Mastering Probability and Conditional Probability for the SAT

Introduction

Welcome to our comprehensive guide on probability and conditional probability for the SAT! These concepts form a crucial part of the Problem-Solving and Data Analysis section of the SAT Math test. Understanding probability will not only help you succeed on the SAT but also develop critical thinking skills that are valuable in many real-world situations. This lesson is designed for students with no prior knowledge of probability, taking you from the basic concepts to solving complex SAT-level problems step by step.

What are Probability and Conditional Probability?

Probability is a mathematical measure of the likelihood that an event will occur. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

Basic Probability Formula:

Probability of an event =

For example, the probability of rolling a 3 on a fair six-sided die is because there is 1 favorable outcome (rolling a 3) out of 6 possible outcomes.

Conditional Probability is the probability of an event occurring given that another event has already occurred. It's denoted as , which reads as "the probability of event A given that event B has occurred."

Conditional Probability Formula:

Where:

  • is the conditional probability of A given B
  • is the probability of both A and B occurring (intersection)
  • is the probability of B occurring

Conditional probability helps us update our beliefs about the likelihood of events based on new information.

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How to Use Probability and Conditional Probability

Step 1: Identify the Sample Space

The sample space is the set of all possible outcomes of an experiment. For example, when flipping a coin, the sample space is {Heads, Tails}.

Step 2: Determine Favorable Outcomes

Identify the outcomes that satisfy the condition you're looking for. For example, if you're calculating the probability of drawing a king from a standard deck of cards, the favorable outcomes are the 4 kings in the deck.

Step 3: Calculate Basic Probability

Use the formula:

For example, the probability of drawing a king from a standard 52-card deck is

Step 4: For Conditional Probability Problems

  1. Identify the condition (the given event B)
  2. Determine the new sample space (only outcomes where B occurs)
  3. Calculate the probability of event A within this new sample space

Step 5: Use the Conditional Probability Formula When Needed

Additional Probability Rules:

Addition Rule (for mutually exclusive events):

Addition Rule (for non-mutually exclusive events):

Multiplication Rule (for independent events):

Multiplication Rule (for dependent events):

SAT-Specific Tips:

  • Draw diagrams or tables to organize information
  • Use fractions rather than decimals for clearer work
  • Check if events are independent or dependent
  • Look for keywords like "given," "if," or "knowing that" which signal conditional probability

Probability and Conditional Probability Worksheet

Basic Probability Practice

  1. A fair six-sided die is rolled. Find the probability of rolling:
    a) A number greater than 4
    b) An even number
    c) A number less than 7

  2. A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. If a marble is drawn at random, find the probability of drawing:
    a) A red marble
    b) A blue or green marble
    c) A yellow marble

Conditional Probability Practice

  1. A standard deck of 52 cards is shuffled. Find the probability of:
    a) Drawing a king
    b) Drawing a king given that the card is a face card
    c) Drawing a heart given that the card is red

  2. In a class of 30 students, 18 are taking math, 15 are taking science, and 10 are taking both. If a student is selected at random, find the probability that:
    a) The student is taking math given that they are taking science
    b) The student is taking science given that they are taking math

Combined Probability Practice

  1. A survey found that 65% of people like chocolate ice cream, 45% like vanilla ice cream, and 25% like both. Find the probability that a randomly selected person:
    a) Likes either chocolate or vanilla ice cream
    b) Likes chocolate ice cream given that they like vanilla ice cream
    c) Likes neither chocolate nor vanilla ice cream

  2. Two fair dice are rolled. Find the probability that:
    a) The sum is 7
    b) The sum is 7 given that at least one die shows a 4
    c) Both dice show the same number

Solutions are provided at the end of this worksheet.

Probability and Conditional Probability Examples

Example 1

Basic Probability Example

Question: A bag contains 4 red balls and 6 blue balls. If a ball is drawn at random, what is the probability of drawing a red ball?

Solution:

  • Total number of balls = 4 + 6 = 10
  • Number of red balls = 4
  • Probability of drawing a red ball =

The probability of drawing a red ball is or 0.4 or 40%.

Example 2

Conditional Probability Example

Question: In a high school, 60% of students take Spanish and 30% of students take both Spanish and French. What is the probability that a student takes French, given that the student takes Spanish?

Solution:

  • Let S = event that student takes Spanish
  • Let F = event that student takes French
  • We need to find P(F|S)
  • Given: P(S) = 0.6 and P(S ∩ F) = 0.3
  • Using the conditional probability formula:

The probability that a student takes French, given that the student takes Spanish, is 0.5 or 50%.

Example 3

Probability with Cards Example

Question: A standard deck of 52 cards is shuffled. What is the probability of drawing a face card (jack, queen, or king)?

Solution:

  • Total number of cards in the deck = 52
  • Number of face cards = 12 (4 jacks, 4 queens, 4 kings)
  • Probability of drawing a face card =

The probability of drawing a face card is or approximately 0.231 or 23.1%.

Example 4

Probability with Independent Events Example

Question: A fair coin is tossed 3 times. What is the probability of getting exactly 2 heads?

Solution:

  • Each coin toss is independent
  • Total number of possible outcomes = (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)
  • Favorable outcomes (exactly 2 heads) = 3 (HHT, HTH, THH)
  • Probability =

Alternatively, using combinations:

  • Number of ways to choose which 2 tosses will be heads =
  • Probability of each specific sequence =
  • Probability =

The probability of getting exactly 2 heads is or 0.375 or 37.5%.

Example 5

Conditional Probability with Venn Diagrams Example

Question: In a survey of 200 college students, 120 own a laptop, 85 own a tablet, and 65 own both. If a student is selected at random, what is the probability that the student owns a laptop given that the student owns a tablet?

Solution:

  • Let L = event that student owns a laptop
  • Let T = event that student owns a tablet
  • We need to find P(L|T)
  • Number of students who own a tablet = 85
  • Number of students who own both a laptop and a tablet = 65
  • Using the conditional probability formula:

The probability that a randomly selected student owns a laptop, given that the student owns a tablet, is or approximately 76.5%.

Example 6

SAT-Style Probability Example

Question: A box contains 5 white balls and 3 black balls. Two balls are drawn without replacement. What is the probability that both balls are white?

Solution:

  • Total number of balls = 5 + 3 = 8
  • Number of ways to select 2 balls from 8 balls =
  • Number of ways to select 2 white balls from 5 white balls =
  • Probability of drawing 2 white balls =

Alternatively, using the multiplication rule for dependent events:

  • Probability of first ball being white =
  • Probability of second ball being white given first ball is white =
  • Probability of both balls being white =

The probability that both balls are white is or approximately 0.357 or 35.7%.

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Common Misconceptions

1. Confusing Independent and Dependent Events

Many students incorrectly assume that all events are independent. In reality, many real-world scenarios involve dependent events, where the occurrence of one event affects the probability of another event.

Example: When drawing cards without replacement, each draw changes the composition of the deck, making the events dependent.

2. The Gambler's Fallacy

This is the belief that if an event has occurred more frequently than normal in the past, it's less likely to happen in the future (or vice versa).

Example: Thinking that after flipping 5 heads in a row, a tail is "due" or more likely on the next flip. In reality, each flip of a fair coin has a 50% chance of being heads, regardless of previous flips.

3. Misunderstanding Conditional Probability

Students often confuse P(A|B) with P(B|A), which are generally not equal.

Example: The probability of having a disease given a positive test result P(Disease|Positive) is not the same as the probability of a positive test result given that you have the disease P(Positive|Disease).

4. Addition Rule Errors

Incorrectly adding probabilities without accounting for overlap between events.

Example: If P(A) = 0.5 and P(B) = 0.3, the probability of either A or B occurring is not simply 0.5 + 0.3 = 0.8 if the events can occur simultaneously. You must subtract the probability of both occurring: P(A or B) = P(A) + P(B) - P(A and B).

5. Sample Space Confusion

Not correctly identifying the total number of possible outcomes, especially in complex scenarios.

Example: When calculating the probability of getting a sum of 7 when rolling two dice, some students count only the 6 possible outcomes for each die rather than the 36 possible combinations.

6. Overlooking the "At Least One" Scenario

Struggling to calculate the probability of "at least one" occurrence of an event.

Example: To find the probability of getting at least one 6 when rolling three dice, it's often easier to calculate 1 - P(no 6's) rather than adding P(exactly one 6) + P(exactly two 6's) + P(exactly three 6's).

Practice Questions for Probability and Conditional Probability

Question 1

A company manufactures smartphones and tablets. Quality control testing shows that 5% of smartphones and 3% of tablets are defective. If the company produces twice as many smartphones as tablets, what is the probability that a randomly selected device is defective?

A) 3.5%
B) 4.0%
C) 4.3%
D) 5.0%

Solution:
Let's denote:

  • S = event that device is a smartphone
  • T = event that device is a tablet
  • D = event that device is defective

Given information:

  • P(D|S) = 0.05 (5% of smartphones are defective)
  • P(D|T) = 0.03 (3% of tablets are defective)
  • P(S) = 2/3 (smartphones are twice as many as tablets)
  • P(T) = 1/3

Using the law of total probability:
P(D) = P(D|S) × P(S) + P(D|T) × P(T)
P(D) = 0.05 × (2/3) + 0.03 × (1/3)
P(D) = 0.0333... + 0.01
P(D) = 0.0433... or approximately 4.3%

The answer is C) 4.3%.

Question 2

In a certain game, a player rolls two fair six-sided dice. If the sum of the dice is 7 or 11, the player wins. What is the probability that the player wins?

A) 1/6
B) 2/9
C) 5/18
D) 1/3

Solution:
When rolling two dice, there are 6 × 6 = 36 possible outcomes, all equally likely.

Let's count the favorable outcomes:

  • Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
  • Sum of 11: (5,6), (6,5) → 2 outcomes

Total favorable outcomes = 6 + 2 = 8

Probability of winning = 8/36 = 2/9

The answer is B) 2/9.

Probability and Conditional Probability Questions

Question 1

A bag contains 3 red marbles, 4 blue marbles, and 5 green marbles. If two marbles are drawn without replacement, what is the probability that both marbles are the same color?

A) 7/33
B) 8/33
C) 9/33
D) 10/33

Solution:
Total number of marbles = 3 + 4 + 5 = 12
Total number of ways to select 2 marbles from 12 = C(12,2) = 66

Favorable outcomes:

  • Two red marbles: C(3,2) = 3
  • Two blue marbles: C(4,2) = 6
  • Two green marbles: C(5,2) = 10

Total favorable outcomes = 3 + 6 + 10 = 19

Probability = 19/66 = 19/66

Simplifying: 19/66 = (19/11)/(66/11) = 19/11 ÷ 6 = 19/66

The answer is not among the given options. Let's double-check our work...

Actually, the correct calculation is:
Probability = (3 + 6 + 10)/66 = 19/66

Reducing: 19/66 = 19/66

This equals approximately 0.288 or about 29%.

Since none of the given options match exactly, we need to convert to the same denominator:
7/33 ≈ 14/66
8/33 ≈ 16/66
9/33 ≈ 18/66
10/33 ≈ 20/66

The closest answer is 10/33, which is option D.

Question 2

In a survey of 150 students, 70 take mathematics, 60 take physics, and 40 take both mathematics and physics. If a student is selected at random, what is the probability that the student takes mathematics or physics or both?

A) 0.4
B) 0.5
C) 0.6
D) 0.9

Solution:
Let M = event that student takes mathematics
Let P = event that student takes physics

Given:

  • P(M) = 70/150
  • P(P) = 60/150
  • P(M ∩ P) = 40/150

We need to find P(M ∪ P), the probability that a student takes mathematics or physics or both.

Using the addition rule for non-mutually exclusive events:
P(M ∪ P) = P(M) + P(P) - P(M ∩ P)
P(M ∪ P) = 70/150 + 60/150 - 40/150
P(M ∪ P) = (70 + 60 - 40)/150
P(M ∪ P) = 90/150 = 0.6

The answer is C) 0.6.

Probability and Conditional Probability Learning Checklist

  • I can calculate the probability of a simple event using the formula: P(Event) = Number of favorable outcomes / Total number of possible outcomes.

  • I understand the difference between independent and dependent events and can identify them in a problem.

  • I can apply the addition rule for both mutually exclusive events [P(A or B) = P(A) + P(B)] and non-mutually exclusive events [P(A or B) = P(A) + P(B) - P(A and B)].

  • I can apply the multiplication rule for both independent events [P(A and B) = P(A) × P(B)] and dependent events [P(A and B) = P(A) × P(B|A)].

  • I understand conditional probability and can calculate P(A|B) using the formula: P(A|B) = P(A ∩ B) / P(B).

  • I can solve probability problems involving cards, dice, coins, and other random selection scenarios.

  • I can create and interpret Venn diagrams to solve probability problems involving multiple events.

  • I can calculate probabilities for scenarios involving sampling with and without replacement.

  • I can solve "at least one" probability problems using the complement method: P(at least one) = 1 - P(none).

  • I can identify and avoid common probability misconceptions like the Gambler's Fallacy and confusion between P(A|B) and P(B|A).

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