🎯 Session 2.7 — Independence: when one event doesn’t influence the other

What does independence mean?

Two events A and B are independent if the probability of one is unchanged by knowing the other. Any of these equivalent statements can be used:

(1) P(A | B) = P(A)
(2) P(B | A) = P(B)
(3) P(A ∩ B) = P(A) · P(B)

From P(A|B)=P(A∩B)/P(B): if P(A|B)=P(A), multiply both sides by P(B) ⇒ P(A∩B)=P(A)P(B). Same for P(B|A).

Independence no influence vs. Mutually exclusive can’t both happen

Mutually Exclusive (Disjoint)

A and B are mutually exclusive if P(A ∩ B) = 0.

They cannot occur together (e.g., one coin toss: A = “Heads”, B = “Tails”).

A B No overlap ⇒ P(A∩B)=0
Important: If P(A)>0 and P(B)>0 and A,B are mutually exclusive, then
P(A∩B) = 0,   but   P(A)P(B) > 0. Independence would require P(A∩B) = P(A)P(B). ⇒ A and B are not independent.

Independent

A and B are independent if P(A ∩ B) = P(A) · P(B).

Knowing one does not change the chance of the other (e.g., two different fair coin tosses).

A B A∩B Overlap area ∝ P(A)P(B)
Quick tests
• If P(A∩B)=P(A)P(B) ⇒ Independent (can overlap; usually do).
• If P(A∩B)=0 and P(A),P(B)>0 ⇒ Mutually exclusive and therefore dependent.
• Both independent and mutually exclusive can only happen if P(A)=0 or P(B)=0.

🔁 Example 2.21 — Sampling with replacement

Question: A bin has 50 parts with 3 defectives. Draw twice with replacement. Let A = “1st draw defective”, B = “2nd draw defective”. Are A and B independent? Find P(A ∩ B).

  1. P(A) = 3/50.
  2. P(B | A) = 3/50 (bin is reset) = P(B).
  3. Therefore A ⟂ B, and P(A ∩ B) = P(A)·P(B) = (3/50)(3/50) = 9/2500 ≈ 0.0036.

Replacement keeps probabilities the same across draws → independence.

🚫 Example 2.22 — Sampling without replacement

Question: Same bin, but now draw twice without replacement. Let A = “1st defective”, B = “2nd defective”. Are A and B independent? What is P(B)?

  1. P(A) = 3/50.
  2. P(B | A) = 2/49 (one defective removed) ≠ 3/50 = P(B) ⇒ dependent.
  3. Total probability for P(B):
    P(B) = P(B|A)P(A) + P(B|A′)P(A′) = (2/49)(3/50) + (3/49)(47/50) = 3/50.

Conditional chance changes (dependence), but averaged over the first draw, the marginal chance for the second still works out to 3/50.

Why is P(B) still 3/50?
• Symmetry: Any of the 50 positions is equally likely to hold one of the 3 defectives.
• Total probability: Weighted average of conditionals balances back to 3/50.
• Key idea: Marginals can match the population even when events are dependent.

🔌 Example 2.23 — Series circuit (AND)

Question: Two devices in series. The circuit works only if both work. Assume independence. P(L)=0.80, P(R)=0.90. Find P(circuit works).

P(circuit) = P(L ∩ R) = P(L)P(R|L). With independence, P(R|L)=P(R).
Therefore P(circuit) = 0.80 × 0.90 = 0.72.

Series = all must succeed ⇒ reliabilities multiply (under independence).

🔀 Example 2.24 — Parallel circuit (OR)

Question: Two devices in parallel. The circuit works if at least one works. Assume independence. P(T)=0.95, P(B)=0.90. Find P(circuit works).

P(T ∪ B) = P(T)+P(B)−P(T ∩ B) and with independence P(T ∩ B)=P(T)P(B).
⇒ P(circuit) = 1 − (1−P(T))(1−P(B)) = 1 − (0.05)(0.10) = 0.995.

Parallel fails only if all branches fail; subtract from 1.

💡 Example 2.25 — Advanced circuit (3 parallel paths across columns)

Three parallel columns: Left has three 0.9 in parallel; Middle has two 0.95 in parallel; Right has one 0.99. System works if at least one column works.

a 0.9 0.9 0.9 0.95 0.95 0.99 b Left column (parallel) Middle column (parallel) Right column
Columns are in parallel across the network, so the system fails only if all columns fail simultaneously:
P(circuit works) = 1 − (1−P(L))(1−P(M))(1−P(R))
= 1 − (0.001)(0.0025)(0.01) = 1 − 0.000000025 = 0.999999975.
Common pitfall: Multiplying the three success probabilities (0.999×0.9975×0.99) assumes all three columns must work. Here we need “at least one works,” so use 1 − product of failures.
What if each column were series inside?

Left: 0.9³ = 0.729, Middle: 0.95² = 0.9025, Right: 0.99. Still parallel across columns:

P(circuit) = 1 − (1−0.729)(1−0.9025)(1−0.99) ≈ 0.999736

Redundancy across columns makes failure extremely rare because every column would need to fail at the same time.

🧪 Practice (self-check)

Q: Using the “parallel inside columns” setup above (Left: three 0.9, Middle: two 0.95, Right: one 0.99), compute P(circuit works).

Left: 1 − 0.1³ = 0.999; Middle: 1 − 0.05² = 0.9975; Right: 0.99.
P(circuit) = 1 − (0.001)(0.0025)(0.01) = 0.999999975.
Interpretation: Only a simultaneous failure of all three columns stops the circuit — extremely unlikely here.

📢 Student Q&A

Q1: Why is P(B) still 3/50 in the sampling without replacement example?

A1: Because each draw position is equally likely to be any of the 50 items (symmetry), so the marginal defective rate stays 3/50 even though P(B|A) changes.

Q2: What’s the difference between independence and mutually exclusive events?

A2: Independence: knowing one doesn’t change the other (P(A∩B)=P(A)P(B)). Mutually exclusive: can’t occur together (P(A∩B)=0). With positive probabilities, mutually exclusive events are never independent.

Q3: In a series circuit, why do we multiply probabilities?

A3: P(L∩R)=P(L)P(R|L). Under independence P(R|L)=P(R) ⇒ multiply.

Q4: In a parallel circuit, why use 1 − product of failures?

A4: Because the system works if at least one branch works; the complement “all fail” is easier to compute and then subtract from 1.

Q5: Does every draw (1st, 2nd, 3rd, …) in the urn problem have the same chance to be defective?

A5: Yes—symmetry. Dependence shows up in conditional probabilities (like P(B|A)), not in the marginal per position.

Q6:

A6:

Q7:

A7:

Q8:

A8:

Q9:

A9:

Q10:

A10: