Chapter 13 · Probability

Exercise 13.3 · Q9: Board election: new product introduced by the second group, $P = \frac29$

Exercise 13.3Bayes' theorem · Section 13.5

The question

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Exam answer

Let $E_1$ ⁠: the first group wins, $E_2$ ⁠: the second group wins, and $A$ ⁠: the new product is introduced.

$E_1$ and $E_2$ form a partition of the sample space (exactly one of the two groups wins).

$P(E_1) = 0.6 = \frac35$ and $P(E_2) = 0.4 = \frac25$

$P(A|E_1) = 0.7 = \frac{7}{10}$ and $P(A|E_2) = 0.3 = \frac{3}{10}$

By the theorem of total probability,

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac35\cdot\frac{7}{10} + \frac25\cdot\frac{3}{10} \\ &= \frac{21}{50} + \frac{6}{50} = \frac{27}{50} \end{aligned}

By Bayes' theorem,

\begin{aligned} P(E_2|A) &= \frac{P(E_2)P(A|E_2)}{P(A)} \\ &= \frac{6}{50} \div \frac{27}{50} \\ &= \frac{6}{50} \times \frac{50}{27} \end{aligned}

$\therefore$ The probability that the new product was introduced by the second group is $\dfrac{6}{27} = \dfrac29$ (dividing by 3).

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

The one idea this problem is built on

The new product can follow a win by either group. The first group is more likely to win and much more likely to introduce the product, so most new products would come from the first group. Bayes' theorem gives the second group's share: its route to a new product divided by both routes together.

Given, to find, and the plan

Given: The first group wins with probability 0.6 and the second with probability 0.4. A new product is introduced with probability 0.7 if the first group wins and 0.3 if the second group wins. A new product has been introduced.
To find: The probability that it was introduced by the second group, that is, that the second group won, given the new product: $P(E_2|A)$ ⁠.
Plan: (1) Name the two causes and the result; check the partition. (2) Turn the decimals into fractions. (3) Total probability for $P(A)$ ⁠. (4) Bayes' theorem.

Basics you need

1Decimals as fractions

$0.6 = \frac{6}{10} = \frac35$ and $0.7 = \frac{7}{10}$ ⁠. Fractions keep the arithmetic exact. Basics: classical probability

2Conditional probability (Section 13.2)

"If the first group wins, the probability of introducing a new product is 0.7" is $P(A|E_1) = 0.7$ ⁠: a probability inside the case $E_1$ ⁠. Lesson: conditional probability

3Total probability and Bayes' theorem (Section 13.5)

For a partition $E_1, E_2$ ⁠: $P(A) = \sum P(E_i)P(A|E_i)$ and $P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(A)}$ ⁠. Lesson: Bayes' theorem

4Simplifying a fraction

Divide the top and the bottom by a common factor: $\frac{6}{27} = \frac{6 \div 3}{27 \div 3} = \frac29$ ⁠. Basics: dividing by a fraction

Solution

Name the events

Let
- $E_1$ ⁠: the first group wins;
- $E_2$ ⁠: the second group wins;
- $A$ ⁠: the new product is introduced.
$E_1$ and $E_2$ form a partition of the sample space: only one group can win the position, one of the two groups does win (the two probabilities add to $0.6 + 0.4 = 1$ ⁠), and each has a nonzero chance.

Who wins is the cause; the new product is the result we observe. "The new product introduced was by the second group" means that the second group won, given that the product was introduced.

Partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: Bayes' theorem
The four probabilities as fractions
$\begin{aligned} P(E_1) &= 0.6 = \frac35, \\ P(E_2) &= 0.4 = \frac25, \end{aligned}$ $\begin{aligned} P(A|E_1) &= 0.7 = \frac{7}{10}, \\ P(A|E_2) &= 0.3 = \frac{3}{10}. \end{aligned}$
"Respectively" pairs 0.6 with the first group and 0.4 with the second. The values 0.7 and 0.3 are each stated for one winner ("if the first group wins", "if the second group wins"), so they are conditional probabilities. Note that 0.7 and 0.3 need not add to 1: they belong to different cases.

A probability given that an event has happened is a conditional probability. Lesson: conditional probability
The chance of a new product

Since $E_1$ and $E_2$ form a partition, by the theorem of total probability
$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac35\cdot\frac{7}{10} + \frac25\cdot\frac{3}{10} \\ &= \frac{21}{50} + \frac{6}{50} = \frac{27}{50}. \end{aligned}$

The tree for Question 9. The rose path (second group wins, then the new product) has product $\frac{6}{50}$ ⁠; the green path (first group wins, then the new product) has $\frac{21}{50}$ ⁠. Together, $P(A) = \frac{27}{50}$ ⁠.

The new product follows a win by exactly one group, so the two path probabilities add. Both have denominator 50, so the next step only compares the numerators 6 and 27.

Multiply along a path, add the paths. Basics: tree diagrams
Second group, given the new product: Bayes' theorem

By Bayes' theorem,
$\begin{aligned} P(E_2|A) &= \frac{P(E_2)P(A|E_2)}{P(A)} \\ &= \frac{6}{50} \div \frac{27}{50} \\ &= \frac{6}{50} \times \frac{50}{27} = \frac{6}{27} = \frac29. \end{aligned}$
The top is the second group's path to a new product; the bottom is both paths. The 50s cancel, leaving $\frac{6}{27}$ ⁠; dividing the top and the bottom by 3 gives $\frac29$ ⁠.

Dividing by a fraction is multiplying by its flip. Basics: dividing by a fraction
A quick sense check

For the first group, $P(E_1|A) = \frac{21}{50} \div \frac{27}{50} = \frac{21}{27} = \frac79$ ⁠, and $\frac29 + \frac79 = 1$ ⁠.

The new product lowers the second group's chance from $0.4$ to $\frac29 \approx 0.222$ ⁠, because the second group is less likely to introduce it (0.3 against 0.7).

Given the new product, exactly one group won, so the two updated probabilities must add to 1.

The posterior probabilities of a partition add to 1. Lesson: Bayes' theorem

Result

The probability that the new product was introduced by the second group is $\frac29$ ⁠. (Textbook answer: $\frac29$ ⁠.)

Check it with numbers

Imagine the election held 100 times. The first group wins about 60 times and introduces the product in $0.7 \times 60 = 42$ of them; the second group wins about 40 times and introduces it in $0.3 \times 40 = 12$ ⁠. So the product appears $42 + 12 = 54$ times, 12 of them after a second-group win: $\frac{12}{54} \approx 0.2222$ ⁠, the same as $\frac29 \approx 0.2222$ ⁠.

Common mistakes

Answering 0.3. That is $P(A|E_2)$ ⁠, the chance of a new product if the second group wins. The question asks the reverse.
Answering 0.4. That is the second group's chance of winning before we know about the new product.
Expecting 0.7 and 0.3 to add to 1. They are conditional probabilities in two different cases; only $P(E_1) + P(E_2)$ must be 1.
Stopping at $\frac{6}{27}$ ⁠. It is correct but not in lowest terms; divide by 3 to get $\frac29$ ⁠.

What to take away

Read each sentence for its condition: "if the first group wins, … 0.7" is $P(A|E_1)$ ⁠, not $P(A)$ ⁠.
When the products share a denominator (here 50), Bayes' theorem reduces to comparing numerators: $\frac{6}{6 + 21}$ ⁠.
Book Example 15 uses the same two-case shape with a strike and a construction job (Examples 15 to 21); Question 8 is the same calculation with machines.

Every fact used, and where it comes from

Fact	Used in	Source
Partition of a sample space	Step 1	Section 13.5.1 (Lesson: Bayes' theorem)
Conditional probability $P(A\|E)$	Step 2	Section 13.2 (Lesson: conditional probability)
Theorem of total probability	Step 3	Section 13.5.2; used in Example 15 (Lesson: Bayes' theorem)
Bayes' theorem	Steps 4, 5	Section 13.5 (Examples 15 to 21)
Dividing by a fraction; lowest terms	Step 4	Arithmetic (Basics: dividing by a fraction)

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Exercise 13.3 · Q9: Board election: new product introduced by the second group, P=29P = \frac29P=92​

Basics you need

Solution

Name the events

The four probabilities as fractions

The chance of a new product

Second group, given the new product: Bayes' theorem

A quick sense check

Every fact used, and where it comes from

Exercise 13.3 · Q9: Board election: new product introduced by the second group, $P = \frac29$