SAT Math 11: Probability and Evaluating Statistical Claims

Event

Any collection or combination of outcomes from a sample space, usually represented by an uppercase letter such as $A$ or $B$.

What is the multiplication rule for independent events?

If $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B)$.

Probability

A numerical measure of how likely an outcome or event is to occur. Probabilities range from $0$ to $1$, inclusive.

Chance experiment

An experiment performed under controlled conditions whose result is not predetermined, such as repeatedly flipping a coin.

Outcome

A possible result of an experiment. For one coin flip, the outcomes are heads and tails.

Sample space

The set of all possible outcomes of an experiment, commonly denoted by $S$. It can be represented by listing outcomes, a tree diagram, or a Venn diagram.

Two fair coins are flipped. What is the sample space, and why are $HT$ and $TH$ different outcomes?

The sample space is $\{HH,HT,TH,TT\}$. $HT$ means heads first and tails second, while $TH$ means tails first and heads second; order matters.

How are the values $P(A)=0$, $P(A)=0.5$, and $P(A)=1$ interpreted?

$P(A)=0$ means $A$ is impossible, $P(A)=1$ means $A$ is certain, and $P(A)=0.5$ means $A$ and its nonoccurrence are equally likely.

Long-term relative frequency

The proportion of repetitions producing a particular outcome or event. As the number of repetitions becomes very large, this observed relative frequency tends to approach the event's theoretical probability.

Law of large numbers

As the number of repetitions of a chance experiment increases, the observed relative frequency tends to get closer to the theoretical probability. It does not imply that short-run results must match theoretical probabilities exactly.

Equally likely outcomes

Outcomes are equally likely when each has the same probability, as with the six faces of a fair die.

For equally likely outcomes, how is the probability of an event calculated?

$P(A)=\dfrac{\text{number of outcomes in }A}{\text{total number of outcomes in }S}$.

A fair dime and a fair nickel are tossed. What is the probability of getting exactly one head?

The equally likely sample space is $\{HH,HT,TH,TT\}$. Exactly one head occurs in $HT$ and $TH$, so $P(\text{exactly one head})=\frac{2}{4}=\frac{1}{2}$.

A fair six-sided die is rolled. What is the probability of rolling at least 5?

The favorable outcomes are $\{5,6\}$, so $P(\text{at least 5})=\frac{2}{6}=\frac{1}{3}$.

Biased or unfair experiment

An experiment is biased when its outcomes are not equally likely. For example, a biased coin may produce heads more often than tails in the long run.

What is the difference between an empirical probability and a theoretical probability?

An empirical probability is based on observed relative frequency from data, whereas a theoretical probability is based on a model or mathematical reasoning. Empirical results generally approach the theoretical value with many repetitions.

Union: the event $A$ OR $B$

The union $A\cup B$ contains outcomes in $A$, in $B$, or in both. Outcomes shared by both events are included only once.

Intersection: the event $A$ AND $B$

The intersection $A\cap B$ contains only outcomes that belong to both $A$ and $B$ simultaneously.

Complement of an event

The complement $A'$ contains every outcome in the sample space that is not in $A$. Its probability satisfies $P(A')=1-P(A)$.

What relationship must hold between an event and its complement?

$P(A)+P(A')=1$. The event and its complement are mutually exclusive and together make up the entire sample space.

What is the addition rule for the probability of $A$ OR $B$?

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. The intersection is subtracted because it is counted twice in $P(A)+P(B)$.

When can the addition rule be simplified to $P(A\cup B)=P(A)+P(B)$?

It can be simplified when $A$ and $B$ are mutually exclusive, because then $P(A\cap B)=0$.

Mutually exclusive events

Events are mutually exclusive if they cannot occur at the same time. Equivalently, they have no outcomes in common and $P(A\cap B)=0$.

A fair die is rolled. Let $A$ be odd and $B$ be even. What is the relationship between $A$ and $B$?

$A$ and $B$ are complements: $A'=B$, $A\cap B=\varnothing$, and $P(A)+P(B)=1$. They are also mutually exclusive.

Two fair coins are flipped. Are the events 'at most one tail' and 'two faces are the same' mutually exclusive?

No. They both include the outcome $HH$, so their intersection is not empty and has positive probability.

Two fair coins are flipped. Are the events 'all tails' and 'heads on the first flip' mutually exclusive?

Yes. The first event is $\{TT\}$ and the second is $\{HH,HT\}$; they share no outcomes, so their intersection has probability $0$.

Two fair coins are flipped. What is the probability of getting at least one tail?

The event is $\{HT,TH,TT\}$. Thus $P(\text{at least one tail})=\frac{3}{4}$, equivalently $1-P(\text{two heads})=1-\frac{1}{4}$.

Conditional probability

The probability that event $A$ occurs given that event $B$ has already occurred, written $P(A\mid B)$. Conditioning restricts attention to the outcomes in $B$.

What is the formula for conditional probability $P(A\mid B)$?

Provided $P(B)>0$, $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$.

How does conditioning affect the sample space?

A condition such as $B$ reduces the relevant sample space from $S$ to the outcomes in $B$. Probabilities are then computed relative to $B$, not the original sample space.

How can a data table be used to calculate conditional probability?

For $P(A\mid B)$, use the count in the intersection $A\cap B$ as the numerator and the total count for condition $B$ as the denominator. The condition determines the relevant subgroup.

A fair die is rolled. Let $A=\{2,3\}$ and $B=\{2,4,6\}$. Find $P(A\mid B)$.

Once $B$ is known, the possible outcomes are $\{2,4,6\}$. Only $2$ is also in $A$, so $P(A\mid B)=\frac{1}{3}$.

A fair die is rolled. Let $A$ be an odd result and $C=\{3,5\}$. Find $P(C\mid A)$.

Conditioning on $A$ restricts the outcomes to $\{1,3,5\}$. Two of these are in $C$, so $P(C\mid A)=\frac{2}{3}$.

Independent events

Events $A$ and $B$ are independent when knowing that one occurred does not change the probability of the other.

How can independence of two events be verified?

Any one of these equivalent conditions is sufficient: $P(A\mid B)=P(A)$, $P(B\mid A)=P(B)$, or $P(A\cap B)=P(A)P(B)$, with relevant conditional probabilities defined.

If $P(A\mid B)=P(A)$, what does this imply?

It implies that $A$ and $B$ are independent, because occurrence of $B$ does not change the probability of $A$.

If $P(A)=0.6$, $P(B)=0.5$, and $P(A\cap B)=0.3$, are $A$ and $B$ independent?

Yes. Since $P(A)P(B)=(0.6)(0.5)=0.3=P(A\cap B)$, the events are independent.

If $P(A)=0.4$, $P(B)=0.2$, and $P(A\cap B)=0.08$, are $A$ and $B$ independent?

Yes. $P(A)P(B)=(0.4)(0.2)=0.08=P(A\cap B)$.

If $P(A)=0.75$, $P(B)=0.30$, and $P(A\cap B)=0.225$, are $A$ and $B$ mutually exclusive?

No. Mutually exclusive events must have $P(A\cap B)=0$, but here the intersection probability is $0.225$.

What is the key difference between independent and mutually exclusive events?

Independence concerns whether one event changes the probability of another; mutual exclusivity concerns whether the events can occur together. They are not synonymous.

Can two nontrivial events be both independent and mutually exclusive?

Only in a degenerate case where at least one event has probability $0$. If mutually exclusive events both have positive probability, then $P(A\cap B)=0$ but $P(A)P(B)>0$, so they are not independent.

If $P(A\cap B)=0$ and both $P(A)$ and $P(B)$ are positive, what can be concluded?

$A$ and $B$ are mutually exclusive but not independent, because $P(A\cap B)=0\ne P(A)P(B)$.

With-replacement sampling

After an item is selected, it is returned before the next selection. The probabilities remain unchanged from draw to draw, so selections are generally independent.

Without-replacement sampling

Selected items are not returned before later selections. The composition of the population changes, so successive selections are generally dependent.

Why are successive draws from a deck generally dependent when cards are not replaced?

The first draw changes the cards remaining in the deck, which changes the probabilities for the second draw. Therefore, knowledge of the first draw affects later probabilities.

How does the multiplication principle determine the number of outcomes for sequential experiments?

If one stage has $m$ possible outcomes and a second stage has $n$ possible outcomes for each first-stage result, the combined experiment has $mn$ ordered outcomes.