Chapter 13 · Probability

Miscellaneous · Q9: Two subsystems: A fails given B failed (0.5), and A fails alone (0.05)

Miscellaneous ExerciseConditional probability · Section 13.2

The question

An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) = 0.2

P(B fails alone) = 0.15

P(A and B fail) = 0.15

Evaluate the following probabilities

(i) P(A fails|B has failed) (ii) P(A fails alone)

Exam answer

Let $A$ ⁠: subsystem A fails, and $B$ ⁠: subsystem B fails. Then

$P(A) = 0.2 = \frac15$ ⁠, $P(A \cap B) = 0.15 = \frac{3}{20}$

and P(B fails alone) $= P(A' \cap B) = 0.15 = \frac{3}{20}$ ⁠.

(i) B fails either together with A or alone, and these cannot happen together, so

\begin{aligned} P(B) &= P(A \cap B) + P(A' \cap B) \\ &= \frac{3}{20} + \frac{3}{20} = \frac{3}{10} \end{aligned}

\begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} \\ &= \frac{3}{20} \div \frac{3}{10} = \frac{3}{20} \times \frac{10}{3} \end{aligned}

$\therefore$ P(A fails | B has failed) $= \frac12 = 0.5$ ⁠.

(ii) A fails alone means A fails and B does not. As $A$ is the union of the disjoint events $A \cap B$ and $A \cap B'$ ⁠,

\begin{aligned} P(A \cap B') &= P(A) - P(A \cap B) \\ &= \frac15 - \frac{3}{20} = \frac{4 - 3}{20} \end{aligned}

$\therefore$ P(A fails alone) $= \dfrac{1}{20} = 0.05$ ⁠.

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

The one idea this problem is built on

Translate each phrase into an event. "B fails alone" is "B fails and A does not", $A' \cap B$ ⁠; "A fails alone" is $A \cap B'$ ⁠. Then the failures of B split into two disjoint pieces (with A, without A), and so do the failures of A. Adding or subtracting pieces gives every probability we need.

Given, to find, and the plan

Given: P(A fails) = 0.2, P(B fails alone) = 0.15, P(A and B fail) = 0.15.
To find: (i) P(A fails | B has failed). (ii) P(A fails alone).
Plan: (1) Name the events and write each given phrase as a set; convert the decimals to fractions. (2) Find $P(B)$ by adding its two pieces. (3) Use $P(A|B) = \frac{P(A \cap B)}{P(B)}$ ⁠. (4) Subtract to get A alone. (5) Check with the Venn diagram.

Basics you need

1Events as sets

"A fails" is an event $A$ ⁠; "B does not fail" is the complement $B'$ ⁠; "and" is $\cap$ ⁠. Basics: events as sets

2Splitting an event into disjoint pieces

$B = (A \cap B) \cup (A' \cap B)$ and $A = (A \cap B) \cup (A \cap B')$ ⁠, with each pair of pieces disjoint, so their probabilities add. Basics: Venn diagrams

3Conditional probability (Section 13.2)

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ ⁠, provided $P(B) \neq 0$ ⁠. Lesson: conditional probability

Part (i): P(A fails | B has failed)

Name the events and translate the data

Let $A$ ⁠: subsystem A fails, and $B$ ⁠: subsystem B fails. Then
$\begin{aligned} P(A) &= 0.2 = \frac15, \\ P(A \cap B) &= 0.15 = \frac{3}{20}. \end{aligned}$
"B fails alone" means B fails and A does not fail:
$P(A' \cap B) = 0.15 = \frac{3}{20}.$

The four regions: A alone $0.2 - 0.15 = 0.05$ ⁠, both $0.15$ ⁠, B alone $0.15$ ⁠, neither $1 - 0.35 = 0.65$ ⁠. B fails in the two right-hand regions, $0.15 + 0.15 = 0.3$ ⁠.

"Alone" is the key word: it excludes the other subsystem. So "B fails alone" is not $P(B)$ ⁠; it is only the part of $B$ outside $A$ ⁠. Writing $0.2 = \frac{2}{10} = \frac15$ and $0.15 = \frac{15}{100} = \frac{3}{20}$ keeps the arithmetic exact.

Events as sets; complements. Basics: events as sets
Find $P(B)$

Whenever B fails, A either fails too or does not, so $B = (A \cap B) \cup (A' \cap B)$ ⁠, and the two pieces are disjoint:
$\begin{aligned} P(B) &= P(A \cap B) + P(A' \cap B) \\ &= \frac{3}{20} + \frac{3}{20} = \frac{6}{20} = \frac{3}{10}. \end{aligned}$
The conditional probability in the next step divides by $P(B)$ ⁠, which is not given directly; it is the sum of the two given pieces.

Disjoint pieces add. Basics: Venn diagrams
Apply the definition
$\begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} \\ &= \frac{3}{20} \div \frac{3}{10} \\ &= \frac{3}{20} \times \frac{10}{3} = \frac12 = 0.5. \end{aligned}$
Given that B has failed, we look only inside $B$ (probability $\frac{3}{10}$ ⁠); the part of it where A also fails has probability $\frac{3}{20}$ ⁠, exactly half.

Section 13.2, Definition 1. Lesson: conditional probability

Part (ii): P(A fails alone)

Take the "both fail" piece away from $A$

"A fails alone" is $A \cap B'$ ⁠. Since $A = (A \cap B) \cup (A \cap B')$ with disjoint pieces,
$\begin{aligned} P(A \cap B') &= P(A) - P(A \cap B) \\ &= \frac15 - \frac{3}{20} \\ &= \frac{4}{20} - \frac{3}{20} = \frac{1}{20} = 0.05. \end{aligned}$
Every failure of A happens with or without a failure of B. The failures of A with B have probability $\frac{3}{20}$ ⁠, so those without B have what is left of $\frac15$ ⁠.

$P(A \cap B') = P(A) - P(A \cap B)$ ⁠. Basics: Venn diagrams
What the numbers say about the subsystems

On its own, A fails with probability 0.2. Once B has failed, A fails with probability 0.5. So the failures are not independent: $P(A)\,P(B) = \frac15 \times \frac{3}{10} = \frac{3}{50} = 0.06$ ⁠, but $P(A \cap B) = 0.15$ ⁠.

A failure of B makes a failure of A much more likely, as would happen if the two subsystems share a power supply. This is why we must use the given $P(A \cap B)$ and not multiply $P(A)$ by $P(B)$ ⁠.

Independence test $P(A \cap B) = P(A)\,P(B)$ (Section 13.4, Definition 3). Lesson: independent events

Result

(i) P(A fails | B has failed) $= \frac12 = 0.5$ ⁠. (ii) P(A fails alone) $= \frac{1}{20} = 0.05$ ⁠.

(Textbook answer: (i) 0.5 (ii) 0.05.)

Check it with numbers

Picture 100 assemblies. A fails in 20 of them; both fail in 15; B fails alone in 15. So B fails in $15 + 15 = 30$ ⁠, and A fails in 15 of these: $\frac{15}{30} = 0.5$ ⁠. A fails alone in $20 - 15 = 5$ ⁠: $\frac{5}{100} = 0.05$ ⁠. Neither fails in $100 - 5 - 15 - 15 = 65$ assemblies.

Common mistakes

Using 0.15 as $P(B)$ ⁠. 0.15 is "B fails alone"; B also fails in the 0.15 of cases where both fail, so $P(B) = 0.3$ ⁠.
Getting $\frac{0.15}{0.15} = 1$ in (i). That divides by "B fails alone" instead of by $P(B)$ ⁠.
Answering 0.2 in (ii). 0.2 is P(A fails), which includes the cases where B fails as well.

What to take away

"Alone" means "and not the other": $A \cap B'$ ⁠, $A' \cap B$ ⁠.
Split an event into disjoint pieces to find a probability that is not given directly.
Draw the four-region Venn diagram; every region follows by adding or subtracting.

Every fact used, and where it comes from

Fact	Used in	Source
Events as sets; complement $B'$	Part (i), Step 1; Part (ii), Step 1	Sets, Class XI (Basics: events as sets)
Disjoint events add	Part (i), Step 2; Part (ii), Step 1	Class XI axioms (Basics: Venn diagrams)
$P(E\|F) = \frac{P(E \cap F)}{P(F)}$ ⁠, $P(F) \neq 0$	Part (i), Step 3	Section 13.2, Definition 1 (Lesson: conditional probability)
Independence test	Part (ii), Step 2	Section 13.4, Definition 3 (Lesson: independent events)

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Miscellaneous · Q9: Two subsystems: A fails given B failed (0.5), and A fails alone (0.05)

Basics you need

Part (i): P(A fails | B has failed)

Name the events and translate the data

Find P(B)P(B)P(B)

Apply the definition

Part (ii): P(A fails alone)

Take the "both fail" piece away from AAA

What the numbers say about the subsystems

Every fact used, and where it comes from

Find $P(B)$

Take the "both fail" piece away from $A$