Chapter 13 · Probability

Example 13: If E and F are independent, so are E and $F'$

ExamplesIndependent events · Section 13.4

The question

Prove that if E and F are independent events, then so are the events E and F′.

Exam answer

Since $E$ and $F$ are independent,

P(E\cap F) = P(E)\,P(F) \quad\ldots(1)

$E\cap F$ and $E\cap F'$ are mutually exclusive, and $E = (E\cap F)\cup(E\cap F')$ ⁠.

So, by the addition rule for mutually exclusive events,

P(E) = P(E\cap F) + P(E\cap F')

\begin{aligned} P(E\cap F') &= P(E) - P(E\cap F) \\ &= P(E) - P(E)\,P(F) \\ &\qquad [\text{by }(1)] \\ &= P(E)\,[1 - P(F)] \end{aligned}

Since $1 - P(F) = P(F')$ ⁠,

P(E\cap F') = P(E)\,P(F')

$\therefore\ E$ and $F'$ are independent events. Hence proved.

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

The one idea this problem is built on

Split $E$ into the part inside $F$ and the part outside $F$ ⁠. The two parts do not overlap, so their probabilities add up to $P(E)$ ⁠. The part inside $F$ has probability $P(E)\,P(F)$ by independence, so the part outside $F$ gets what is left:

\begin{aligned} &P(E) - P(E)\,P(F) \\ &= P(E)\,[1 - P(F)] \\ &= P(E)\,P(F'). \end{aligned}

That is exactly the product test for $E$ and $F'$ ⁠.

Given, to find, and the plan

Given: $E$ and $F$ are independent: $P(E \cap F) = P(E)\,P(F)$ ⁠.
To show: $E$ and $F'$ are independent: $P(E \cap F') = P(E)\,P(F')$ ⁠.
Plan: Write $E$ as two mutually exclusive pieces; add their probabilities; substitute the given product; factor out $P(E)$ ⁠; recognise $1 - P(F)$ as $P(F')$ ⁠.

Basics you need

1Independent events (Section 13.4)

$E$ and $F$ are independent if $P(E \cap F) = P(E)\,P(F)$ ⁠. To prove two events independent, show this product rule holds. Lesson: independent events

2Two events cut S into four regions

$E \cap F$ ⁠, $E \cap F'$ ⁠, $E' \cap F$ and $E' \cap F'$ do not overlap, and $E$ is made of the first two. Basics: Venn diagrams

3Adding probabilities

If $X$ and $Y$ are mutually exclusive, $P(X \cup Y) = P(X) + P(Y)$ ⁠; and $P(F') = 1 - P(F)$ ⁠. Basics: rules of probability

Solution

Write down what is given and what must be shown

Since $E$ and $F$ are independent,
$P(E \cap F) = P(E)\,P(F) \quad\ldots(1)$
We must show the same kind of equation with $F'$ in place of $F$ ⁠:
$P(E \cap F') = P(E)\,P(F').$
Independence is defined by the product rule (Definition 3, Section 13.4). So " $E$ and $F'$ are independent" means exactly this equation; proving it is the whole task. Equation (1) is the only information we have, so it must be used somewhere.

Independent means $P(E \cap F) = P(E)\,P(F)$ ⁠. Lesson: independent events
Split E into two pieces

Every outcome in $E$ is either in $F$ or not in $F$ ⁠. So
$E = (E \cap F) \cup (E \cap F'),$
and the two pieces $E \cap F$ and $E \cap F'$ are mutually exclusive (no outcome is in both $F$ and $F'$ ⁠).

The shaded circle is $E$ ⁠. The circle $F$ cuts it into the lens $E \cap F$ and the crescent $E \cap F'$ ⁠. The two pieces do not overlap and together make up $E$ ⁠.

This is the same picture the book draws (Fig 13.3). Cutting $E$ along the edge of $F$ is useful because one piece, $E \cap F$ ⁠, is the piece equation (1) talks about, and the other piece, $E \cap F'$ ⁠, is the one we need.

Two events cut S into four regions. Basics: Venn diagrams
Add the probabilities of the pieces

By the addition rule for mutually exclusive events,
$P(E) = P(E \cap F) + P(E \cap F').$
Subtract $P(E \cap F)$ from both sides:
$P(E \cap F') = P(E) - P(E \cap F).$
For events with no common outcome, the probability of "one or the other" is the plain sum; nothing is counted twice. This turns the set equation of Step 2 into an equation between numbers.

$P(X \cup Y) = P(X) + P(Y)$ when $X \cap Y = \phi$ ⁠. Basics: rules of probability
Use independence and factor
$\begin{aligned} P(E \cap F') &= P(E) - P(E \cap F) \\ &= P(E) - P(E)\,P(F) \\ &\qquad [\text{by }(1)] \\ &= P(E)\,[1 - P(F)] \\ &= P(E)\,P(F'). \end{aligned}$
So $E$ and $F'$ are independent.

Equation (1) replaces $P(E \cap F)$ by a product. Both terms then contain $P(E)$ ⁠, so it comes out as a common factor. What is left in the bracket, $1 - P(F)$ ⁠, is the probability that $F$ does not happen, which is $P(F')$ ⁠. The last line is the product rule for $E$ and $F'$ ⁠, which is what we had to show.

Complement rule: $P(F') = 1 - P(F)$ ⁠. Basics: rules of probability
The other two pairs (the book's Note)

The same result, used twice more, gives the Note after this example:
- Swap the names of the events: if $F$ and $E$ are independent, then so are $F$ and $E'$ ⁠. So $E'$ and $F$ are independent.
- Now apply the result to the independent pair $E'$ and $F$ ⁠: then $E'$ and $F'$ are independent.
The proof used nothing special about $E$ or $F$ ⁠, only that they are independent. So it can be applied to any independent pair, including pairs that already have a dash. Independence is a two-way relation ( $P(E \cap F) = P(F \cap E)$ ⁠), so the order of the two events does not matter.

Independence of $E$ and $F$ is the same as independence of $F$ and $E$ ⁠. Lesson: independent events

Result

$P(E \cap F') = P(E)\,P(F')$ ⁠, so if $E$ and $F$ are independent, then $E$ and $F'$ are independent. Hence proved.

(Textbook answer: Hence, E and F′ are independent.)

Check it with numbers

A die. Throw a fair die; let $E = \{2, 4, 6\}$ (even) and $F = \{1, 2\}$ ⁠. Then $P(E) = \frac{3}{6}$ ⁠, $P(F) = \frac{2}{6}$ and $E \cap F = \{2\}$ ⁠, so $P(E \cap F) = \frac{1}{6} = \frac{3}{6} \times \frac{2}{6}$ ⁠: $E$ and $F$ are independent.

Now $F' = \{3, 4, 5, 6\}$ and $E \cap F' = \{4, 6\}$ ⁠, so $P(E \cap F') = \frac{2}{6} \approx 0.3333$ ⁠, and $P(E)\,P(F') = \frac{3}{6} \times \frac{4}{6} = \frac{12}{36} \approx 0.3333$ ⁠. The two agree, as the proof says they must.

Common mistakes

Assuming what is to be proved. Writing $P(E \cap F') = P(E)\,P(F')$ in the first line proves nothing; it must come out at the end of a chain that starts from $P(E \cap F')$ ⁠.
Confusing independent with mutually exclusive. $E \cap F$ and $E \cap F'$ are mutually exclusive, so their probabilities add. $E$ and $F$ themselves are independent, so their probabilities multiply. Do not mix up the two rules.
Writing $P(F') = P(F) - 1$ ⁠. The complement rule is $P(F') = 1 - P(F)$ ⁠.

What to take away

If $E$ and $F$ are independent, then so are $E$ and $F'$ ⁠, $E'$ and $F$ ⁠, and $E'$ and $F'$ ⁠. Knowing that $F$ did not happen tells you no more about $E$ than knowing that it did.
To prove a probability identity, split an event into mutually exclusive pieces and add.
This result is used in Example 14 and in questions such as Exercise 13.2, Question 11, where $P(A \text{ and not } B) = P(A)\,P(B')$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Independent if $P(E \cap F) = P(E)\,P(F)$	Steps 1, 4	Definition 3, Section 13.4 (Lesson: independent events)
$E = (E \cap F) \cup (E \cap F')$ ⁠, pieces mutually exclusive	Step 2	Fig 13.3, Section 13.4 (Basics: Venn diagrams)
Addition rule for mutually exclusive events	Step 3	Class XI, Probability (Basics: rules of probability)
$P(F') = 1 - P(F)$	Step 4	Class XI, Probability (Basics: rules of probability)
$E'$ and $F$ ⁠, $E'$ and $F'$ independent	Step 5	Note after Example 13, Section 13.4

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Example 13: If E and F are independent, so are E and F′F'F′

Basics you need

Solution

Write down what is given and what must be shown

Split E into two pieces

Add the probabilities of the pieces

Use independence and factor

The other two pairs (the book's Note)

Every fact used, and where it comes from

Example 13: If E and F are independent, so are E and $F'$