, so a run like this is not unusual. After a streak of even results the next roll is still even with probability 1/21/2 in that model, because independent rolls keep no record of earlier outcomes.

Check yourself

  1. A model has 4040 equally likely outcomes. A diagram of events AA and BB shows 1010 outcomes in AA only, 66 in both, 88 in BB only and 1616 in neither. Use the addition rule to find P(A∪B)P(A \cup B), then find the probability that neither event occurs.
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The event AA contains 10+6=1610 + 6 = 16 outcomes and BB contains 8+6=148 + 6 = 14, so P(A)=0.4P(A) = 0.4, P(B)=0.35P(B) = 0.35 and P(A∩B)=6/40=0.15P(A \cap B) = 6/40 = 0.15. The addition rule gives P(A∪B)=0.4+0.35−0.15=0.6P(A \cup B) = 0.4 + 0.35 - 0.15 = 0.6, which matches the direct count (10+6+8)/40=24/40(10 + 6 + 8)/40 = 24/40. The complement rule then gives 1−0.6=0.41 - 0.6 = 0.4 for neither event, the 1616 outcomes outside both. Adding 0.4+0.35=0.750.4 + 0.35 = 0.75 without the correction would count the 66 shared outcomes twice.

  1. Events AA and BB in some model have P(A)=0.6P(A) = 0.6 and P(B)=0.5P(B) = 0.5. Can they be disjoint? What is the smallest possible value of P(A∩B)P(A \cap B)?
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If they were disjoint, the addition rule would give P(A∪B)=0.6+0.5=1.1P(A \cup B) = 0.6 + 0.5 = 1.1, which is impossible for a probability. In general P(A∪B)≤1P(A \cup B) \le 1, so the addition rule gives

P(A∩B)=P(A)+P(B)−P(A∪B)≥0.6+0.5−1=0.1.\begin{aligned} P(A \cap B) &= P(A) + P(B) \\[0.4em] &\quad - P(A \cup B) \\[0.4em] &\ge 0.6 + 0.5 - 1 \\[0.4em] &= 0.1. \end{aligned}

The bound 0.10.1 can be reached. Take 1010 equally likely outcomes 1,…,101, \dots, 10 with A={1,…,6}A = \{1, \dots, 6\} and B={6,…,10}B = \{6, \dots, 10\}. Then P(A)=0.6P(A) = 0.6 and P(B)=0.5P(B) = 0.5, and the overlap {6}\{6\} has probability 0.10.1. So the smallest possible value is 0.10.1, and the two events cannot be disjoint.

  1. A process model gives each part a defect probability of 0.030.03, independently from part to part. An inspector finds 99 defective parts among 200200. Find the relative frequency of defects. Does it show that the model probability is wrong, and does it make the next part more likely to be sound?
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The relative frequency is 9/200=0.0459/200 = 0.045. It describes this run of 200 parts, while 0.030.03 is a probability in the model, so the two numbers answer different questions. Under the model a run of 200 parts has 200×0.03=6200 \times 0.03 = 6 defects on average, but the count varies from run to run, and under the model 99 or more defects among 200200 parts has probability about 0.150.15. A count this high is not unusual under the model, so this run does not by itself show that the model is wrong. It does not show that the model is right either, and weighing such evidence is a task for the inference lessons later in the course. Under the model the next part is still defective with probability 0.030.03, because independent parts do not make up for earlier defects.

How this lesson connects