Chapter 13 · Probability

Exercise 13.2 · Q1: $P(A \cap B)$ for independent A and B

Exercise 13.2Independent events · Section 13.4

The question

If $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$ ⁠, find $P(A\cap B)$ if $A$ and $B$ are independent events.

Exam answer

Since $A$ and $B$ are independent events,

\begin{aligned} P(A\cap B) &= P(A)\,P(B) \\ &= \frac{3}{5}\times\frac{1}{5} \end{aligned}

$\therefore\ P(A\cap B) = \dfrac{3}{25}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Two events are independent exactly when the probability that both happen is the product of their separate probabilities: $P(A\cap B) = P(A)\,P(B)$ ⁠. That is the book's Definition 3. So once we are told that $A$ and $B$ are independent, we may simply multiply.

Given, to find, and the plan

Given: $P(A) = \frac35$ ⁠, $P(B) = \frac15$ ⁠, and $A$ ⁠, $B$ are independent.
To find: $P(A\cap B)$ ⁠, the probability that $A$ and $B$ both happen.
Plan: (1) Write down what "independent" means. (2) Multiply the two probabilities. (3) Check the answer against the other form of the definition, $P(A|B) = P(A)$ ⁠.

Basics you need

1Events are sets

$A\cap B$ is the event " $A$ and $B$ both happen": the outcomes that lie in both sets. Basics: events as sets

2Independent events

$A$ and $B$ are independent if $P(A\cap B) = P(A)\,P(B)$ (Definition 3, Section 13.4). Equivalently, knowing that one happened does not change the probability of the other: $P(A|B) = P(A)$ (Definition 2). Lesson: independent events

3Multiplying fractions

Multiply the tops and multiply the bottoms: $\frac{a}{b}\times\frac{c}{d} = \frac{ac}{bd}$ ⁠.

Solution

Write down what "independent" gives us

Because $A$ and $B$ are independent events,
$P(A\cap B) = P(A)\,P(B).$
This is the definition of independence in the form we need. The question gives $P(A)$ and $P(B)$ and asks for $P(A\cap B)$ ⁠, and this one equation connects exactly those three numbers. Without the word "independent" we could not find $P(A\cap B)$ from $P(A)$ and $P(B)$ alone.

Definition 3, Section 13.4: $E$ and $F$ are independent if $P(E\cap F) = P(E)\,P(F)$ ⁠. Lesson: independent events
Multiply the two probabilities
$\begin{aligned} P(A\cap B) &= \frac{3}{5}\times\frac{1}{5} \\ &= \frac{3\times 1}{5\times 5} \\ &= \frac{3}{25}. \end{aligned}$
The fraction is already in lowest terms, since $3$ and $25$ have no common factor.

We substitute the given values into the rule of Step 1. The answer $\frac{3}{25}$ is smaller than both $\frac35$ and $\frac15$ ⁠, as it must be: "both happen" can never be more likely than either one on its own, because $A\cap B$ is a part of $A$ and a part of $B$ ⁠.

Product of fractions: tops times tops over bottoms times bottoms. The product rule itself is Definition 3, Section 13.4. Lesson: independent events
Check with the other form of independence

By the definition of conditional probability,
$\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} \\ &= \frac{3}{25}\div\frac{1}{5} \\ &= \frac{3}{25}\times 5 = \frac{3}{5}, \end{aligned}$
which is exactly $P(A)$ ⁠. So knowing that $B$ happened leaves the probability of $A$ unchanged, as independence promises.

Definition 2 says independence means $P(A|B) = P(A)$ ⁠. Our answer passes that test too, which confirms that we used the product rule correctly and not, for example, a sum.

Definition 1, Section 13.2: $P(E|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, provided $P(F) \ne 0$ ⁠. Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction

Result

$P(A\cap B) = P(A)\,P(B) = \frac{3}{5}\times\frac{1}{5} = \frac{3}{25}$ ⁠.

(Textbook answer: $\frac{3}{25}$ ⁠.)

Check it with numbers

In decimals, $P(A) = 0.6$ and $P(B) = 0.2$ ⁠, so $0.6\times 0.2 = 0.12$ ⁠, and $\frac{3}{25} = 0.12$ ⁠. Picture 100 equally likely outcomes: 60 lie in $A$ ⁠, and if one fifth of those also lie in $B$ ⁠, that is $12$ outcomes, which is $12$ out of $100$ ⁠.

Common mistakes

Adding instead of multiplying. $\frac35 + \frac15 = \frac45$ is not $P(A\cap B)$ ⁠. Adding belongs to "or" (and then the overlap must be taken away), not to "and".
Confusing independent with mutually exclusive. Mutually exclusive events have $P(A\cap B) = 0$ ⁠. Independent events with non-zero probabilities always overlap: here $P(A\cap B) = \frac{3}{25} > 0$ ⁠.
Multiplying without the word "independent". $P(A\cap B) = P(A)\,P(B)$ is not true for all events. It is true here only because the question says the events are independent.

What to take away

For independent events, "and" means multiply: $P(A\cap B) = P(A)\,P(B)$ ⁠.
The probability of "both" is never larger than either single probability.
A quick check of independence: $P(A|B)$ should come back equal to $P(A)$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$A$ and $B$ independent $\iff P(A\cap B) = P(A)\,P(B)$	Steps 1, 2	Definition 3, Section 13.4 (Lesson: independent events)
Independence as $P(E\|F) = P(E)$	Step 3	Definition 2, Section 13.4
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Step 3	Definition 1, Section 13.2
Independent events with non-zero probabilities are not mutually exclusive	Common mistakes	Remarks (ii), Section 13.4

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Exercise 13.2 · Q1: P(A∩B)P(A \cap B)P(A∩B) for independent A and B

Basics you need

Solution

Write down what "independent" gives us

Multiply the two probabilities

Check with the other form of independence

Every fact used, and where it comes from

Exercise 13.2 · Q1: $P(A \cap B)$ for independent A and B