Chapter 13 · Probability

Example 14: Independent A, B: at least one is $1 - P(A')P(B')$

ExamplesIndependent events · Section 13.4

The question

If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by $1 - P(A')\,P(B')$

Exam answer

Since $A$ and $B$ are independent, $P(A\cap B) = P(A)\,P(B)$ ⁠.

By the addition theorem of probability,

\begin{aligned} &P(\text{at least one of A and B}) \\ &= P(A\cup B) \\ &= P(A) + P(B) - P(A\cap B) \\ &= P(A) + P(B) - P(A)\,P(B) \\ &= P(A) + P(B)\,[1 - P(A)] \\ &= P(A) + P(B)\,P(A') \end{aligned}

Writing $P(A) = 1 - P(A')$ ⁠,

\begin{aligned} &P(\text{at least one of A and B}) \\ &= 1 - P(A') + P(B)\,P(A') \\ &= 1 - P(A')\,[1 - P(B)] \end{aligned}

$\therefore$ P(at least one of A and B) $= 1 - P(A')\,P(B')$ ⁠, as $1 - P(B) = P(B')$ ⁠. Hence proved.

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

The one idea this problem is built on

"At least one of $A$ and $B$ " is the event $A \cup B$ ⁠. The addition theorem gives its probability; independence turns the overlap term $P(A \cap B)$ into the product $P(A)\,P(B)$ ⁠; and some factoring with complements gives the neat form $1 - P(A')\,P(B')$ ⁠.

The form makes sense: at least one happens unless neither happens, and "neither" has probability $P(A')\,P(B')$ for independent events.

Given, to find, and the plan

Given: $A$ and $B$ are independent: $P(A \cap B) = P(A)\,P(B)$ ⁠.
To show: P(at least one of $A$ ⁠, $B$ ⁠) $= 1 - P(A')\,P(B')$ ⁠.
Plan: Write "at least one" as $A \cup B$ ⁠; use the addition theorem; replace $P(A \cap B)$ by $P(A)\,P(B)$ ⁠; factor, and write $1 - P(A)$ as $P(A')$ and $1 - P(B)$ as $P(B')$ ⁠.

Basics you need

1"At least one" is a union

" $A$ or $B$ or both happen" is the event $A \cup B$ ⁠. Basics: events as sets

2The addition theorem

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ ⁠: add, but do not count the overlap twice. Basics: rules of probability

3Independent events (Section 13.4)

$P(A \cap B) = P(A)\,P(B)$ ⁠; and then $A'$ and $B'$ are independent too (Note after Example 13). Lesson: independent events

4Complements

$P(A') = 1 - P(A)$ ⁠, so also $P(A) = 1 - P(A')$ ⁠. Basics: rules of probability

Solution

Write "at least one" as a union and use the addition theorem
$\begin{aligned} &P(\text{at least one of } A, B) \\ &= P(A \cup B) \\ &= P(A) + P(B) - P(A \cap B). \end{aligned}$
"At least one of $A$ and $B$ " means $A$ happens, or $B$ happens, or both: that is the union. Adding $P(A)$ and $P(B)$ counts the outcomes in both events twice, so $P(A \cap B)$ is subtracted once.

The addition theorem. Basics: rules of probability
Use independence

Since $A$ and $B$ are independent, $P(A \cap B) = P(A)\,P(B)$ ⁠. So
$\begin{aligned} P(A \cup B) &= P(A) + P(B) \\ &\quad - P(A)\,P(B). \end{aligned}$
This is the only place where independence is used. For events that are not independent the overlap term cannot be written as a product, and the final formula would be false.

Definition 3, Section 13.4. Lesson: independent events
Factor out P(B), then use complements
$\begin{aligned} P(A \cup B) &= P(A) + P(B)\,[1 - P(A)] \\ &= P(A) + P(B)\,P(A') \\ &= 1 - P(A') + P(B)\,P(A') \\ &= 1 - P(A')\,[1 - P(B)] \\ &= 1 - P(A')\,P(B'). \end{aligned}$
This is the required result.

Line 1 takes $P(B)$ out of the last two terms. Line 2 writes $1 - P(A)$ as $P(A')$ ⁠. Line 3 writes the first term $P(A)$ as $1 - P(A')$ ⁠, so that $P(A')$ appears twice. Line 4 takes $-P(A')$ out of the two terms that contain it, so $-P(A') + P(B)\,P(A')$ becomes $-P(A')\,[1 - P(B)]$ ⁠. Line 5 writes $1 - P(B)$ as $P(B')$ ⁠. These are the book's lines, one change per line.

Complement rule: $P(A') = 1 - P(A)$ ⁠. Basics: rules of probability
A shorter way: "at least one" fails only when neither happens

The shaded region is $A \cup B$ ⁠, "at least one". Everything else in $S$ is $A' \cap B'$ ⁠, "neither". So $P(A \cup B) = 1 - P(A' \cap B')$ ⁠.

"At least one of $A$ ⁠, $B$ " is the complement of "neither $A$ nor $B$ ", which is $A' \cap B'$ ⁠. Since $A$ and $B$ are independent, so are $A'$ and $B'$ (Note after Example 13). Hence
$\begin{aligned} P(A \cup B) &= 1 - P(A' \cap B') \\ &= 1 - P(A')\,P(B'). \end{aligned}$
This is the same result in two lines. It shows where the formula comes from: $P(A')\,P(B')$ is the chance that both fail, and "at least one" is everything else. It relies on the Note after Example 13; the algebra of Steps 1 to 3, which is the book's method, does not need it.

$(A \cup B)' = A' \cap B'$ (De Morgan's law). Basics: Venn diagrams

Result

For independent events $A$ and $B$ ⁠, P(at least one of $A$ ⁠, $B$ ⁠) $= P(A \cup B)$ $= 1 - P(A')\,P(B')$ ⁠. Hence proved.

(Textbook answer: P(at least one of A and B) = 1 − P(A′) P(B′).)

Check it with numbers

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

At least one happens for the faces in $A \cup B = \{1, 2, 4, 6\}$ ⁠, so $P(A \cup B) = \frac{4}{6} \approx 0.6667$ ⁠.

The formula gives $1 - P(A')\,P(B') = 1 - \frac{3}{6} \times \frac{4}{6}$ ⁠, which is $1 - \frac{12}{36} = \frac{24}{36} \approx 0.6667$ ⁠. The two agree.

Common mistakes

Writing $P(A \cup B) = P(A) + P(B)$ ⁠. That is only for mutually exclusive events; independent events with nonzero probabilities do overlap, and the overlap must be subtracted.
Writing $1 - P(A)\,P(B)$ ⁠. $1 - P(A)\,P(B)$ is the probability that not both happen. "At least one" is $1 - P(A')\,P(B')$ ⁠, with the complements.
Using the formula without independence. $P(A' \cap B') = P(A')\,P(B')$ needs $A$ and $B$ independent.

What to take away

For independent events, $P(\text{at least one}) = 1 - P(\text{none})$ ⁠, and $P(\text{none})$ is a product of complements. The same idea works for three or more independent events.
Questions that say "the problem is solved" (by at least one person) use this formula; see Exercise 13.2, Question 14 and Question 12.

Every fact used, and where it comes from

Fact	Used in	Source
"At least one" is $A \cup B$	Step 1	Class XI, Probability (Basics: events as sets)
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$	Step 1	Class XI, Probability (Basics: rules of probability)
Independent: $P(A \cap B) = P(A)\,P(B)$	Step 2	Definition 3, Section 13.4 (Lesson: independent events)
$P(A') = 1 - P(A)$	Step 3	Class XI, Probability (Basics: rules of probability)
$A'$ and $B'$ independent when $A$ ⁠, $B$ are	Step 4	Note after Example 13, Section 13.4
$(A \cup B)' = A' \cap B'$	Step 4	De Morgan's law, Class XI, Sets (Basics: Venn diagrams)

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Example 14: Independent A, B: at least one is 1−P(A′)P(B′)1 - P(A')P(B')1−P(A′)P(B′)

Basics you need

Solution

Write "at least one" as a union and use the addition theorem

Use independence

Factor out P(B), then use complements

A shorter way: "at least one" fails only when neither happens

Every fact used, and where it comes from

Example 14: Independent A, B: at least one is $1 - P(A')P(B')$