Chapter 13 · Probability

Exercise 13.3 · Q10: Die, then coin: threw 1 to 4, given exactly one head, $P = \frac{8}{11}$

Exercise 13.3Bayes' theorem · Section 13.5

The question

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Exam answer

Let $E_1$ ⁠: she gets 5 or 6, $E_2$ ⁠: she gets 1, 2, 3 or 4, and $A$ ⁠: she obtains exactly one head.

$E_1$ and $E_2$ form a partition of the sample space (the die shows exactly one number from 1 to 6).

$P(E_1) = \frac26 = \frac13$ and $P(E_2) = \frac46 = \frac23$

If $E_1$ occurs, the coin is tossed three times. Exactly one head occurs in HTT, THT, TTH out of 8 equally likely outcomes, so $P(A|E_1) = \frac38$ ⁠.

If $E_2$ occurs, the coin is tossed once, so $P(A|E_2) = \frac12$ ⁠.

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) \\ &= \frac13\cdot\frac38 + \frac23\cdot\frac12 \\ &= \frac18 + \frac13 \\ &= \frac{3}{24} + \frac{8}{24} = \frac{11}{24} \end{aligned}

By Bayes' theorem,

\begin{aligned} P(E_2|A) &= \frac{P(E_2)P(A|E_2)}{P(A)} \\ &= \frac{8}{24} \div \frac{11}{24} \\ &= \frac{8}{24} \times \frac{24}{11} \end{aligned}

$\therefore$ The probability that she threw 1, 2, 3 or 4 with the die is $\dfrac{8}{11}$ ⁠.

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

The one idea this problem is built on

"Exactly one head" can happen after either throw of the die, but it is more likely after a single toss ( $\frac12$ ⁠) than after three tosses ( $\frac38$ ⁠), and the single toss also follows the more likely throw. Split by what the die showed, find the chance of exactly one head in each case, and use Bayes' theorem to turn the question round.

Given, to find, and the plan

Given: A fair die is thrown. On 5 or 6 a coin is tossed three times; on 1, 2, 3 or 4 a coin is tossed once. Exactly one head was obtained.
To find: The probability that the die showed 1, 2, 3 or 4: $P(E_2|A)$ ⁠.
Plan: (1) Name the events; check the partition. (2) $P(E_1)$ ⁠, $P(E_2)$ from the die. (3) Count outcomes for $P(A|E_1)$ ⁠; $P(A|E_2)$ from one toss. (4) Total probability, then Bayes' theorem.

Basics you need

1Equally likely outcomes

A fair die gives each of 1 to 6 with probability $\frac16$ ⁠; three tosses of a fair coin give $2 \times 2 \times 2 = 8$ equally likely outcomes. Then $P(E) = \frac{n(E)}{n(S)}$ ⁠. Basics: classical probability

2Listing outcomes

Each toss has 2 results, so three tosses have $2^3 = 8$ outcomes, from HHH to TTT. Basics: multiply the choices

3Conditional probability (Section 13.2)

$P(A|E_1)$ is the chance of exactly one head when we know the coin is tossed three times. Lesson: conditional probability

4Total 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

Solution

Name the events

Let
- $E_1$ ⁠: the girl gets 5 or 6 on the die;
- $E_2$ ⁠: the girl gets 1, 2, 3 or 4 on the die;
- $A$ ⁠: she obtains exactly one head.
$E_1$ and $E_2$ form a partition of the sample space: the die shows exactly one number, which is either in $\{5, 6\}$ or in $\{1, 2, 3, 4\}$ ⁠, and both events can happen.

The die decides how many times the coin is tossed, so it is the cause; "exactly one head" is the result we observe. For one toss, "exactly one head" simply means the toss is a head.

Partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: Bayes' theorem
The chances for the die
$\begin{aligned} P(E_1) &= \frac26 = \frac13, \\ P(E_2) &= \frac46 = \frac23. \end{aligned}$
The die is fair, so each face has probability $\frac16$ ⁠; $E_1$ contains 2 faces and $E_2$ contains 4.

$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: classical probability
Exactly one head in each case

If $E_1$ occurs, the coin is tossed three times. The 8 equally likely outcomes are

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT,

and exactly one head appears in HTT, THT and TTH. So $P(A|E_1) = \frac38$ ⁠.

If $E_2$ occurs, the coin is tossed once, and exactly one head means a head: $P(A|E_2) = \frac12$ ⁠.

Knowing the case tells us which experiment the coin goes through. Three tosses can give 0, 1, 2 or 3 heads, and only 3 of the 8 outcomes have exactly one.

Count the outcomes in the event and divide by all of them. Basics: multiply the choices
The chance of exactly one head

Since $E_1$ and $E_2$ form a partition, by the theorem of total probability
$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac13\cdot\frac38 + \frac23\cdot\frac12 \\ &= \frac18 + \frac13 \\ &= \frac{3}{24} + \frac{8}{24} = \frac{11}{24}. \end{aligned}$

The tree for Question 10. The rose path (1 to 4, then a head on the single toss) has product $\frac13 = \frac{8}{24}$ ⁠; the green path (5 or 6, then exactly one head in three tosses) has $\frac18 = \frac{3}{24}$ ⁠. Together, $P(A) = \frac{11}{24}$ ⁠.

Exactly one head comes after exactly one kind of throw, so the two path probabilities add. The common denominator 24 makes the next step a comparison of 8 with 11.

Multiply along a path, add the paths. Basics: tree diagrams
Threw 1 to 4, given exactly one head: Bayes' theorem

Here $P(E_2)P(A|E_2) = \frac23\cdot\frac12 = \frac13 = \frac{8}{24}$ ⁠. By Bayes' theorem,
$\begin{aligned} P(E_2|A) &= \frac{P(E_2)P(A|E_2)}{P(A)} \\ &= \frac{8}{24} \div \frac{11}{24} \\ &= \frac{8}{24} \times \frac{24}{11} = \frac{8}{11}. \end{aligned}$
The top is the "1 to 4" path to exactly one head; the bottom is both paths. Of every 11 parts of the probability of exactly one head, 8 come from the single toss.

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

For the other case, $P(E_1|A) = \frac{3}{24} \div \frac{11}{24} = \frac{3}{11}$ ⁠, and $\frac{8}{11} + \frac{3}{11} = 1$ ⁠.

Before the coin, the chance of 1 to 4 was $\frac23 \approx 0.667$ ⁠; after seeing exactly one head it is $\frac{8}{11} \approx 0.727$ ⁠. It goes up because exactly one head is more likely after a single toss ( $\frac12$ ⁠) than after three ( $\frac38$ ⁠).

Given exactly one head, the die showed either 5 or 6 or else 1 to 4, 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 she threw 1, 2, 3 or 4 with the die, given exactly one head, is $\frac{8}{11}$ ⁠. (Textbook answer: $\frac{8}{11}$ ⁠.)

Check it with numbers

Picture 48 rounds of the game. About 16 start with 5 or 6, and exactly one head comes in $\frac38 \times 16 = 6$ of them; about 32 start with 1 to 4, and a head comes in $\frac12 \times 32 = 16$ of them. So exactly one head appears $6 + 16 = 22$ times, 16 of them after 1 to 4: $\frac{16}{22} \approx 0.7273$ ⁠, the same as $\frac{8}{11} \approx 0.7273$ ⁠.

Common mistakes

Taking $\frac13$ or $\frac12$ for $P(A|E_1)$ ⁠. Three tosses have 8 outcomes, and exactly one head occurs in 3 of them: $\frac38$ ⁠.
Answering $\frac23$ ⁠. That is $P(E_2)$ ⁠, the chance before the head is seen.
Counting "at least one head" instead of "exactly one". HHT, for example, has two heads and does not count.
Adding $\frac38 + \frac12$ ⁠. Each conditional probability must first be multiplied by the chance of its case.

What to take away

When the cause decides which experiment happens next, find the result's probability separately inside each experiment.
Small sample spaces, such as three coin tosses, are easiest to list in full and count.
Book Example 17 (three boxes of coins) also mixes a choice with a second experiment (Examples 15 to 21); Question 6 is a coin version with three causes.

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)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Steps 2, 3	Classical probability, Class XI (Basics: classical probability)
Three coin tosses have 8 outcomes	Step 3	Section 13.2 lists them (Basics: multiply the choices)
Conditional probability $P(A\|E)$	Step 3	Section 13.2 (Lesson: conditional probability)
Theorem of total probability	Step 4	Section 13.5.2 (Lesson: Bayes' theorem)
Bayes' theorem	Steps 5, 6	Section 13.5 (Examples 15 to 21)

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Exercise 13.3 · Q10: Die, then coin: threw 1 to 4, given exactly one head, P=811P = \frac{8}{11}P=118​

Basics you need

Solution

Name the events

The chances for the die

Exactly one head in each case

The chance of exactly one head

Threw 1 to 4, given exactly one head: Bayes' theorem

A quick sense check

Every fact used, and where it comes from

Exercise 13.3 · Q10: Die, then coin: threw 1 to 4, given exactly one head, $P = \frac{8}{11}$