Use two-way tables and conditional probability

Your feedback matters:

Lesson progressPractice problems 0/4

Difficulty: Intermediate
Estimated time: 27 minutes
Domains: Problem Solving and Data Analysis
Techniques: Conditional-probabilityTwo-way-tablesConditioned-denominatorComplementsUnknown-counts

What you’ll learn

Spot the wording that limits a pick to one group.
Use a row, column, stated group or region as the new total.
Find $P(A\mid B)$ as the outcomes in $B$ that are also $A$ , over all the outcomes in $B$ .
Use a complement inside that smaller group.
Find a missing count from a conditional probability.

Prerequisites

Why this matters on the SAT

Let the condition choose the denominator

SAT probability questions often tuck a condition into a short phrase. That phrase changes who can be picked, and that changes the answer.

Solution to the example

The phrase among the volunteers who submitted a survey means you’re picking only from the $55$ volunteers in that column. Of those $55$ , $18$ attended coding. So

P(\text{coding}\mid\text{submitted}) =\frac{18}{55}.

Choice C is correct. B divides by all $96$ volunteers, and A is the same fraction simplified. But $96$ includes the $41$ volunteers who didn’t submit a survey, and the condition has already ruled them out.

SAT example

The table summarizes $96$ volunteers at a community event.

Workshop	Submitted a survey	Did not submit a survey	Total
Coding	$18$	$12$	$30$
Design	$21$	$19$	$40$
Science	$16$	$10$	$26$
Total	$55$	$41$	$96$

Among the volunteers who submitted a survey, one volunteer is selected at random. What is the probability that the selected volunteer attended the coding workshop?

A
$\frac{3}{16}$
B
$\frac{18}{96}$
C
$\frac{18}{55}$
D
$\frac{30}{55}$

Restrict first, then count

A condition is something you already know about the outcome before you find the probability. It shrinks the sample space, the set of outcomes that could happen, down to one subgroup.

You’ll see this written as

P(A\mid B)

and read as “the probability of $A$ given $B$ .” The vertical bar means given that. $B$ is the condition, so $B$ sets the denominator:

P(A\mid B) = \frac{\text{outcomes in both }A\text{ and }B} {\text{all outcomes in }B}.

So the order is: restrict first, then count. Use this three-part scan:

Name the condition. Which group do you already know the outcome is in?
Find that group’s total. That’s your denominator.
Count the target inside that group. That’s your numerator.

The condition can show up in several forms:

given that the student is in grade 11
of those who attended the event
among the devices that passed inspection
selected from the morning group
a table, list or region that only shows one group to begin with

With no condition, the denominator is the original total. With a condition, it’s the subgroup’s total, and the numerator has to come from inside that same subgroup.

Common mistake:

A target count can look right and still come from the wrong group. So start with the condition. Underline the given-that or among-the phrase, mark the row, column or region it names, and ignore everything outside it. Only then count the favorable outcomes.

Example: Condition on a table row

In a two-way table, one phrase names the condition and another names the target. The cell where they cross gives the favorable count.

Worked example

The table summarizes how $90$ students travel to school and whether they participate in an after-school club.

Travel method	Participates in the club	Does not participate in the club	Total
Bus	$14$	$22$	$36$
Walk	$18$	$12$	$30$
Bicycle	$9$	$15$	$24$
Total	$41$	$49$	$90$

One student is selected at random, given that the student travels by bus. What is the probability that the selected student does not participate in the club?

A
$\frac{11}{45}$
B
$\frac{22}{90}$
C
$\frac{7}{18}$
D
$\frac{11}{18}$

Step 1

Keep only the bus row

You already know the student travels by bus, so only the bus row is left. Its total, $36$ , is the denominator.

Step 2

Find the target inside the row

In the bus row, $22$ students don’t participate in the club. That’s the numerator.

Step 3

Divide and simplify

P(\text{not in club}\mid\text{bus}) = \frac{22}{36} = \frac{11}{18}.

Choice D is correct. The other choices show the usual slips. A and B divide by all $90$ students. C equals $\frac{14}{36}$ , so it uses the right row but counts the students who do join the club.

Check your understanding:

Now flip it. Using the same table, what is $P(\text{bus}\mid\text{participates in the club})$ ?

Notice restrictions that are already built in

Not every conditional question says given that. Sometimes the title or the first sentence has already narrowed the data for you.

The display below includes only the $48$ applications submitted by juniors.

Final decision	Number of junior applications
Accepted	$18$
Waitlisted	$12$
Denied	$18$
Total	$48$

If you pick one application from this display at random, the probability that it was not denied is

\frac{18+12}{48} = \frac{30}{48} = \frac{5}{8}.

The denominator is $48$ because every application shown is already a junior’s.

Use complements inside the group

An event and its complement, everything that isn’t the event, still fill the whole sample space. Now the sample space is the conditioned group:

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

For the junior applications, that gives the same answer a second way:

P(\text{not denied}\mid\text{junior}) =1-\frac{18}{48} =\frac{30}{48} =\frac{5}{8}.

Both ways stay inside the $48$ junior applications.

Check your understanding:

Among $70$ concertgoers seated in the balcony, $28$ arrived before the doors opened. If one balcony concertgoer is selected at random, what is the probability that the person did not arrive before the doors opened?

Condition on an area

A condition can also limit a random point to one part of a figure. Then you compare the favorable area inside that part with the whole area of that part.

The community garden below is divided into $60$ equal-area regions. The outlined east section contains $24$ regions, and $15$ of those are shaded.

Once you know the point is in the east section, only its

24

equal-area regions are possible.

Given that a randomly selected point is in the east section, the probability that it’s in a shaded region is

P(\text{shaded}\mid\text{east}) = \frac{\text{shaded area inside east}} {\text{total area of east}} = \frac{15}{24} = \frac{5}{8}.

The shaded regions outside the east section don’t count in the numerator, and nothing outside the east section counts in the denominator.

Common mistake:

When some shaded pieces sit outside the outline, it’s easy to count them too. The condition draws a new boundary, so cover the rest of the figure in your mind and count both the shaded area and the total area inside the outline only. Counting pieces works here because every piece has the same area.

Recover an unknown conditioned count

Sometimes a conditional probability gives you one count and hides another. If you keep track of what the numerator and denominator stand for, you can write an equation.

At a clinic, $24$ of the appointments booked online were rescheduled. For an appointment picked at random from those booked online, the probability that it was rescheduled is $\frac{3}{8}$ . Let $n$ be the total number of appointments booked online.

The condition is booked online, so $n$ is the denominator:

\frac{24}{n}=\frac{3}{8}.

Cross-multiply:

\begin{aligned} 3n&=24(8)\\[1.4em] 3n&=192\\[1.4em] n&=64. \end{aligned}

So $64$ appointments were booked online. The $\frac{3}{8}$ describes only the online appointments, which is why you never needed the clinic’s total.

Check your understanding:

Among $45$ evening attendees, the probability that a selected attendee is a student is $\frac{4}{9}$ . How many of the evening attendees are students?

Work these by hand

You’ll usually solve these by hand, with the three-part scan from earlier: name the condition, find its total, and count the target inside it. Simplify or convert the answer only if the question asks you to.

Desmos can work out $\frac{22}{36}$ or solve $\frac{24}{n}=\frac{3}{8}$ . What it can’t do is decide that the bus row has replaced the full table, or that the east section has replaced the whole garden. That choice is the real question.

If the arithmetic is awkward, the Bluebook calculator can cut down on errors. Still, write your setup so the denominator you chose stays visible.

Practice problems

The first problem uses a single row. After that, each one adds a twist: a restriction built into the display, a group you build from the story, then two conditions at once.

Condition on a row

Practice problem

The table summarizes participation in three library programs.

Program	Completed feedback form	Did not complete feedback form	Total
Coding	$15$	$9$	$24$
Writing	$18$	$12$	$30$
Art	$10$	$20$	$30$
Total	$43$	$41$	$84$

If one participant in the writing program is selected at random, what is the probability that the participant completed a feedback form?

Calculator loads as you approach

Pick the denominator from the writing row before you divide.

Use a complement inside the group

Practice problem

A competition report lists the awards received by all $50$ finalists.

Award	Number of finalists
Gold	$8$
Silver	$12$
Bronze	$20$
No award	$10$

If one finalist is selected at random, what is the probability that the finalist did not receive a bronze award?

Your answer

Calculator loads as you approach

A complement, counted inside the finalists.

Build the group from the story

Practice problem

A community program has $180$ volunteers. Of the $105$ volunteers who are available on weekdays, $\frac{3}{5}$ have first-aid certification. Of the remaining volunteers, $\frac{2}{5}$ have first-aid certification.

If one volunteer is selected at random from among those with first-aid certification, what is the probability that the selected volunteer is available on weekdays?

Your answer

Calculator loads as you approach

Find both certified counts before you divide.

Two conditions at once

Practice problem

The table summarizes the ages of $100$ visitors to a nature center and how they arrived.

Age	Walked	Arrived by bicycle	Arrived by bus	Total
18 to 29 years	$14$	$9$	$7$	$30$
30 to 49 years	$12$	$11$	$17$	$40$
50 years or older	$8$	$5$	$17$	$30$
Total	$34$	$25$	$41$	$100$

One visitor is selected at random from among those who were at least $30$ years old and did not arrive by bus. What is the probability that the selected visitor was $50$ years old or older?

Your answer

Calculator loads as you approach

Mark the rows and columns that meet both conditions before you add.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

A condition shrinks the sample space before you find the probability: restrict first, then count.
In $P(A\mid B)$ , $B$ is the condition, and it supplies the denominator.
Phrases like given that, of those who and among the signal a condition.
A table, display, story or figure can already be limited to one group, even without the word given.
Keep both the favorable count and the total inside the condition.
For a complement, subtract inside the group.
To find a missing count, set favorable-in-group over total-in-group equal to the probability.
Work these by hand, because choosing the group matters more than the division.

Next lesson

Estimate populations and margin of error

Use a sample statistic to estimate a population and interpret how margin of error affects precision.

Start next lesson

Practice

Practice this lesson

323 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

Start practice