Checking whether events are independent
Events are independent only if P(A and B) = P(A) × P(B) — check, don't assume.
The common mistake
Multiplying P(A) × P(B) for events that affect each other, such as drawing two cards without putting the first one back.
Instead: Ask: does knowing A happened change the chance of B? If it does, use the conditional probability P(B | A) for the second step. To test independence from data, check whether P(A and B) equals P(A) × P(B), or whether P(B | A) equals P(B).
Worked example with Guide me
Two cards are drawn from a 52-card deck without replacement. What is P(both are aces)?
- Sugar · What's the chance the first card is an ace?Student · 4/52.
- Sugar · After one ace is gone, does the second chance change?Student · Yes — 3 aces left out of 51, so 3/51.
- Sugar · Now multiply those.Student · 4/52 × 3/51 = 12/2652.
Answer: 1/221 — not (4/52)², because the draws aren't independent.
Try 3 practice questions
1. P(A) = 0.4, P(B) = 0.5, P(A and B) = 0.2. Independent?
2. P(A) = 0.3, P(B) = 0.6, P(A and B) = 0.25. Independent?
3. A bag has 5 red and 5 blue. Draw two without replacement. P(red, red)?
Quick answers
- What is the most common mistake with checking whether events are independent?
- Multiplying P(A) × P(B) for events that affect each other, such as drawing two cards without putting the first one back.
- How do I get checking whether events are independent right?
- Ask: does knowing A happened change the chance of B? If it does, use the conditional probability P(B | A) for the second step. To test independence from data, check whether P(A and B) equals P(A) × P(B), or whether P(B | A) equals P(B).
- Can you show a worked example?
- Two cards are drawn from a 52-card deck without replacement. What is P(both are aces)? 1/221 — not (4/52)², because the draws aren't independent.
Related concepts
Save this as a miss — practice it before your test
Sugar brings it back for review until it sticks. Free, no card.
Save this as a miss — practice it before your test