Independence

Statistical Independence

Definition:

Two events \(A\) and \(B\) are independent if knowing that one of them occurred gives no additional information about the probability of the other occurring.

• Formally: \(P(A \mid B) = P(A) \quad \text{and} \quad P(B \mid A) = P(B)\)

This means that the occurrence of one event does not affect the probability of the other.

Key Consequence of Independence

If \(A\) and \(B\) are independent:

\[ P(A \cap B) = P(A) \cdot P(B) \]

Proof using conditional probability:

We know:

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]

If \(A\) and \(B\) are independent, then \(P(A \mid B) = P(A)\), so:

\[ P(A) = \frac{P(A \cap B)}{P(B)} \Rightarrow P(A \cap B) = P(A) \cdot P(B) \]

Example: Deck of Cards

For a single card draw:

• Let \(A\): card is a Spade → \(P(A) = \frac{13}{52} = \frac{1}{4}\)
• Let \(B\): card is a Queen → \(P(B) = \frac{4}{52} = \frac{1}{13}\)

Then:

• \(A \cap B\): card is the Queen of Spades → \(P(A \cap B) = \frac{1}{52}\)
• Since \(\frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52}\), \(A\) and \(B\) are independent.

Selecting either of the cards won’t affect the probability of the other one.

Graphically, you can imagine the full deck as a grid of suits and ranks. The Spades form 1/4 of the set, and the Queens form 1/13. Their intersection, the Queen of Spades is just one card out of 52.

Example: Non-Independent Event

For a single card draw:

• \(A\): selecting a black card → 26 black cards (Spades + Clubs)
  • \(P(A) = \frac{26}{52} = \frac{1}{2}\)
• \(B\): selecting a spade → 13 cards
  • \(P(B) = \frac{13}{52} = \frac{1}{4}\)
• \(P(A \cap B) = P(\text{card is a spade}) = \frac{13}{52} = \frac{1}{4}\)

Since \(P(A) \cdot P(B) = \frac{1}{8} \neq \frac{1}{4} = P(A \cap B)\), the events are dependent. This makes intuitive sense: if you know a card is black, you've increased the probability it's a spade from 1/4 to 1/2 (since spades are half of all black cards).

Independence in Practice: Reliability

Series Configuration

• n components in series (e.g., Christmas lights):
  • If any component fails, the system fails.
  • Let failure probability of one component be pp
  • Probability of success: \(P(\text{success}) = (1 - p)^n\)
  • Therefore, \(P(\text{failure}) = 1 - (1 - p)^n\)

Example:

If \(p = 0.05\), \(n = 10\):

\(P(\text{failure}) = 1 - (0.95)^{10} \approx 0.40\)

So, even components with 95% reliability fail 40% of the time in series.

Parallel Configuration

• n components in parallel:
  • System fails only if all components fail.
  • Probability of failure: \(P(\text{failure}) = p^n\)

Example:

If \(p = 0.05\), \(n = 10\):

\(P(\text{failure}) = (0.05)^{10} \approx 10^{-13}\)

So, parallel configurations are vastly more reliable under the assumption of independence.