SAT Probability & Conditional Probability — SATMath800

SATMath800 • Problem-Solving & Data Analysis

Probability & Conditional Probability

Read the representation first. Identify the correct sample space. Then calculate the probability. On the SAT, the hardest part is often not the arithmetic—it is deciding what the denominator should be.

What this lesson covers

Basic and complementary probability
Relative frequency
One-way and two-way tables
Conditional probability
Tree diagrams and area models
Independent and dependent events
Multi-step probability
Finding unknown frequencies from probabilities
SAT wording traps

The SAT habit

Before calculating, ask: “What outcomes are possible after I apply the condition?”

\[P(A)=\frac{\text{favorable outcomes}}{\text{total outcomes}}\]

For conditional probability, the condition usually changes the denominator.

1. Basic Probability

When all outcomes are equally likely, probability is the fraction of outcomes that satisfy the event.

\[P(A)=\frac{\text{favorable outcomes}}{\text{total outcomes}}\]

SAT-style example: A box contains 7 blue markers and 5 black markers. If one marker is selected at random, the probability that it is blue is \(7/(7+5)=7/12\).

Check: A probability must be between 0 and 1, inclusive.

2. Complementary Probability

The complement of an event is the event that it does not occur.

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

If \(P(A)=0.72\), then \(P(\text{not }A)=1-0.72=0.28\).

SAT trap: “At least one” is often easier to calculate using the complement: 1 − P(none).

3. Relative Frequency from Data

When probability is estimated from observed data, use relative frequency.

\[\text{relative frequency}=\frac{\text{frequency of the event}}{\text{total observations}}\]

Preferred study time	Students
Morning	18
Afternoon	27
Evening	15
Total	60

The observed probability that a randomly selected student prefers afternoon study is \(27/60=0.45\).

4. Reading Probability from a Two-Way Table

A two-way table separates observations by two categorical variables. First identify the event, then determine which total belongs in the denominator.

	Uses app	Does not use app	Total
Grade 11	32	18	50
Grade 12	24	26	50
Total	56	44	100

The probability that a randomly selected student uses the app is \(56/100=0.56\).

Key idea: If the question does not impose a condition, the denominator is usually the overall total.

5. Conditional Probability: The Denominator Changes

\(P(A\mid B)\) means the probability of \(A\) given that \(B\) has already occurred.

\[P(A\mid B)=\frac{P(A\text{ and }B)}{P(B)}\]

	Completed	Did not complete	Total
Morning group	21	9	30
Afternoon group	16	14	30
Total	37	23	60

If a student is known to be in the morning group, the relevant sample space contains only 30 students:

\[P(\text{completed}\mid\text{morning})=\frac{21}{30}=0.70\]

Classic SAT mistake: Do not divide by 60. The condition “morning group” has already restricted the sample space to 30 students.

6. “Given That” Language

Conditional probability may be written without the notation \(P(A\mid B)\).

Given that

“A student is known to be a senior.”

Among

“Among students who chose option A…”

If known

“If the selected student is in group B…”

\[\boxed{\text{Condition}\Rightarrow\text{new sample space}}\]

7. Addition Rule: “A or B”

For events that cannot happen together, add their probabilities.

\[P(A\text{ or }B)=P(A)+P(B)\]

More generally, if two events can overlap:

\[P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\]

The subtraction prevents the overlap from being counted twice.

8. Multiplication Rule: “A and B”

For two events occurring in sequence:

\[P(A\text{ and }B)=P(A)P(B\mid A)\]

If the events are independent, \(P(B\mid A)=P(B)\), so:

\[P(A\text{ and }B)=P(A)P(B)\]

Language cue: “and” often points toward multiplication; “or” often points toward addition. Always check whether the events overlap or depend on one another.

9. Independent vs. Dependent Events

Independent

Knowing that one event occurred does not change the probability of the other.

\[P(B\mid A)=P(B)\]

Dependent

The first event changes the probabilities for the second event.

\[P(B\mid A)\ne P(B)\]

Drawing an item and not replacing it is a common reason successive selections are dependent.

10. Tree Diagrams: Follow the Branches

A tree diagram separates a multi-step experiment into branches. Multiply along a path to find the probability of that path.

Multiply along a path. If several paths produce the desired outcome, add their path probabilities.

11. Area Models

An area model can represent a probability as part of a whole region. It is useful when the problem provides percentages or proportions.

The regions represent complementary probabilities: \(P(A)=0.60\) and \(P(\text{not }A)=0.40\).

12. Finding an Unknown Number from Probabilities

Some SAT probability questions give the total number of objects and the probabilities of several categories, then ask for the number in the category that remains.

\[P(A)+P(B)+P(C)=1\]

If the outcomes are mutually exclusive and exhaustive, first find the missing probability:

\[P(C)=1-P(A)-P(B)\]

Then convert that probability into a count by multiplying by the total:

\[\text{number in C}=P(C)\times\text{total}\]

SAT shortcut: You can also calculate the counts for the known categories and subtract them from the total. Both methods should give the same answer.

Method 1 • Find the missing probability

Step 1: Add the known probabilities.

Step 2: Subtract from 1.

Step 3: Multiply by the total.

Method 2 • Find the known counts

Step 1: Multiply each known probability by the total.

Step 2: Add those counts.

Step 3: Subtract from the total.

Worked example: A group contains 300 people. The probability that a randomly selected person belongs to Group A is 0.40, and the probability of belonging to Group B is 0.35. Everyone belongs to A, B, or C.

\[P(C)=1-0.40-0.35=0.25\]

\[0.25\times300=75\]

So 75 people belong to Group C.

13. A Reliable SAT Probability Strategy

Observe → Identify → Define the sample space → Calculate → Verify

1. Observe

Look at the table, diagram, graph, or description first.

2. Identify

Determine exactly what event the question asks about.

3. Sample space

Ask whether a condition restricts the possible outcomes.

4. Calculate

Use the appropriate count, ratio, product, or complement.

5. Verify

Check that the probability is between 0 and 1.

6. Re-read

Make sure the answer matches the wording of the condition.

Advanced Practice

Original SATMath800 questions. The visuals are part of the problem: read the representation before calculating.

Q1 • Easy • Frequency Table

Observed Probability

Preferred study location	Students
Library	24
Home	36
School	20
Other	10

One student is selected at random. What is the probability that the student prefers studying at home?

A) \(0.20\)

B) \(0.36\)

C) \(0.40\)

D) \(0.60\)

Show solution

There are 36 home-preferring students out of 90: \(36/90=0.40\). Answer: C.

Q2 • Easy • Complement

Not the Event

A certain event has probability \(0.18\). What is the probability that the event does not occur?

A) \(0.18\)

B) \(0.64\)

C) \(0.72\)

D) \(0.82\)

Show solution

Use the complement: \(1-0.18=0.82\). Answer: D.

Q3 • Medium • Two-Way Table

Overall Probability

	Completed	Did not complete	Total
Group A	28	12	40
Group B	24	16	40
Total	52	28	80

A student is selected at random. What is the probability that the student completed the program?

A) \(0.28\)

B) \(0.35\)

C) \(0.52\)

D) \(0.65\)

Show solution

No condition is imposed, so use the overall total: \(52/80=0.65\). Answer: D.

Q4 • Medium • Conditional Probability

Find the Correct Denominator

	Passed	Did not pass	Total
Class X	27	13	40
Class Y	18	22	40
Total	45	35	80

A student is known to be in Class X. What is the probability that the student passed?

A) \(27/80\)

B) \(18/40\)

C) \(27/40\)

D) \(45/80\)

Show solution

The condition restricts the sample space to Class X: \(27/40\). Answer: C.

Q5 • Medium • Conditional Comparison

Compare Rates

	Uses feature	Does not use feature	Total
First-year	42	18	60
Second-year	35	15	50

For which group is the probability of using the feature greater?

A) First-year

B) Second-year

C) The probabilities are equal

D) Cannot be determined

Show solution

Compare rates: \(42/60=0.70\) and \(35/50=0.70\). Answer: C.

Q6 • Medium • Dependent Events

Two Green Tokens

A container has 5 green tokens and 3 yellow tokens. One token is selected and is not replaced. A second token is then selected. What is the probability that both selected tokens are green?

A) \(5/16\)

B) \(5/14\)

C) \(25/64\)

D) \(1/2\)

Show solution

After a green token is selected, 4 green tokens remain among 7: \(\frac58*\frac47=\frac5{14}\). Answer: B.

Q7 • Medium/Hard • Tree Diagram

Follow the Path

What is the probability of event A occurring and event B occurring?

A) \(0.15\)

B) \(0.30\)

C) \(0.50\)

D) \(0.80\)

Show solution

Follow the A then B branch: \(0.6*0.5=0.30\). Answer: B.

Q8 • Hard • Reverse Percentage Probability

Recover the Missing Count

	Selected	Not selected	Total
Group A	?	24	60
Group B	18	22	40
Total			100

The probability that a randomly selected member of Group A was selected is \(0.60\). How many members of Group A were selected?

A) 24

B) 30

C) 36

D) 40

Show solution

Let the missing count be \(x\). Then \(x/60=0.60\), so \(x=36\). Answer: C.

Q9 • Hard • Addition Rule

Overlap Matters

In a group of 100 students, 58 participate in activity A, 37 participate in activity B, and 19 participate in both activities. What is the probability that a randomly selected student participates in at least one activity?

A) \(0.57\)

B) \(0.76\)

C) \(0.95\)

D) \(1.14\)

Show solution

Use inclusion-exclusion: \(58+37-19=76\). Thus the probability is \(0.76\). Answer: B.

Q10 • Hard • Conditional Probability

Condition First, Calculate Second

	Option P	Option Q	Total
Grade 10	18	12	30
Grade 11	24	16	40
Grade 12	30	20	50

A student is selected from the students who chose Option P. What is the probability that the student is in Grade 12?

A) \(30/50\)

B) \(30/72\)

C) \(30/120\)

D) \(50/120\)

Show solution

The condition restricts the sample space to the P column: \(18+24+30=72\). Therefore \(30/72=5/12\). Answer: B.

Q11 • Hard • Multi-Step Probability

Two Different Paths

What is the probability of a success, regardless of which route is taken?

A) \(0.20\)

B) \(0.28\)

C) \(0.50\)

D) \(0.58\)

Show solution

Success can occur along two paths: \((0.4)(0.7)=0.28\) and \((0.6)(0.5)=0.30\). Add them: \(0.58\). Answer: D.

Q12 • Hard • Conditional Probability from Data

Conditional Probability from Relative Frequencies

	Uses method A	Uses method B	Total
Under 30 min	28	12	40
30–60 min	24	36	60
Over 60 min	18	42	60
Total	70	90	160

Among the people who use method B, what is the probability that a randomly selected person spends more than 60 minutes?

A) \(42/60\)

B) \(42/90\)

C) \(60/160\)

D) \(90/160\)

Show solution

The condition restricts the sample space to the 90 people who use method B. Of those, 42 spend more than 60 minutes: \(42/90=7/15\). Answer: B.

Q13 • Medium • Missing Probability & Frequency

Find the Third Category

At a community center, there are 240 members. Each member participates in exactly one of three programs: A, B, or C. If a member is selected at random, the probability of selecting a member in Program A is 0.35, and the probability of selecting a member in Program B is 0.25. How many members participate in Program C?

A) 84

B) 90

C) 96

D) 108

Show solution

The three programs account for all members, so the probability for Program C is:

\[P(C)=1-0.35-0.25=0.40\]

Therefore, the number of members in Program C is:

\[0.40\times240=96\]

Answer: C.

Q14 • Medium • Probability to Count

Recover the Missing Group

A library has 480 registered members. Each member is classified as a student, faculty member, or community member. The probability that a randomly selected member is a student is 0.45, and the probability that the member is a faculty member is 0.30. How many members are community members?

A) 96

B) 120

C) 144

D) 168

Show solution

First find the probability of being a community member:

\[P(\text{community})=1-0.45-0.30=0.25\]

Then convert the probability to a number of members:

\[0.25\times480=120\]

Answer: B.

Q15 • Hard • Probability, Count & Verification

Use the Probabilities to Complete the Distribution

Category	Probability	Number
A	0.28	—
B	0.17	—
C	0.25	—
D	?	126
Total	1.00	420

The table represents all 420 participants in a program. The categories are mutually exclusive and include every participant. Which value is the probability for Category D?

A) 0.25

B) 0.28

C) 0.30

D) 0.35

Show solution

Add the known probabilities:

\[0.28+0.17+0.25=0.70\]

The remaining probability is:

\[P(D)=1-0.70=0.30\]

Verify with the given count: \(0.30\times420=126\), which matches the table.

Answer: C.

SAT Probability Checklist

Before calculating

What is the event?
What is the sample space?
Is there a condition?
Are the events independent?

After calculating

Is the probability between 0 and 1?
Did I use the correct denominator?
Did I count an overlap twice?
Does the answer match the wording?

New SAT Pattern Added

Questions 13–15 add a distinct probability skill: using given probabilities and a total count to determine an unknown frequency, including the complementary-probability method and a table-based verification.

Latest Updates

Categories

What this lesson covers

The SAT habit

1. Basic Probability

2. Complementary Probability

3. Relative Frequency from Data

4. Reading Probability from a Two-Way Table

5. Conditional Probability: The Denominator Changes

6. “Given That” Language

Given that

Among

If known

7. Addition Rule: “A or B”

8. Multiplication Rule: “A and B”

9. Independent vs. Dependent Events

Independent

Dependent

10. Tree Diagrams: Follow the Branches

11. Area Models

12. Finding an Unknown Number from Probabilities

Method 1 • Find the missing probability

Method 2 • Find the known counts

13. A Reliable SAT Probability Strategy

Observe → Identify → Define the sample space → Calculate → Verify

1. Observe

2. Identify

3. Sample space

4. Calculate

5. Verify

6. Re-read

Advanced Practice

Observed Probability

Not the Event

Overall Probability

Find the Correct Denominator

Compare Rates

Two Green Tokens

Follow the Path

Recover the Missing Count

Overlap Matters

Condition First, Calculate Second

Two Different Paths

Conditional Probability from Relative Frequencies

Find the Third Category

Recover the Missing Group

Use the Probabilities to Complete the Distribution

SAT Probability Checklist

Before calculating

After calculating

New SAT Pattern Added

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