Event Independence

Events $A$ and $B$ are independent if:

Product of probabilities is the formal definition of independence. Equivalently: $P(A|B)=P(A)$ - knowing $B$ does not change the probability of $A$. Disjointness ($A\cap B=\emptyset$) is a different concept.

🪙 Two coin flips

Classic independence check

$A$ = "first flip is heads" $B$ = "second flip is heads" Sample space: {HH, HT, TH, TT} - all equally likely. $P(A) = 2/4 = 1/2$ (HH and HT) $P(B) = 2/4 = 1/2$ (HH and TH) $P(A \cap B) = 1/4$ (only HH) **Check:** $P(A) \cdot P(B) = 1/2 \cdot 1/2 = 1/4 = P(A \cap B)$ ✓ The events are independent! The outcome of the first flip has no effect on the second.

For independent events A and B, P(A|B) = P(A). What does this mean in practice?

$P(A|B) = P(A)$ literally means: "the probability of A given B" = "the probability of A with no conditions". Information about B is **useless** for predicting A. It doesn't matter whether B happened or not - our estimate of P(A) doesn't change. This is the essence of independence: events "don't know" about each other.

🎲 Mutually exclusive, but DEPENDENT

Rolling a die

$A$ = "rolled 1", $B$ = "rolled 6" **Mutually exclusive?** ✅ Yes - rolling both 1 and 6 simultaneously is impossible. $P(A \cap B) = 0$ **Independent?** ❌ NO! If $A$ occurred (rolled 1), then $B$ definitely didn't: $P(B|A) = 0 \neq P(B) = 1/6$ Knowing $A$ **completely determines** $B$ - that's the strongest possible dependence!

🎲 Independent and COMPATIBLE

"Even" and "greater than 2"

$A$ = "rolled even" = {2, 4, 6}, $P(A) = 1/2$ $B$ = "rolled greater than 2" = {3, 4, 5, 6}, $P(B) = 4/6 = 2/3$ **Compatible?** ✅ Yes - $A \cap B$ = {4, 6}, $P(A \cap B) = 2/6 = 1/3$ **Independent?** Let's check: $P(A) \cdot P(B) = 1/2 \cdot 2/3 = 1/3 = P(A \cap B)$ ✅ Yes, independent! Knowing "greater than 2" doesn't change the probability of even.

If events cannot occur simultaneously, they are independent (don't affect each other)

Mutually exclusive events (other than trivial cases) are ALWAYS dependent - and very strongly so!

If $A$ and $B$ are mutually exclusive, then $P(A \cap B) = 0$. But $P(A) \cdot P(B) > 0$ (if both events are possible). So $P(A \cap B) \neq P(A) \cdot P(B)$ - that is dependence. Moreover, it is maximum negative dependence: if one occurred, the other is guaranteed not to have. Mutual exclusivity is the opposite of independence, not a synonym.

We roll a die. A = "even" = {2,4,6}, B = "less than 4" = {1,2,3}. Are these events independent?

$A = \{2,4,6\}$, $P(A) = 3/6 = 1/2$ $B = \{1,2,3\}$, $P(B) = 3/6 = 1/2$ $A \cap B = \{2\}$, $P(A \cap B) = 1/6$ $P(A) \cdot P(B) = 1/2 \cdot 1/2 = 1/4$ But $P(A \cap B) = 1/6 \neq 1/4$ The events are **dependent**! Knowing "less than 4" reduces the probability of even (only 2 out of {1,2,3} is even).

🪙 5 heads in a row

Simple multiplication

What is the probability of getting 5 heads in a row with a fair coin? Flips are **independent**, $P(\text{heads}) = 1/2$. $$P(\text{5 heads}) = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \approx 3.1\%$$ Rare, but not impossible. On average this happens once every 32 series of 5 flips.

🔧 Reliability of a series system

Real-world application

A system of 3 components works only if **all three** are working. Reliability: $R_1 = 0.95$, $R_2 = 0.98$, $R_3 = 0.99$ Failures are **independent**. $$P(\text{system works}) = 0.95 \cdot 0.98 \cdot 0.99 \approx 0.921$$ Even with high component reliability (~95-99%), the system only works 92% of the time! **Engineering takeaway:** for critical systems, use **parallel** redundancy (duplication).

The probability of rain on any given day is 30% (independently). What is the probability of rain on at least one day out of 7?

$P(\text{no rain on a day}) = 0.7$ $P(\text{no rain all week}) = 0.7^7 \approx 0.082$ $P(\text{at least 1 rainy day}) = 1 - 0.082 \approx \mathbf{91.8\%}$ **The complement trick** - it is often easier to compute $P(\text{none})$ than to enumerate all the cases for "at least one".

🎡 Roulette after 5 blacks

Flawed vs. correct reasoning

**Flawed:** "Black came up 5 times! The law of large numbers says there must be balance. I'm betting on red!" **Correct:** Every spin is **independent**. The roulette wheel doesn't remember previous results. $P(\text{red on spin 6}) = 18/37 \approx 48.6\%$ Exactly the same probability as on **any** spin. "Balance" is restored not by red becoming more likely, but by the black fraction washing out over a long series.

A coin has landed heads 10 times in a row. What is the probability of tails on the 11th flip?

For a **fair** coin, each flip is independent. The previous 10 heads have no effect on the 11th flip. $P(\text{tails on flip 11}) = 50\%$ **However**, 10 heads in a row is a very rare event $(1/1024)$. If it happened, it's reasonable to **question** whether the coin is fair! Perhaps $P(\text{tails}) \neq 50\%$ for this particular coin. The answer "depends on whether it's fair" would be correct if the question were about a real-world situation. But for an **ideal** fair coin - 50%.

☂️ Umbrella and sunglasses

Classic example of conditional independence

$A$ = "Alice brought an umbrella" $B$ = "Bob is wearing sunglasses" $C$ = "It is sunny today" **Unconditionally**, $A$ and $B$ are **dependent** - both depend on the weather. If Alice brought an umbrella, it's probably cloudy, and Bob probably isn't wearing sunglasses. **Given "it is sunny"**, $A$ and $B$ may become **independent**: each person makes their own decision, and the weather is already known. $$P(A \cap B | C) = P(A|C) \cdot P(B|C)$$

Two students' exam results are unconditionally dependent (both prepared from the same materials). Under what condition might they become independent?

The results **correlate** through a shared hidden factor - level of preparation (and exam difficulty). If we **know** each student's level of preparation, the results become conditionally independent: each sits their own exam, given their known level. $$P(A \geq 4, B \geq 4 | \text{levels}) = P(A \geq 4 | \text{level of A}) \cdot P(B \geq 4 | \text{level of B})$$ This is the principle of **d-separation** in Bayesian networks: conditioning on a common "ancestor" makes the descendants independent.

A shooter hits a target with probability 0.8. Shots are independent. What is the probability of hitting at least once in 3 shots?

$P(\text{miss}) = 0.2$ $P(\text{3 misses in a row}) = 0.2^3 = 0.008$ $P(\text{at least 1 hit}) = 1 - 0.008 = \mathbf{0.992 = 99.2\%}$ Almost certainly hits at least once!

Two dice are rolled. Are these events independent: A = "the first die shows 6" and B = "the sum of the dice is 7"?

$P(A) = 1/6$ (first die shows 6) $P(B) = 6/36 = 1/6$ - pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) $A \cap B$: first die = 6 **and** sum = 7 → second die = 1 That is the single pair (6, 1): $P(A \cap B) = 1/36$ **Check:** $P(A) \cdot P(B) = 1/6 \cdot 1/6 = 1/36 = P(A \cap B)$ ✅ The events are **independent**! Surprising but true.

A system consists of 4 components: two connected in series (both must work), then that block is connected in parallel with a second identical block. Each component has reliability 0.9. Failures are independent. What is the system's reliability?

**Step 1:** Reliability of one series block: $R_{\text{block}} = 0.9 \cdot 0.9 = 0.81$ **Step 2:** Probability of block failure: $Q_{\text{block}} = 1 - 0.81 = 0.19$ **Step 3:** For the parallel connection, the system works if at least one block works: $R_{\text{system}} = 1 - Q_{\text{block}}^2 = 1 - 0.19^2 = 1 - 0.0361 = \mathbf{0.964}$ Parallel redundancy boosted reliability from 81% to 96.4%!

A server is replicated across three independent instances, each failing on a given day with probability $0.05$. What is the probability that at least one instance stays up for the day?

At least one survives iff not all fail. With independent failures $P(\text{all fail}) = 0.05^3$, so the answer is $1 - 0.05^3 = 0.999875$. Parallel redundancy sharply boosts fault tolerance.