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The product rule as a test, at least one through the complement, and why disjoint events are the opposite of independent.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will test two events for independence by comparing the probability of their intersection with the product of their probabilities, use independence to compute intersections and the probability of at least one, and say when independence is being assumed rather than checked. You will also tell independence apart from disjointness, which for events of positive probability is its opposite.
Every chained calculation so far conditioned each stage on what had already happened. Independence is the name for the case where that conditioning changes nothing — so this is not a new rule but a named special case, and naming it is what makes its misuse visible.
$A$ and $B$ are independent when $P(A \cap B) = P(A)P(B)$. They are disjoint, or mutually exclusive, when $A \cap B = \varnothing$. Several events are mutually independent when the product rule holds for every sub-collection of them, and pairwise independent when it holds for every pair — which is strictly weaker.
Independence is defined by an equation: $P(A \cap B) = P(A)P(B)$. When $P(B) > 0$ this is the same as $P(A \mid B) = P(A)$, which is the reading worth carrying: knowing $B$ happened tells you nothing about $A$. It is symmetric, so it also tells you nothing the other way.
In practice independence arrives two ways. Sometimes it is checked, by computing both sides. More often it is assumed, from the physics of the situation — two dice on a table, two machines in different rooms, sampling with replacement — and then it is a modelling decision that should be written down as one.
Disjoint and independent are not two words for the same idea; for events of positive probability they are opposites. Disjoint means one happening rules the other out, so learning that the first happened changes the second's probability to zero — which is the strongest possible dependence.
For more than two events, the product rule must hold for every sub-collection. Pairwise independence is not enough: three events can agree in pairs and fail as a triple, and the standard example is two fair coins with $C$ the event that they match.
Another way: story
Two machines in different rooms, each failing on $10\%$ of days. If nothing connects them, both fail on $1\%$ of days. If they share a power supply, the figure could be anything up to $10\%$ — and the difference between $1\%$ and $10\%$ is a tenfold error produced entirely by assuming independence instead of arguing for it.
Another way: steps
Two fair coins. Let $A$ be the first is heads, $B$ the second is heads, and $C$ the two match.
| Event | Probability |
|---|---|
| $A$, $B$, $C$ | $1/2$ each |
| $A \cap B$, $A \cap C$, $B \cap C$ | $1/4$ each |
| $A \cap B \cap C$ | $1/4$ |
Every pair multiplies correctly: $1/4 = 1/2 \times 1/2$. The triple does not: $1/4 \neq 1/8$. And that is unsurprising once said in words — any two of these three determine the third completely.
*Reading independent as cannot both happen. The single most expensive confusion in this subject. Disjoint events with positive probability are maximally dependent*.
Inferring independence from an absence of an obvious connection. Two insurance claims in the same street, two loans in the same industry, two machines on the same supply: nothing visible connects them and they are not independent.
*Adding for at least one.* $P(A \cup B)$ is $P(A) + P(B)$ only for disjoint events. Under independence use $1 - (1 - P(A))(1 - P(B))$.
Taking pairwise independence for mutual independence. The product rule must hold for every sub-collection, and checking the pairs is not checking the triple.
One fair die. $A$ is even, $B$ is at most two. $P(A) = 1/2$, $P(B) = 1/3$.
Write all three down first.
$A \cap B = \{2\}$, so $P(A \cap B) = 1/6$.
The overlap, counted.
$1/2 \times 1/3 = 1/6$: independent, on one die, with no physical separation at all.
The arithmetic decides, not the story.
One die. $A$ is even, $B$ is shows a five. These cannot both happen.
Disjoint.
$P(A \cap B) = 0$, but $P(A)P(B) = 1/2 \times 1/6 = 1/12$.
The product test fails.
And $P(A \mid B) = 0 \neq 1/2 = P(A)$: learning $B$ changed $A$ as much as it possibly could.
Maximal dependence, not independence.
At least one works is the complement of both fail.
Complement first.
Both fail with probability $0.1 \times 0.1 = 0.01$, by independence.
The product rule, used where it is assumed.
So the system works with probability $1 - 0.01 = 0.99$.
Of $100$ people, $40$ belong to $A$, $50$ belong to $B$ and $35$ belong to both. Work the test through.
| Probability | Independent? | |
|---|---|---|
| $P(A) \times P(B)$ | — | |
| $P(A \cap B)$ | — | |
| Verdict | — |
Consider the first of two cards drawn without replacement being an ace, and the second being an ace. How do the two events stand to each other?
Two independent alarms are fitted. The first sounds with probability $3/5$ and the second with probability $3/10$. What is the probability that at least one sounds?
Answer:
$A$ and $B$ are independent, with $P(A) = 3/5$ and $P(B) = 1/5$. Give the two probabilities.
Intersection: p. Union: q.
Four sentences from a report. Mark the two that use *independent* to mean something it does not mean.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$A$ and $B$ are independent, $P(A) = 2/5$ and $P(B) = 7/10$. Put these four probabilities in order, smallest first.
Number the steps in order (write the number in the box):
You can apply the product test, compute at least one through the complement, and say whether independence in a problem was checked or assumed. Say in your own words why two events that cannot both happen are the furthest thing from independent.
9. Your turn: two independent components each work with probability $0.9$; find the probability the system works when it needs at least one of them, step 3