Probability and Two-Way Tables, SAT math lesson

A teacher pulls one name out of a hat holding the names of all $30$ students in a class. Nobody knows whose name comes out, yet everyone can say how likely it is that it belongs to someone who plays soccer. Probability turns that sense of “how likely” into a number between $0$ and $1$ . On the SAT the hard part is rarely the division. It is choosing the right group to divide by.

Key terms

Sample space

Definition The sample space of a random process is the set of all its possible outcomes, listed so that every run of the process gives exactly one of them.

In plain words The sample space is the full list of everything that could happen, with nothing left out and nothing listed twice.

Example Rolling one fair six-sided die has the sample space $\{1, 2, 3, 4, 5, 6\}$ . Rolling two dice and recording both numbers has $6 \cdot 6 = 36$ outcomes, such as $(1, 4)$ and $(4, 1)$ , which count as different outcomes because the dice are different.

Event

Definition An event is a set of outcomes, that is, a subset of the sample space. The event occurs when the outcome of the process belongs to that set.

In plain words An event is a description such as “the roll is even”, together with the outcomes that fit it.

Example For one die, the event “the roll is even” is $\{2, 4, 6\}$ and the event “the roll is greater than $4$ ” is $\{5, 6\}$ . The outcome $6$ belongs to both events.

Probability

Definition The probability $P(A)$ of an event $A$ is a number with $0 \le P(A) \le 1$ , where $P(A) = 0$ for an impossible event and $P(A) = 1$ for a certain one. When the sample space $S$ has finitely many outcomes that are all equally likely,

P(A) = \frac{\text{outcomes in } A}{\text{outcomes in } S}.

This counting formula needs equally likely outcomes. A random choice of one person from a group, made so that each person has the same chance, gives equally likely outcomes, and then the formula counts people.

In plain words When every outcome has the same chance, probability is favorable over total. Any answer below $0$ or above $1$ means a mistake.

Example For one fair die the probability of an even roll is $\frac{3}{6} = \frac{1}{2}$ . Now roll two fair dice and add them. The possible sums are $2$ through $12$ , eleven values, but the answer for a sum of $7$ is not $\frac{1}{11}$ , since the sums are not equally likely. Count the $36$ equally likely ordered pairs instead. Six of them, from $(1, 6)$ to $(6, 1)$ , give $7$ , so

P(\text{sum is } 7) = \frac{6}{36} = \frac{1}{6},

while only $(1, 1)$ gives a sum of $2$ , so that probability is $\frac{1}{36}$ .

Relative frequency

Definition If a process is observed $n$ times and an event occurs $k$ of those times, the relative frequency of the event is $\frac{k}{n}$ . Relative frequency from data is used as an estimate of the probability of the event, and in a table of counts it is the probability for a random choice from the people or items counted.

In plain words Relative frequency is “how often it happened” divided by “how many tries”. With lots of tries it is a reasonable stand in for the probability.

Example A factory inspects $400$ bolts and finds $6$ defective. The relative frequency of a defect is

\frac{6}{400} = 0.015.

Using that rate, about $0.015 \cdot 5000 = 75$ bolts in a batch of $5{,}000$ are expected to be defective. That is an estimate, not a promise.

Two-way table

Definition A two-way table counts a group by two categorical variables at once. Each row is a category of one variable, each column is a category of the other, each inner cell counts the members in both its row and its column, and the row totals, column totals and grand total add the cells. For a member chosen at random from the whole group, the probability of a cell or a row or a column is its count divided by the grand total.

In plain words A two-way table sorts everyone into exactly one inner cell. Before dividing, decide which total is the reference group the question describes.

Example A school surveyed $200$ students in grades $11$ and $12$ about whether they play a musical instrument.

	Instrument	No instrument	Total
Grade $11$	$42$	$78$	$120$
Grade $12$	$18$	$62$	$80$
Total	$60$	$140$	$200$

One of the $200$ students is chosen at random. The probability that the student plays an instrument is $\frac{60}{200} = \frac{3}{10}$ , and the probability that the student is in grade $12$ and plays an instrument is $\frac{18}{200} = \frac{9}{100}$ . A question that picks only among grade $12$ students has a different reference group, the $80$ students in that row, and belongs to Conditional Probability.

Complement

Definition The complement of an event $A$ , written $A^c$ or “not $A$ ”, is the set of all outcomes in the sample space that are not in $A$ . Since $A$ and its complement do not overlap and together make up the whole sample space,

P(\text{not } A) = 1 - P(A).

In plain words Either the event happens or it does not. Sometimes the “does not” side is much easier to count.

Example In the instrument table, the probability that a randomly chosen student does not play an instrument is

1 - \frac{3}{10} = \frac{7}{10},

which matches the column total $\frac{140}{200}$ . For three fair coins, the probability of at least one head is $1 - \frac{1}{8} = \frac{7}{8}$ , since the only outcome with no heads is all three tails.

Union

Definition The union of events $A$ and $B$ , written $A \cup B$ or ” $A$ or $B$ ”, is the event that $A$ occurs, $B$ occurs, or both occur. For any two events,

\begin{aligned} P(A \text{ or } B) &= P(A) + P(B) \\[0.4em] &\quad - P(A \text{ and } B). \end{aligned}

The overlap is subtracted once because it was counted in both $P(A)$ and $P(B)$ . When $A$ and $B$ cannot happen together, the rule shrinks to $P(A) + P(B)$ .

In plain words “Or” includes the people who belong to both groups, but each of them must be counted only once.

Example In the instrument table, find the probability that a randomly chosen student is in grade $12$ or plays an instrument. There are $80$ grade $12$ students and $60$ players, and $18$ students are both, so

\frac{80 + 60 - 18}{200} = \frac{122}{200} = \frac{61}{100}.

Adding $80 + 60 = 140$ without the subtraction would count those $18$ students twice. As a check, the cells in the union are $42 + 18 + 62 = 122$ .

Uniform area probability

Definition If a point is chosen uniformly at random from a region of positive area, meaning that equal areas within the region are equally likely to contain it, then the probability that it lands in a subregion $R$ is

P(\text{point in } R) = \frac{\text{area of } R}{\text{total area}}.

In plain words With a random point, area plays the role of counting. The bigger the share of the picture, the bigger the chance.

Example A square board is $8$ inches on each side, and a circle of radius $2$ inches is painted on it. A dart lands at a uniformly random point of the board. The probability that it lands inside the circle is

\frac{\pi \cdot 2^2}{8^2} = \frac{4\pi}{64} = \frac{\pi}{16},

which is about $0.196$ . As long as the whole circle lies on the board, its position does not matter, only its area.

Check yourself

Of $150$ people surveyed, $64$ own a dog, $45$ own a cat and $20$ own both. If one of the $150$ people is chosen at random, what is the probability that the person owns a dog or a cat? (A) $\frac{109}{150}$ (B) $\frac{89}{150}$ (C) $\frac{69}{150}$ (D) $\frac{20}{150}$

Show the answer

The correct choice is (B). By the union rule the count is $64 + 45 - 20 = 89$ , so the probability is $\frac{89}{150}$ . Choice (A) adds $64$ and $45$ and counts the $20$ owners of both pets twice. Choice (C) leaves out the people who own both, but “or” includes them. Choice (D) is the probability of owning both.

One of the $240$ people at a concert is chosen at random. The probability that the person holds a student ticket is $0.35$ . How many of the $240$ people do not hold a student ticket?

Show the answer

The answer is $156$ . The probability of not holding a student ticket is $1 - 0.35 = 0.65$ , and $0.65 \cdot 240 = 156$ . The tempting answer $84$ is the number of people who do hold a student ticket.

A circle with radius $5$ is drawn inside a square with side length $10$ , touching all four sides. A point is chosen uniformly at random inside the square. What is the probability that the point lies outside the circle? (A) $\frac{\pi}{4}$ (B) $1 - \frac{\pi}{4}$ (C) $1 - \frac{\pi}{10}$ (D) $\frac{4 - \pi}{\pi}$

Show the answer

The correct choice is (B). The circle’s area is $25\pi$ and the square’s area is $100$ , so the probability of landing inside the circle is $\frac{25\pi}{100} = \frac{\pi}{4}$ and the complement is $1 - \frac{\pi}{4}$ , about $0.215$ . Choice (A) is the probability of landing inside. Choice (C) uses the circumference $10\pi$ in place of the area. Choice (D) divides the area outside the circle, $100 - 25\pi$ , by the circle’s area $25\pi$ instead of by the square’s area, so it compares the two parts rather than the part with the whole.

Probability and Two-Way Tables

Theory, part 83

Practice, part 84

Key terms

Sample space

Event

Probability

Relative frequency

Two-way table

Complement

Union

Uniform area probability

Check yourself

How this lesson connects

Builds on

Used later in