Chapter 13 · Probability

Example 15: Construction job on time: total probability over a strike, $0.488$

ExamplesTheorem of total probability · Section 13.5

The question

A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.

Exam answer

Let $A$ ⁠: the job is completed on time, and $B$ ⁠: there is a strike.

$B$ and $B'$ form a partition of the sample space (there is either a strike or no strike).

$P(B) = 0.65 = \frac{13}{20}$

$P(B') = 1 - \frac{13}{20} = \frac{7}{20}$

$P(A|B) = 0.32 = \frac{8}{25}$

$P(A|B') = 0.80 = \frac45$

By the theorem of total probability,

\begin{aligned} P(A) &= P(B)P(A|B) \\ &\quad + P(B')P(A|B') \\ &= \frac{13}{20}\cdot\frac{8}{25} + \frac{7}{20}\cdot\frac45 \\ &= \frac{26}{125} + \frac{7}{25} \\ &= \frac{26}{125} + \frac{35}{125} \end{aligned}

$\therefore$ The probability that the job is completed on time is $\dfrac{61}{125} = 0.488$ ⁠.

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

The one idea this problem is built on

Split by the cause, then add. We know the chance of finishing on time in each situation (strike, no strike), and how likely each situation is. The job finishes on time either with a strike or without one, never both, so

\begin{aligned} &P(\text{on time}) \\ &= P(\text{strike and on time}) \\ &\quad + P(\text{no strike and on time}). \end{aligned}

Each piece is a product (multiplication rule). This is the theorem of total probability.

Given, to find, and the plan

Given: P(strike) = 0.65; P(on time | no strike) = 0.80; P(on time | strike) = 0.32.
To find: P(the job is completed on time).
Plan: (1) Name the events and check the partition. (2) Write every probability as a fraction. (3) Apply the theorem of total probability. (4) Simplify and give the decimal.

Basics you need

1Reading "if" as a condition

"0.80 that the job will be completed on time if there is no strike" is $P(\text{on time} \mid \text{no strike})$ ⁠. Lesson: conditional probability

2Complement

$P(B') = 1 - P(B)$ ⁠: "no strike" has probability $1 - 0.65$ ⁠. Basics: rules of probability

3Decimals as fractions

$0.65 = \frac{65}{100} = \frac{13}{20}$ ⁠; divide top and bottom by their highest common factor. Basics: working with fractions

4Theorem of total probability (Section 13.5.2)

For a partition $\{B, B'\}$ ⁠: $P(A) = P(B)P(A|B) + P(B')P(A|B')$ ⁠. Lesson: total probability and Bayes' theorem

Solution

Name the events

Let
- $A$ ⁠: the construction job is completed on time;
- $B$ ⁠: there is a strike, so $B'$ ⁠: there is no strike.
We have to find $P(A)$ ⁠. The events $B$ and $B'$ form a partition of the sample space: they do not overlap, together they cover every possibility, and both have nonzero probability.

The strike is the "cause" that changes the chance of finishing on time; finishing on time is the event we want. An event and its complement always form a partition when neither is impossible (Section 13.5.1), so the theorem of total probability applies.

$\{B, B'\}$ is a partition of $S$ ⁠. Lesson: total probability
Every number as a fraction
$\begin{aligned} P(B) &= 0.65 = \frac{65}{100} = \frac{13}{20}, \\ P(B') &= 1 - \frac{13}{20} = \frac{7}{20}, \\ P(A|B) &= 0.32 = \frac{32}{100} = \frac{8}{25}, \\ P(A|B') &= 0.80 = \frac{80}{100} = \frac45. \end{aligned}$
The question gives decimals; fractions keep every later step exact. "0.32 that the job will be completed on time if there is a strike" is about jobs with a strike, so it is $P(A|B)$ ⁠; the 0.80 is for no strike, $P(A|B')$ ⁠. The book writes $P(B') = 1 - 0.65 = 0.35$ ⁠, which is $\frac{35}{100} = \frac{7}{20}$ ⁠.

Cancel by the highest common factor: $\frac{65}{100} = \frac{13}{20}$ (divide by 5). Basics: working with fractions
Apply the theorem of total probability

Since $B$ and $B'$ form a partition, by the theorem of total probability
$\begin{aligned} P(A) &= P(B)P(A|B) \\ &\quad + P(B')P(A|B') \\ &= \frac{13}{20}\cdot\frac{8}{25} + \frac{7}{20}\cdot\frac45 \\ &= \frac{104}{500} + \frac{28}{100} \\ &= \frac{26}{125} + \frac{7}{25}. \end{aligned}$

The tree for Example 15. The two green paths end with the job on time: strike then on time ( $\frac{26}{125}$ ⁠) and no strike then on time ( $\frac{7}{25}$ ⁠). Their sum is $P(A)$ ⁠.

The job is on time in two ways that cannot both happen: with a strike, or without one. Each way is a path in the tree, and its probability is the product along the path (multiplication rule, $P(B \cap A) = P(B)P(A|B)$ ⁠). The two paths are mutually exclusive, so they add.

Multiply along a path, add the paths. Basics: tree diagrams
Add the fractions
$\begin{aligned} P(A) &= \frac{26}{125} + \frac{35}{125} \\ &= \frac{61}{125} = 0.488. \end{aligned}$
To add, give both fractions the denominator 125: $\frac{7}{25} = \frac{35}{125}$ ⁠. The fraction $\frac{61}{125}$ cannot be simplified (61 is prime and does not divide 125), and since $125 \times 8 = 1000$ ⁠, it is exactly $\frac{488}{1000} = 0.488$ ⁠, the book's value.

Same denominator, then add the numerators. Basics: working with fractions

Result

The probability that the construction job will be completed on time is $\frac{61}{125} = 0.488$ ⁠.

(Textbook answer: $0.488$ ⁠.)

Check it with numbers

Picture 10000 such jobs. A strike hits 65% of them, 6500 jobs, and 32% of those, 2080, still finish on time. The other 3500 jobs have no strike, and 80% of them, 2800, finish on time. So $2080 + 2800 = 4880$ of the 10000 jobs finish on time: $\frac{4880}{10000} = 0.488$ ⁠, the same as $\frac{61}{125} = 0.488$ ⁠.

The answer lies between the two conditional values 0.32 and 0.80, closer to 0.32 because a strike is the more likely case. A total probability is always such a weighted average.

Common mistakes

Swapping the conditional probabilities. 0.80 belongs to "no strike" and 0.32 to "strike". Read which condition each number comes with.
Adding the conditional probabilities. $0.80 + 0.32$ is more than 1. Each must first be multiplied by the probability of its own condition.
Forgetting $P(B')$ ⁠. The question does not give the chance of no strike; it is $1 - 0.65 = 0.35$ ⁠.

What to take away

When a probability depends on which of several situations occurs, find each situation's probability, multiply by the conditional probability, and add: the theorem of total probability.
The next examples turn this round: given that the event happened, which cause was it? That is Bayes' theorem, starting with Example 16. An exercise like this one is Exercise 13.3, Question 1.

Every fact used, and where it comes from

Fact	Used in	Source
An event and its complement form a partition	Step 1	Section 13.5.1 (Lesson: total probability)
$P(B') = 1 - P(B)$ ⁠; decimals as fractions	Step 2	Class XI (Basics: rules of probability)
Conditional probability $P(A\|B)$	Step 2	Section 13.2 (Lesson: conditional probability)
Theorem of total probability	Step 3	Section 13.5.2 (Lesson: total probability)
Multiplication rule $P(B \cap A) = P(B)P(A\|B)$	Step 3	Section 13.3 (Lesson: multiplication rule)

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Example 15: Construction job on time: total probability over a strike, 0.4880.4880.488

Basics you need

Solution

Name the events

Every number as a fraction

Apply the theorem of total probability

Add the fractions

Every fact used, and where it comes from

Example 15: Construction job on time: total probability over a strike, $0.488$