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AP Statistics 2.6 Conditional Probability Study Notes - New Syllabus

AP Statistics 2.6 Conditional Probability Study Notes – New Syllabus

AP Statistics 2.6 Conditional Probability Study Notes – As per latest AP Statistics Syllabus.

LEARNING OBJECTIVES

  • 2.6.A Calculate conditional probabilities.

ESSENTIAL KNOWLEDGE:

  • 2.6.A.1 The probability that event A will occur given that event B has occurred is called a conditional probability and is written as \( P(A\mid B) \). Conditional probability is defined as

    \( P(A\mid B)=\frac{P(A\cap B)}{P(B)} \)
  • 2.6.A.2 The general multiplication rule states that the probability that events A and B will occur is equal to the probability that event A will occur multiplied by the conditional probability that event B will occur given that event A has occurred. The multiplication rule is defined as

    \( P(A\cap B)=P(A)\cdot P(B\mid A) \)

AP Statistics – Concise Summary Notes – All Topics

2.6.A.1 Conditional Probability

Conditional probability is the probability that an event \(A\) occurs given that another event \(B\) has already occurred.

Knowing that event \(B\) has occurred changes the sample space because only the outcomes in event \(B\) are now possible.

The conditional probability of event \(A\) given event \(B\) is written as

\(\mathrm{ P(A \mid B)} \)

which is read as

“The probability of A given B.”

Conditional Probability Formula

If \(P(B)>0\), then

\(\mathrm{ P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}} \)

Where:

  • \(P(A\mid B)\) = Probability that event \(A\) occurs given that event \(B\) has occurred.
  • \(P(A\cap B)\) = Probability that both \(A\) and \(B\) occur.
  • \(P(B)\) = Probability that event \(B\) occurs.

Understanding the Formula

When we know that event \(B\) has already occurred, the original sample space is reduced to only the outcomes in \(B\).

Among those outcomes, we determine how many also belong to event \(A\).

Therefore, conditional probability is simply

Probability of both events occurring ÷ Probability that the given event occurred.


Example 1: Rolling a Die

A fair six-sided die is rolled once.

Let

  • \(A\): Rolling an even number
  • \(B\): Rolling a number greater than 3

Find \(P(A\mid B)\).

Step 1: Write each event.

\(A=\{2,4,6\}\)

\(B=\{4,5,6\}\)

Step 2: Find the intersection.

\(A\cap B=\{4,6\}\)

Step 3: Calculate the probabilities.

\(P(A\cap B)=\dfrac26\)

\(P(B)=\dfrac36\)

Step 4: Apply the conditional probability formula.

\(P(A\mid B)=\dfrac{\frac26}{\frac36}=\dfrac23\)

Interpretation

Given that the die roll is greater than 3, there is a \(\dfrac23\) probability that it is even.


Example 2: Selecting a Card

One card is drawn from a standard deck.

Let

  • \(A\): Drawing a King
  • \(B\): Drawing a Face Card

Step 1: Find the probabilities.

\(P(A\cap B)=\dfrac4{52}\)

\(P(B)=\dfrac{12}{52}\)

Step 2: Apply the formula.

\(P(A\mid B)=\dfrac{\frac4{52}}{\frac{12}{52}}=\dfrac13\)

Interpretation

Given that the selected card is a face card, the probability that it is a King is \(\dfrac13\).


When to Use Conditional Probability

Look for words such as:

  • given
  • knowing that
  • assuming that
  • among
  • of those who…

These phrases indicate that the sample space has changed and a conditional probability should be calculated.

Important AP Exam Notes

  • Conditional probability is written as \(P(A\mid B)\).
  • Read it as “the probability of A given B.”
  • The denominator is always the probability of the given event, \(P(B)\).
  • The sample space changes because event \(B\) has already occurred.
  • Conditional probability is only defined when \(P(B)>0\).

Common AP Exam Mistakes

IncorrectCorrect
Using \(P(A)\) as the denominator.Use the probability of the given event, \(P(B)\).
Ignoring the word “given.”“Given” indicates a conditional probability.
Using the original sample space.Restrict the sample space to event \(B\).

 Example

A standard deck of 52 cards is shuffled thoroughly. One card is selected at random.

Given that the selected card is a Heart, find the probability that it is a Face Card.

▶️ Answer / Explanation

Step 1: Define the events.

  • \(A\): Drawing a Face Card
  • \(B\): Drawing a Heart

Step 2: Find the probabilities.

There are 3 face cards in the Hearts suit (Jack, Queen, King).

\(P(A\cap B)=\dfrac3{52}\)

\(P(B)=\dfrac{13}{52}\)

Step 3: Apply the conditional probability formula.

\(P(A\mid B)=\dfrac{\frac3{52}}{\frac{13}{52}}=\dfrac3{13}\)

Answer

Given that the selected card is a Heart, the probability that it is a Face Card is

\(\dfrac3{13}\).

2.6.A.2 General Multiplication Rule

The General Multiplication Rule is used to calculate the probability that two events occur together.

It states that the probability of both events \(A\) and \(B\) occurring equals the probability that \(A\) occurs multiplied by the probability that \(B\) occurs given that \(A\) has already occurred.

The General Multiplication Rule is written as

\( \mathrm{ P(A\cap B)=P(A)\cdot P(B\mid A)} \)

It can also be written as

\(\mathrm{ P(A\cap B)=P(B)\cdot P(A\mid B)} \)

Both formulas are equivalent and produce the same joint probability.

Formula

\( \mathrm{ P(A\cap B)=P(A)\cdot P(B\mid A)} \)

Where:

  • \(P(A)\) = Probability that event \(A\) occurs.
  • \(P(B\mid A)\) = Probability that event \(B\) occurs given that \(A\) has already occurred.
  • \(P(A\cap B)\) = Probability that both events occur.

Why the Rule Works

To find the probability that both events occur:

  1. First calculate the probability that event \(A\) occurs.
  2. Then, assuming event \(A\) has occurred, calculate the probability that event \(B\) also occurs.
  3. Multiply the two probabilities together.

Example 1: Drawing Two Cards Without Replacement

Two cards are drawn from a standard deck without replacement.

Find the probability that the first card is an Ace and the second card is a King.

Step 1: Define the events.

  • \(A\): First card is an Ace.
  • \(B\): Second card is a King.

Step 2: Calculate each probability.

\(P(A)=\dfrac4{52}\)

After drawing an Ace, 51 cards remain, including 4 Kings.

\(P(B\mid A)=\dfrac4{51}\)

Step 3: Apply the General Multiplication Rule.

\(P(A\cap B)=\dfrac4{52}\times\dfrac4{51}=\dfrac{16}{2652}=\dfrac4{663}\approx0.0060\)


Example 2: Rolling a Die

A fair six-sided die is rolled once.

Let

  • \(A\): Rolling an even number.
  • \(B\): Rolling a number greater than 3.

Step 1: Find the probabilities.

\(P(A)=\dfrac36=\dfrac12\)

Among the even numbers \(\{2,4,6\}\), two outcomes are greater than 3.

\(P(B\mid A)=\dfrac23\)

Step 2: Apply the rule.

\(P(A\cap B)=\dfrac12\times\dfrac23=\dfrac13\)

Relationship Between Conditional Probability and the Multiplication Rule

FormulaPurpose
\( P(A\mid B)=\dfrac{P(A\cap B)}{P(B)} \)Find a conditional probability.
\( P(A\cap B)=P(A)\cdot P(B\mid A) \)Find the probability that both events occur.

Important AP Exam Notes

  • The General Multiplication Rule calculates the probability that both events occur.
  • The word “and” usually indicates that the multiplication rule should be used.
  • Use the conditional probability after the first event has occurred.
  • The rule works for both independent and dependent events.
  • For independent events, \(P(B\mid A)=P(B)\), so the rule becomes

\( P(A\cap B)=P(A)\times P(B) \)

Common AP Exam Mistakes

IncorrectCorrect
Multiplying by \(P(B)\) when the events are dependent.Use \(P(B\mid A)\).
Using addition for an “and” event.Use multiplication for “and”.
Ignoring the effect of the first event.Adjust the conditional probability after the first event occurs.

Example

A bag contains 5 red marbles and 3 blue marbles. Two marbles are selected without replacement.

Calculate the probability that both marbles are red.

▶️ Answer / Explanation

Step 1: Define the events.

  • \(A\): The first marble is red.
  • \(B\): The second marble is red.

Step 2: Calculate each probability.

\(P(A)=\dfrac58\)

After removing one red marble, there are 4 red marbles left out of 7 marbles.

\(P(B\mid A)=\dfrac47\)

Step 3: Apply the General Multiplication Rule.

\(P(A\cap B)=\dfrac58\times\dfrac47=\dfrac{20}{56}=\dfrac{5}{14}\approx0.357\)

Answer

The probability that both marbles are red is

\(\dfrac{5}{14}\).

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