Chapter 13 · Probability

Exercise 13.3 · Q6: Three coins: the two-headed coin, given heads, $P = \frac49$

Exercise 13.3Bayes' theorem · Section 13.5

The question

There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?

Exam answer

Let $E_1$ ⁠: the two headed coin is chosen, $E_2$ ⁠: the biased coin is chosen, $E_3$ ⁠: the unbiased coin is chosen, and $A$ ⁠: the coin shows heads.

$E_1$ ⁠, $E_2$ and $E_3$ form a partition of the sample space (exactly one coin is chosen).

$P(E_1) = P(E_2) = P(E_3) = \frac13$

$P(A|E_1) = 1$ ⁠, $P(A|E_2) = 75\% = \frac34$ and $P(A|E_3) = \frac12$

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) \\ &\quad + P(E_3)P(A|E_3) \\ &= \frac13\cdot 1 + \frac13\cdot\frac34 + \frac13\cdot\frac12 \\ &= \frac{4}{12} + \frac{3}{12} + \frac{2}{12} \\ &= \frac{9}{12} = \frac34 \end{aligned}

By Bayes' theorem,

\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac13 \div \frac34 \\ &= \frac13 \times \frac43 \end{aligned}

$\therefore$ The probability that it was the two headed coin is $\dfrac49$ ⁠.

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

The one idea this problem is built on

Heads can come from any of the three coins, but not equally easily: the two-headed coin always gives heads, the biased coin three times in four, the fair coin half the time. Bayes' theorem divides the two-headed coin's route to heads by all three routes together. With three causes, the partition and the sum simply have three terms.

Given, to find, and the plan

Given: Three coins: two-headed (heads on both faces), biased (heads 75% of the time), unbiased. One is chosen at random and tossed; it shows heads.
To find: The probability that the coin was the two-headed one: $P(E_1|A)$ ⁠.
Plan: (1) Name the three causes and the result; check the partition. (2) List $P(E_i)$ and $P(A|E_i)$ as fractions. (3) Total probability with three terms. (4) Bayes' theorem.

Basics you need

1Partition into three events (Section 13.5.1)

$E_1, E_2, E_3$ form a partition when no two can happen together, one of them must happen, and each has nonzero probability. Lesson: Bayes' theorem

2Reading each coin as a conditional probability

"Comes up heads 75% of the time" describes that coin only: $P(\text{heads} \mid \text{biased}) = \frac34$ ⁠. A coin with heads on both faces gives heads for certain. Lesson: conditional probability

3Total probability and Bayes' theorem with $n$ causes

$P(A) = \sum_{j} P(E_j)P(A|E_j)$ and $P(E_i|A) = \frac{P(E_i)P(A|E_i)}{P(A)}$ ⁠; here $n = 3$ ⁠. Basics: tree diagrams

4Adding and dividing fractions

Use the common denominator 12 for thirds, quarters and sixths; dividing by $\frac34$ is multiplying by $\frac43$ ⁠. Basics: dividing by a fraction

Solution

Name the events

Let
- $E_1$ ⁠: the two-headed coin is chosen;
- $E_2$ ⁠: the biased coin is chosen;
- $E_3$ ⁠: the unbiased coin is chosen;
- $A$ ⁠: the coin tossed shows heads.
$E_1$ ⁠, $E_2$ and $E_3$ form a partition of the sample space: exactly one coin is chosen, and each has a nonzero chance.

The three coins are the possible causes and "heads" is the observed result. One event per coin keeps the bookkeeping clear.

Partition: pairwise disjoint, exhaustive, nonzero probabilities. Lesson: Bayes' theorem
The six probabilities
$P(E_1) = P(E_2) = P(E_3) = \frac13,$ $\begin{aligned} P(A|E_1) &= 1, \\ P(A|E_2) &= 75\% = \frac{75}{100} = \frac34, \\ P(A|E_3) &= \frac12. \end{aligned}$
"Chosen at random" among three coins gives each coin $\frac13$ ⁠. The two-headed coin cannot show tails, so heads is certain. The biased coin's "75% of the time" is its chance of heads. An unbiased (fair) coin shows heads with probability $\frac12$ ⁠.

A certain event has probability 1; $x\% = \frac{x}{100}$ ⁠. Basics: rules of probability
The chance of heads

Since $E_1$ ⁠, $E_2$ ⁠, $E_3$ form a partition, by the theorem of total probability
$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &= \frac13\cdot 1 + \frac13\cdot\frac34 + \frac13\cdot\frac12 \\ &= \frac13 + \frac14 + \frac16 \\ &= \frac{4}{12} + \frac{3}{12} + \frac{2}{12} = \frac{9}{12} = \frac34. \end{aligned}$

The tree for Question 6, with three branches at the first stage. The rose path (two-headed coin, then heads) has product $\frac13$ ⁠; the green heads paths have $\frac14$ and $\frac16$ ⁠. All three heads paths add to $\frac34$ ⁠.

Heads arrives through exactly one of the three coins, so the three path probabilities add. Thirds, quarters and sixths all go into twelfths.

Multiply along a path, add the paths. Basics: tree diagrams
The two-headed coin, given heads: Bayes' theorem

By Bayes' theorem,
$\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac13 \div \frac34 \\ &= \frac13 \times \frac43 = \frac49. \end{aligned}$
The top is the two-headed coin's path, $\frac13 \times 1 = \frac13$ ⁠; the bottom is all heads paths, $\frac34$ ⁠. The ratio is the share of heads results that come from the two-headed coin.

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

The other two coins: $P(E_2|A) = \frac14 \div \frac34 = \frac13$ and $P(E_3|A) = \frac16 \div \frac34 = \frac29$ ⁠. Then $\frac49 + \frac13 + \frac29 = \frac{4 + 3 + 2}{9} = 1$ ⁠.

Seeing heads raised the two-headed coin from $\frac13$ to $\frac49$ ⁠, lowered the fair coin from $\frac13$ to $\frac29$ ⁠, and left the biased coin at $\frac13$ ⁠, because its chance of heads, $\frac34$ ⁠, equals the overall chance of heads.

Given heads, one of the three coins was certainly used, so the three updated probabilities must add to 1. Each coin's probability is multiplied by $\frac{P(A|E_i)}{P(A)}$ ⁠: more than 1 for the two-headed coin, exactly 1 for the biased coin, less than 1 for the fair coin.

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

Result

The probability that the coin was the two-headed coin is $\frac49$ ⁠. (Textbook answer: $\frac49$ ⁠.)

Check it with numbers

Picture the experiment repeated 1200 times: each coin is chosen about 400 times. Heads comes 400 times from the two-headed coin, $\frac34 \times 400 = 300$ times from the biased coin and $\frac12 \times 400 = 200$ times from the fair coin: $400 + 300 + 200 = 900$ heads. The share from the two-headed coin is $\frac{400}{900} \approx 0.4444$ ⁠, the same as $\frac49 \approx 0.4444$ ⁠.

Common mistakes

Answering 1. That is $P(A|E_1)$ ⁠: the two-headed coin always shows heads. But heads does not prove it was that coin; the other two coins show heads too.
Answering $\frac13$ ⁠. That is the chance before the toss. Seeing heads changes it.
Leaving out a coin in $P(A)$ ⁠. With three causes, the total has three terms.
Treating 75% as $P(E_2 \cap A)$ ⁠. It is the biased coin's own chance of heads; its path is $\frac13 \times \frac34 = \frac14$ ⁠.

What to take away

Bayes' theorem with any number of causes: the path for your cause, divided by the sum of all the paths to the result.
A cause rises after the evidence exactly when it makes the evidence more likely than average, $P(A|E_i) > P(A)$ ⁠.
Book Example 17 (three boxes of coins, Examples 15 to 21) has the same three-way structure; so does Question 7, with unequal first-stage probabilities.

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)
Equally likely choices; a certain event has probability 1	Step 2	Class XI (Basics: rules of probability)
Conditional probability $P(A\|E)$	Step 2	Section 13.2 (Lesson: conditional probability)
Theorem of total probability (three events)	Step 3	Section 13.5.2 (Lesson: Bayes' theorem)
Bayes' theorem	Steps 4, 5	Section 13.5; Example 17 has three boxes (Examples 15 to 21)
Fractions with a common denominator; dividing by a fraction	Steps 3, 4	Arithmetic (Basics: dividing by a fraction)

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Exercise 13.3 · Q6: Three coins: the two-headed coin, given heads, P=49P = \frac49P=94​

Basics you need

Solution

Name the events

The six probabilities

The chance of heads

The two-headed coin, given heads: Bayes' theorem

A quick sense check

Every fact used, and where it comes from

Exercise 13.3 · Q6: Three coins: the two-headed coin, given heads, $P = \frac49$