Chapter 13 · Probability

Exercise 13.1 · Q15: Die, then die or coin: a tail, given at least one die shows 3

Exercise 13.1Conditional probability · Section 13.2

The question

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Exam answer

Write an outcome as $(x, y)$ ⁠: $x$ is the first throw, $y$ the second throw or the coin. Multiplying along the branches of the tree:

for $x = 1, 2, 4, 5$ ⁠: $(x, H)$ and $(x, T)$ ⁠, each with probability $\frac{1}{6}\times\frac{1}{2} = \frac{1}{12}$ ⁠;

for $x = 3, 6$ ⁠: $(x, 1), \ldots, (x, 6)$ ⁠, each with probability $\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}$ ⁠.

Let $E$ ⁠: the coin shows a tail, and $F$ ⁠: at least one die shows a 3.

\begin{aligned} F = \{&(3, 1), (3, 2), (3, 3), \\ &(3, 4), (3, 5), (3, 6), \\ &(6, 3)\} \end{aligned}

P(F) = 7\times\frac{1}{36} = \frac{7}{36}

A tail needs $x = 1, 2, 4$ or $5$ ⁠, and then no die shows 3. So $E\cap F = \phi$ and $P(E\cap F) = 0$ ⁠.

By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= 0\div\frac{7}{36} \end{aligned}

$\therefore\ P(E|F) = 0$ ⁠.

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

The one idea this problem is built on

The coin is tossed only when the first throw is 1, 2, 4 or 5, and in that case the die is thrown just once and does not show 3. So "a tail" and "at least one 3" can never happen together: $E\cap F = \phi$ ⁠. Since $F$ itself is possible, $P(E|F) = \frac{0}{P(F)} = 0$ ⁠. The outcomes are not equally likely, so we work with their probabilities, as the book does in Example 7.

Given, to find, and the plan

Given: A fair die is thrown. If it shows a multiple of 3 (that is, 3 or 6), the die is thrown again; otherwise (1, 2, 4 or 5) a fair coin is tossed.
To find: $P(E|F)$ ⁠, where $E$ ⁠: the coin shows a tail, and $F$ ⁠: at least one die shows a 3.
Plan: (1) List the outcomes with a tree and find their probabilities. (2) Write $F$ and find $P(F)$ ⁠. (3) Show $E\cap F = \phi$ ⁠. (4) Divide.

Basics you need

1Two-stage experiments and trees

Branch for the first throw, then for what follows it; multiply along a path to get the probability of that outcome. Basics: branch, multiply along a path

2Probability of an event

When outcomes are not equally likely, $P(A)$ is the sum of the probabilities of the outcomes in $A$ ⁠. Basics: adding probabilities

3Conditional probability

$P(E|F) = \dfrac{P(E\cap F)}{P(F)}$ ⁠, provided $P(F) \ne 0$ (Definition 1, Section 13.2); it applies to outcomes that are not equally likely too. The book's Example 7 has the same structure with a coin first. Lesson: conditional probability Example 7

Solution

List the outcomes and their probabilities

Write $(x, y)$ ⁠: $x$ is the first throw and $y$ is the result of the second stage. After $x = 1, 2, 4, 5$ the second stage is the coin; after $x = 3, 6$ it is the die. Multiplying along the branches:
$\begin{aligned} &P(x, H) = P(x, T) \\ &= \frac{1}{6}\times\frac{1}{2} = \frac{1}{12} \\ &\quad (x = 1, 2, 4, 5), \end{aligned}$ $\begin{aligned} &P(x, y) = \frac{1}{6}\times\frac{1}{6} = \frac{1}{36} \\ &\quad (x = 3, 6;\ y = 1, \ldots, 6). \end{aligned}$
There are $4\times 2 = 8$ coin outcomes and $2\times 6 = 12$ die outcomes, 20 in all. They are not equally likely, but their probabilities add up to 1: $8\times\frac{1}{12} + 12\times\frac{1}{36} = \frac23 + \frac13 = 1$ ⁠.

Multiply along a path of the tree. Basics: branch, multiply along a path

The 20 outcomes, grouped by the first throw, with the probability of each. Gold: at least one die shows 3, that is $F$ ⁠. Violet: a tail, that is $E$ ⁠. No outcome is both, so $E\cap F = \phi$ ⁠.
Write $F$ and find $P(F)$
$\begin{aligned} F = \{&(3, 1), (3, 2), (3, 3), (3, 4), \\ &(3, 5), (3, 6), (6, 3)\}, \end{aligned}$ $P(F) = 7\times\frac{1}{36} = \frac{7}{36}.$
A 3 can appear on the first throw, giving the 6 outcomes $(3, y)$ ⁠, or on the second throw. A second-throw 3 after a first 3 is $(3, 3)$ ⁠, already listed; after a 6 it is the new outcome $(6, 3)$ ⁠. After 1, 2, 4 or 5 the die is not thrown again, so no 3 can appear. Each of the 7 outcomes has probability $\frac{1}{36}$ ⁠.

Add the probabilities of the outcomes in the event. Basics: adding probabilities
Show that $E$ and $F$ cannot happen together
$E = \{(1, T), (2, T), (4, T), (5, T)\},$ $E\cap F = \phi,\qquad P(E\cap F) = 0.$
A tail needs a coin, and a coin is tossed only after a first throw of 1, 2, 4 or 5. In those outcomes the only die result is that first throw, which is not 3. So no outcome lies in both $E$ and $F$ ⁠.

Disjoint events have an empty intersection. Basics: events as sets
Divide
$P(E|F) = \frac{P(E\cap F)}{P(F)} = \frac{0}{7/36} = 0.$
$P(F) = \frac{7}{36} \ne 0$ ⁠, so the definition applies, and 0 divided by a non-zero number is 0. In words: once we know a die showed 3, the second stage was a die throw, so no coin was tossed and a tail is impossible.

Definition 1, Section 13.2. Lesson: conditional probability

Result

The conditional probability that the coin shows a tail, given that at least one die shows a 3, is $P(E|F) = 0$ ⁠.

(Textbook answer: 0.)

Check it with numbers

Unconditionally, a tail is quite likely: $P(E) = 4\times\frac{1}{12} = \frac13 \approx 0.3333$ ⁠. The given event has probability $P(F) = \frac{7}{36} \approx 0.1944$ ⁠, which is not 0, so the conditional probability is defined; and inside $F$ there is no tail at all, so the answer 0 is exact, not a rounded small number.

Common mistakes

Treating the 20 outcomes as equally likely. A coin outcome such as $(1, T)$ has probability $\frac{1}{12}$ ⁠, three times that of a die outcome such as $(3, 4)$ ⁠. Here the answer is 0 either way, but in general counting outcomes would give wrong probabilities.
Forgetting $(6, 3)$ ⁠. After a 6 the die is thrown again, and a 3 on that throw also makes "at least one die shows a 3" true.
Thinking a tail could still follow a 3. After a 3 (a multiple of 3), the rule says throw the die again, not toss the coin.

What to take away

For a two-stage experiment, draw the tree; multiply along the branches to get each outcome's probability.
If $E\cap F = \phi$ and $P(F) \ne 0$ ⁠, then $P(E|F) = 0$ ⁠: knowing $F$ happened rules $E$ out.

Every fact used, and where it comes from

Fact	Used in	Source
Probabilities along a tree multiply	Step 1	Section 13.2, Example 7 and Fig 13.2 (Basics: branch, multiply along a path)
$P(A)$ is the sum of the probabilities of its outcomes	Steps 1, 2	Class XI, Probability (Basics: adding probabilities)
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$ ⁠, also for outcomes that are not equally likely	Step 4	Section 13.2, Definition 1 and the paragraph before Example 7 (Lesson: conditional probability)

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Exercise 13.1 · Q15: Die, then die or coin: a tail, given at least one die shows 3

Basics you need

Solution

List the outcomes and their probabilities

Write FFF and find P(F)P(F)P(F)

Show that EEE and FFF cannot happen together

Divide

Every fact used, and where it comes from

Write $F$ and find $P(F)$

Show that $E$ and $F$ cannot happen together