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Conditional Probability and Independence

A free Geometry lesson from the “Probability” unit, with a worked example and practice problems including step-by-step solutions.

P(B|A) — read 'probability of B given A' — is the chance B happens assuming A already happened. The formula is P(B|A) = P(A and B) / P(A). Two events are INDEPENDENT when knowing A happened does NOT change the probability of B; equivalently, P(B|A) = P(B), and P(A and B) = P(A) * P(B). If they're not independent, they're DEPENDENT.

What you'll learn

Why it matters: Medical-test interpretation (P(disease | positive test)), insurance pricing, and machine-learning classifiers all use conditional probability.

Worked example

Problem. A bag has 4 red and 6 blue marbles. P(red on second draw GIVEN red on first, without replacement)?

  1. After drawing one red, 3 red and 6 blue remain (9 total).
  2. P(red | first was red) = 3/9 = 1/3.
  3. Connect the result back to Conditional Probability and Independence so the geometric relationship is explicit.
  4. Check the result against the original diagram, units, or proof statement.

Answer: 3/9

Practice problems

1. Roll a fair die. P(showing 4 | result is even) as a fraction.

Show solution
  1. Warm-up: First identify exactly what the question is asking: Roll a fair die. P(showing 4 | result is even) as a fraction.
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. Given even = sample {2, 4, 6}, three outcomes.
  4. Of those, 4 is one. P = 1/3.
  5. Check the result by substituting or estimating: the response should match 1/3 and make sense in the original problem.

Answer: 1/3

2. Roll a fair die. P(showing 6 | result is greater than 3) as a fraction.

Show solution
  1. Warm-up: First identify exactly what the question is asking: Roll a fair die. P(showing 6 | result is greater than 3) as a fraction.
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. Greater than 3 = {4, 5, 6}, three outcomes.
  4. Of those, 6 is one. P = 1/3.
  5. Check the result by substituting or estimating: the response should match 1/3 and make sense in the original problem.

Answer: 1/3

3. If two events are independent, P(B | A) equals:

Choices: P(B) · P(A) * P(B) · P(A)

Show solution
  1. Warm-up: First identify exactly what the question is asking: If two events are independent, P(B | A) equals:
  2. Compare each answer choice with the calculation or rule, and eliminate choices that do not satisfy the condition.
  3. Independence means A does not change B's probability.
  4. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.
  5. Identify the diagram relationship, formula, theorem, or proof reason before calculating.

Answer: P(B)

4. 100 students. 40 play sports; 30 are in band; 12 do both. Find P(band | sports) as a fraction.

Show solution
  1. Core Practice: First identify exactly what the question is asking: 100 students. 40 play sports; 30 are in band; 12 do both. Find P(band | sports) as a fraction.
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. P(band AND sports) / P(sports) = 12 / 40.
  4. Check the result by substituting or estimating: the response should match 12/40 and make sense in the original problem.
  5. Identify the diagram relationship, formula, theorem, or proof reason before calculating.

Answer: 12/40

5. Same survey. Find P(sports | band) as a fraction.

Show solution
  1. Core Practice: First identify exactly what the question is asking: Same survey. Find P(sports | band) as a fraction.
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. P(sports AND band) / P(band) = 12 / 30.
  4. Check the result by substituting or estimating: the response should match 12/30 and make sense in the original problem.
  5. Identify the diagram relationship, formula, theorem, or proof reason before calculating.

Answer: 12/30

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