Independence

Independent vs dependent events

Independence

What is Independence?

Independent Events: Occurrence of one event doesn't affect probability of the other

Formal definition: Events A and B are independent if:

$P(A|B) = P(A)$

Equivalently:

P(B|A) = P(B)
P(A ∩ B) = P(A) × P(B)

Key insight: Knowing one occurred gives no information about the other

Testing for Independence

Method 1: Conditional Probability

Check if P(A|B) = P(A)

Example: Flip coin twice

P(H on flip 2) = 1/2
P(H on flip 2|H on flip 1) = 1/2
Equal → Independent

Method 2: Multiplication Rule

Check if P(A and B) = P(A) × P(B)

Example: Roll die, flip coin

P(6 on die) = 1/6
P(H on coin) = 1/2
P(6 and H) = 1/12 = 1/6 × 1/2 ✓
Independent

Method 3: Two-Way Table

For independence, each cell should equal (row total × column total)/grand total

Independence vs Mutually Exclusive

IMPORTANT: Independent ≠ Mutually Exclusive

Mutually Exclusive: Can't both occur (P(A ∩ B) = 0)
Independent: One doesn't affect other (P(A ∩ B) = P(A) × P(B))

In fact: If P(A) > 0 and P(B) > 0, mutually exclusive events are DEPENDENT!

Why? If A occurs, B definitely can't occur, so P(B|A) = 0 ≠ P(B)

Example:

A: Roll 2 on die
B: Roll 5 on die
Mutually exclusive (can't both happen)
NOT independent (if A occurs, B can't, so they're dependent)

Independence in Practice

Sampling with replacement: Draws are independent
Sampling without replacement: Draws are dependent

Example: Two cards from deck

With replacement:

P(First ace) = 4/52
P(Second ace|First ace) = 4/52 (replaced first card)
P(Second ace) = 4/52
Independent ✓

Without replacement:

P(First ace) = 4/52
P(Second ace|First ace) = 3/51
P(Second ace) ≈ 4/52 (overall, across all possible first cards)
NOT independent (but close if sample is tiny compared to population)

10% Condition

Rule of thumb: If sample size < 10% of population, treat as independent even without replacement

Why? Removing small fraction doesn't appreciably change probabilities

Example: 5 cards from deck

5/52 ≈ 9.6% < 10%
Can approximate as independent (slight error)

Example: 20 cards from deck

20/52 ≈ 38% > 10%
Must account for dependence

Multiplication Rule for Independent Events

If A and B are independent:

$P(A \text{ and } B) = P(A) \times P(B)$

Extends to multiple events:

$P(A_1 \text{ and } A_2 \text{ and } ... \text{ and } A_n) = P(A_1) \times P(A_2) \times ... \times P(A_n)$

Example: Flip coin 3 times, P(HHH) = 1/2 × 1/2 × 1/2 = 1/8

At Least One Calculations

"At least one" problems: Often easier to use complement

P(At least one A) = 1 - P(No A)

Example: Flip coin 3 times, find P(at least one head)

Long way: P(1H) + P(2H) + P(3H) = complicated

Short way: P(At least 1H) = 1 - P(No H) = 1 - P(TTT) = 1 - (1/2)³ = 1 - 1/8 = 7/8

Example: Shoot basketball, 70% success rate, 3 attempts

P(Make at least one) = 1 - P(Miss all 3) = 1 - (0.3)³ = 1 - 0.027 = 0.973

Checking Independence from Two-Way Table

100 students:

| | Male | Female | Total | |-----------|------|--------|-------| | Athlete | 24 | 16 | 40 | | Non-athlete| 36 | 24 | 60 | | Total | 60 | 40 | 100 |

Check independence of Athlete and Male:

Method 1:

P(Athlete) = 40/100 = 0.4
P(Athlete|Male) = 24/60 = 0.4
Equal → Independent ✓

Method 2:

P(Athlete and Male) = 24/100 = 0.24
P(Athlete) × P(Male) = (40/100) × (60/100) = 0.4 × 0.6 = 0.24
Equal → Independent ✓

Method 3: Expected cell count

Expected = (row total × column total)/grand total = (40 × 60)/100 = 24
Actual = 24
Equal → Independent ✓

Real-World Independence

Independent (usually):

Coin flips
Die rolls
Different people's responses (if random sample)
Successive free throws (debatable!)
Rain in New York and LA on same day

Dependent:

Cards without replacement
Success/failure of teammates
Weather on consecutive days
Stock prices over time
Contagious disease among contacts

Independence of Complements

If A and B are independent:

A and B^c are independent
A^c and B are independent
A^c and B^c are independent

Example: Two independent coin flips

If flips are independent, then "First heads" and "Second tails" are also independent

Common Mistakes

❌ Assuming events are independent without checking
❌ Thinking mutually exclusive means independent (opposite!)
❌ Forgetting 10% condition for sampling
❌ Using multiplication rule when events aren't independent
❌ Confusing P(A and B) with P(A) + P(B)

Practice Strategy

Question: Are events independent?
Test: Check if P(A|B) = P(A) or P(A and B) = P(A) × P(B)
Context: Does it make sense? (Replacement? Separate processes?)
Calculate: Use appropriate rule (multiply if independent)

Quick Reference

Definition: P(A|B) = P(A) or P(B|A) = P(B)

Multiplication: P(A and B) = P(A) × P(B) if independent

Complement: P(At least 1) = 1 - P(None)

10% Rule: Treat as independent if n < 0.10N

Key: Independent ≠ Mutually Exclusive

Remember: Independence means events don't influence each other. Always verify independence before using multiplication rule!

📚 Practice Problems

1Problem 1easy

❓ Question:

Roll a fair die twice. Are the two rolls independent? Explain.

💡 Show Solution

Step 1: Define independence Two events A and B are independent if: P(A|B) = P(A), or equivalently P(A and B) = P(A) × P(B)

The outcome of one event doesn't affect the probability of the other.

Step 2: Analyze the die rolls First roll: Any outcome from {1, 2, 3, 4, 5, 6} Second roll: Any outcome from {1, 2, 3, 4, 5, 6}

Does the first roll affect the second roll?

NO! The die has no memory
Second roll probabilities don't change based on first roll

Step 3: Test with specific example Let A = "first roll is 6" Let B = "second roll is 6"

P(A) = 1/6 P(B) = 1/6 P(A and B) = P(both rolls are 6) = 1/6 × 1/6 = 1/36

Check: P(A) × P(B) = 1/6 × 1/6 = 1/36 ✓ P(A and B) = P(A) × P(B), so independent!

Step 4: Check conditional probability P(B|A) = P(second roll is 6 | first roll is 6) = 1/6 P(B) = 1/6

P(B|A) = P(B) ✓

Knowing the first roll doesn't change the probability of the second roll.

Answer: YES, the two rolls are independent. The outcome of the first roll doesn't affect the probability distribution of the second roll. Each roll is a separate, random event with its own probabilities that don't depend on previous rolls.

2Problem 2easy

❓ Question:

Draw two cards from a standard deck WITHOUT replacement. Are the draws independent?

💡 Show Solution

Step 1: Recall independence definition Events are independent if P(A|B) = P(A) Or equivalently: P(A and B) = P(A) × P(B)

Step 2: Set up specific example Let A = "first card is a heart" Let B = "second card is a heart"

Step 3: Calculate P(A) P(A) = 13/52 = 1/4

Step 4: Calculate P(B|A) Given first card is a heart (and removed):

Remaining: 12 hearts out of 51 cards P(B|A) = 12/51

Step 5: Compare P(B|A) with P(B) P(B) without any information = 13/52 = 1/4

But P(B|A) = 12/51 ≈ 0.235

Is 12/51 = 13/52? 12/51 = 12/51 13/52 = 13/52

Cross multiply: 12 × 52 = 624 13 × 51 = 663

624 ≠ 663, so P(B|A) ≠ P(B)

Step 6: Conclusion Since P(B|A) ≠ P(B), the events are NOT independent.

The first draw affects the second draw because:

If first card is a heart, fewer hearts remain (12/51)
If first card is not a heart, more hearts remain (13/51)

Step 7: What if we replaced the card? WITH replacement: P(B|A) = 13/52 = P(B) Then the draws WOULD be independent!

Answer: NO, the draws are NOT independent when sampling without replacement. The outcome of the first draw changes the composition of the deck, which affects the probabilities for the second draw. Without replacement creates dependence between draws.

3Problem 3medium

❓ Question:

In a school, 40% of students are seniors (S) and 30% take AP Statistics (AP). Also, 15% are seniors taking AP Statistics. Are being a senior and taking AP Statistics independent?

💡 Show Solution

Step 1: Identify given information P(S) = 0.40 (seniors) P(AP) = 0.30 (take AP Stats) P(S and AP) = 0.15 (seniors taking AP Stats)

Step 2: Test independence condition For independence: P(S and AP) should equal P(S) × P(AP)

Calculate P(S) × P(AP): P(S) × P(AP) = 0.40 × 0.30 = 0.12

Step 3: Compare P(S and AP) = 0.15 (given) P(S) × P(AP) = 0.12 (calculated)

0.15 ≠ 0.12

Therefore, NOT independent!

Step 4: Verify using conditional probability If independent, P(AP|S) should equal P(AP)

P(AP|S) = P(AP and S) / P(S) = 0.15 / 0.40 = 0.375

P(AP) = 0.30

P(AP|S) = 0.375 ≠ 0.30 = P(AP)

Not independent! ✓

Step 5: Interpret P(AP|S) = 0.375 = 37.5% (seniors taking AP Stats) P(AP) = 0.30 = 30% (all students taking AP Stats)

Seniors are MORE likely to take AP Statistics (37.5% vs 30%) Being a senior and taking AP Stats are positively associated

Step 6: Create table to visualize Take AP Don't Take AP Total Seniors 0.15 0.25 0.40 Non-seniors 0.15 0.45 0.60 Total 0.30 0.70 1.00

If independent, senior/AP cell would be: 0.40 × 0.30 = 0.12 But actual value is 0.15 (more than expected)

Answer: NO, being a senior and taking AP Statistics are NOT independent. P(S and AP) = 0.15 ≠ P(S) × P(AP) = 0.12. Seniors are more likely to take AP Statistics (37.5%) than the overall student body (30%), showing positive association between the two events.

4Problem 4medium

❓ Question:

Events A and B are independent with P(A) = 0.6 and P(B) = 0.4. Find: a) P(A and B) b) P(A or B) c) P(A|B)

💡 Show Solution

Step 1: Use independence to find P(A and B) For independent events: P(A and B) = P(A) × P(B) = 0.6 × 0.4 = 0.24

Step 2: Find P(A or B) using addition rule P(A or B) = P(A) + P(B) - P(A and B) = 0.6 + 0.4 - 0.24 = 1.00 - 0.24 = 0.76

Step 3: Find P(A|B) For independent events: P(A|B) = P(A) = 0.6

Or using formula: P(A|B) = P(A and B) / P(B) = 0.24 / 0.4 = 0.6 ✓

Step 4: Verify the independence We can check that P(B|A) = P(B):

P(B|A) = P(A and B) / P(A) = 0.24 / 0.6 = 0.4 = P(B) ✓

Independence confirmed!

Step 5: Create Venn diagram for visualization A (60%) B (40%) ___ ___ / ____ ____/
| \ / | | 0.36 |0.24| 0.16 | | / \ | _/ _/

Neither: 0.24

Check: 0.36 + 0.24 + 0.16 + 0.24 = 1.00 ✓

Only A: 0.6 - 0.24 = 0.36 Both: 0.24 Only B: 0.4 - 0.24 = 0.16 Neither: 1 - 0.76 = 0.24

Answer: a) P(A and B) = 0.24 b) P(A or B) = 0.76 c) P(A|B) = 0.6

Key insight: For independent events, knowing B occurred doesn't change the probability of A. P(A|B) = P(A) = 0.6.

5Problem 5hard

❓ Question:

A factory has two machines. Machine 1 produces defective items 2% of the time. Machine 2 produces defective items 3% of the time. If the machines operate independently and you randomly select one item from each machine, what is the probability that: a) Both are defective? b) At least one is defective? c) Exactly one is defective?

💡 Show Solution

Step 1: Define events D1 = item from Machine 1 is defective, P(D1) = 0.02 D2 = item from Machine 2 is defective, P(D2) = 0.03

Given: Machines operate independently

Step 2: Find P(both defective) Since independent: P(D1 and D2) = P(D1) × P(D2) = 0.02 × 0.03 = 0.0006

Step 3: Find P(at least one defective) Method 1 - Use complement: P(at least one defective) = 1 - P(neither defective)

P(neither defective) = P(not D1 and not D2) = P(not D1) × P(not D2) [independence] = (1 - 0.02) × (1 - 0.03) = 0.98 × 0.97 = 0.9506

P(at least one defective) = 1 - 0.9506 = 0.0494

Method 2 - List all cases: P(at least one) = P(D1 only) + P(D2 only) + P(both) = (0.02 × 0.97) + (0.98 × 0.03) + (0.02 × 0.03) = 0.0194 + 0.0294 + 0.0006 = 0.0494 ✓

Step 4: Find P(exactly one defective) P(exactly one) = P(D1 only) + P(D2 only) = P(D1 and not D2) + P(not D1 and D2) = [P(D1) × P(not D2)] + [P(not D1) × P(D2)] = (0.02 × 0.97) + (0.98 × 0.03) = 0.0194 + 0.0294 = 0.0488

Step 5: Create probability tree Machine 1 Machine 2 Probability | Defect (0.02)---|--- Defect (0.03) 0.02 × 0.03 = 0.0006 | |--- Not Defect (0.97) 0.02 × 0.97 = 0.0194 | Not Defect (0.98)--|--- Defect (0.03) 0.98 × 0.03 = 0.0294 | |--- Not Defect (0.97) 0.98 × 0.97 = 0.9506

Sum: 0.0006 + 0.0194 + 0.0294 + 0.9506 = 1.0000 ✓

Step 6: Verify all answers sum correctly P(both defective) + P(exactly one) + P(neither) = 1 0.0006 + 0.0488 + 0.9506 = 1.0000 ✓

Answer: a) P(both defective) = 0.0006 or 0.06% b) P(at least one defective) = 0.0494 or 4.94% c) P(exactly one defective) = 0.0488 or 4.88%

Key insight: Independence allows us to multiply probabilities. The multiplication rule for independent events is essential for solving problems with multiple independent trials.

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