Conditional Probability and Independence

Explore sample space, conditional probability, independence.

Lesson 2 of 53 parts Practice Flashcards

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Sample Space

Objective: Explain sample space and use it to solve or interpret a problem.

Probability is favorable outcomes divided by total equally likely outcomes in a finite equally likely model; define what outcomes are being counted. Simple favorable-over-total counts require them.

Worked examples

Try itConditional Probability and Independence — practice 1.1: Among 8 students who chose science, 2 also chose math. Find P(math | science).Show answer

Step 1: Restrict the denominator to 8 science students. Step 2: Count 2 students also choosing math. Step 3: The conditional probability is 1/4.

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Sample Space

Objective: Explain sample space and use it to solve or interpret a problem.

Probability is favorable outcomes divided by total equally likely outcomes in a finite equally likely model; define what outcomes are being counted. Simple favorable-over-total counts require them.

Worked examples

Try itConditional Probability and Independence — practice 1.1: Among 8 students who chose science, 2 also chose math. Find P(math | science).Show answer

Step 1: Restrict the denominator to 8 science students. Step 2: Count 2 students also choosing math. Step 3: The conditional probability is 1/4.

Conditional Probability

Objective: Explain conditional probability and use it to solve or interpret a problem.

P(A|B)=P(A∩B)/P(B) for P(B)>0. Conditioning narrows the reference group to cases in B. The condition B defines the new reference space.

Worked examples

Try itConditional Probability and Independence — practice 2.1: Among 10 students who chose science, 3 also chose math. Find P(math | science).Show answer

Step 1: Restrict the denominator to 10 science students. Step 2: Count 3 students also choosing math. Step 3: The conditional probability is 3/10.

Independence

Objective: Explain independence and use it to solve or interpret a problem.

A and B are independent when P(A|B)=P(A) for P(B)>0, equivalently P(A∩B)=P(A)P(B). Disjoint positive-probability events are not independent. No, one happening rules out the other.

Worked examples

Try itConditional Probability and Independence — practice 3.1: If P(A)=1/4 and P(B)=1/11 independently, find P(A and B).Show answer

Step 1: For independent events multiply probabilities. Step 2: Compute (1/4)(1/11). Step 3: The product is 1/44.

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