Chapter 13 · Probability

Example 7: Coin, then coin or die: die above 4, given at least one tail

ExamplesConditional probability · Section 13.2

The question

Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’.

Exam answer

The sample space is

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

Multiplying the probabilities of the two stages (fair coin, then fair coin or fair die),

\begin{aligned} &P(\{(H,H)\}) = P(\{(H,T)\}) \\ &= \frac{1}{2}\times\frac{1}{2} = \frac{1}{4} \end{aligned}

and for each of the six outcomes $(T,i)$ ⁠,

P(\{(T,i)\}) = \frac{1}{2}\times\frac{1}{6} = \frac{1}{12}

Let $F$ ⁠: there is at least one tail, and $E$ ⁠: the die shows a number greater than 4.

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

E = \{(T,5), (T,6)\}

E\cap F = \{(T,5), (T,6)\}

\begin{aligned} P(F) &= \frac{1}{4} + 6\times\frac{1}{12} \\ &= \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \end{aligned}

\begin{aligned} P(E\cap F) &= \frac{1}{12} + \frac{1}{12} \\ &= \frac{1}{6} \end{aligned}

By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \dfrac{\;\frac{1}{6}\;}{\frac{3}{4}} = \frac{1}{6}\times\frac{4}{3} \end{aligned}

$\therefore\ P(E|F) = \dfrac{2}{9}$ ⁠.

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

The one idea this problem is built on

When the outcomes are not equally likely, add their probabilities instead of counting them. The definition

P(E|F) = \frac{P(E \cap F)}{P(F)}

still works; only $P(F)$ and $P(E \cap F)$ must be found by adding the probabilities of their outcomes, which a tree diagram supplies.

Given, to find, and the plan

Given: A fair coin is tossed. After a head the coin is tossed again; after a tail a fair die is thrown.
To find: $P(E|F)$ ⁠, where $E$ ⁠: the die shows a number greater than 4, and $F$ ⁠: there is at least one tail.
Plan: Draw the tree and list the 8 outcomes with their probabilities; list $F$ ⁠, $E$ and $E \cap F$ ⁠; add probabilities to get $P(F)$ and $P(E \cap F)$ ⁠; divide.

Basics you need

1Tree diagrams

Branch at each stage, multiply the probabilities along a path to get the probability of that outcome, and add paths to get the probability of an event. Basics: tree diagrams

2Probability of an event

The probability of an event is the sum of the probabilities of the outcomes in it (Class XI). Basics: adding probabilities

3Conditional probability

$P(E|F) = \frac{P(E \cap F)}{P(F)}$ when $P(F) \ne 0$ ⁠; the book notes that it applies even when outcomes are not equally likely (Section 13.2). Lesson: conditional probability

Solution

Draw the tree and list the outcomes
$\begin{aligned} S = \{&(H,H), (H,T), (T,1), (T,2), \\ &(T,3), (T,4), (T,5), (T,6)\}. \end{aligned}$
Here $(H,H)$ means both tosses show head, and $(T,i)$ means the first toss shows tail and the die shows $i$ ⁠.

The tree (the book's Fig 13.2). Each head path has probability $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ ⁠; each tail path has $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ ⁠. Every path except $(H,H)$ is in $F$ ⁠; the paths $(T,5)$ and $(T,6)$ are in $E \cap F$ ⁠.

The second stage depends on the first: a head leads to a coin, a tail leads to a die. That is why a head splits into 2 branches but a tail splits into 6, and the outcomes get different probabilities.

Basics: tree diagrams
Find the probability of each outcome
$\begin{aligned} P(\{(H,H)\}) &= P(\{(H,T)\}) \\ &= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}, \end{aligned}$ $P(\{(T,i)\}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \ \text{ for each } i.$
They add up to $\frac{1}{4} + \frac{1}{4} + 6 \times \frac{1}{12} = 1$ ⁠, as they must.

Multiply along each path: the first branch has probability $\frac{1}{2}$ (fair coin); the second has $\frac{1}{2}$ for a coin or $\frac{1}{6}$ for a die. So the eight outcomes are not equally likely: each $(H,\cdot)$ is three times as likely as each $(T,\cdot)$ ⁠.

Basics: multiply along a path
List the events
$\begin{aligned} F = \{&(H,T), (T,1), (T,2), (T,3), \\ &(T,4), (T,5), (T,6)\}, \end{aligned}$ $E = \{(T,5), (T,6)\},$ $E \cap F = \{(T,5), (T,6)\}.$
"At least one tail" includes the tail on the second toss, $(H,T)$ ⁠, and every outcome that starts with a tail; only $(H,H)$ is left out. The die is thrown only after a tail, so $E$ contains just $(T,5)$ and $(T,6)$ ⁠, and both of these have a tail, so $E \cap F = E$ ⁠.

Basics: events as sets
Add probabilities to find $P(F)$ and $P(E \cap F)$
$\begin{aligned} P(F) &= \frac{1}{4} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} \\ &\quad + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} \\ &= \frac{3}{4}, \end{aligned}$ $\begin{aligned} P(E \cap F) &= P(\{(T,5)\}) \\ &\quad + P(\{(T,6)\}) \\ &= \frac{1}{12} + \frac{1}{12} = \frac{1}{6}. \end{aligned}$
The first term of $P(F)$ is $P(\{(H,T)\}) = \frac{1}{4}$ ⁠; the six terms $\frac{1}{12}$ are the outcomes $(T,1)$ to $(T,6)$ ⁠. Adding, $\frac{3}{12} + \frac{6}{12} = \frac{9}{12} = \frac{3}{4}$ ⁠.

Basics: adding probabilities
Divide, using the definition
$\begin{aligned} P(E|F) &= \frac{P(E \cap F)}{P(F)} = \dfrac{\;\frac{1}{6}\;}{\frac{3}{4}} \\ &= \frac{1}{6} \times \frac{4}{3} = \frac{2}{9}. \end{aligned}$
$P(F) = \frac{3}{4} \ne 0$ ⁠, so the definition applies. Dividing by $\frac{3}{4}$ is multiplying by $\frac{4}{3}$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Result

$P(E|F) = \dfrac{2}{9}$ ⁠.

(Textbook answer: $\frac{2}{9}$ ⁠.)

Check it with numbers

$P(F)$ by the complement. $F$ misses only $(H,H)$ ⁠, so $P(F) = 1 - \frac{1}{4} = \frac{3}{4}$ ⁠, as found.

A second route. Given $F$ ⁠, the first toss was a tail with probability $\frac{1/2}{3/4} = \frac{2}{3}$ (the tail paths carry $\frac{1}{2}$ of the $\frac{3}{4}$ ⁠). After a tail the die shows 5 or 6 with probability $\frac{2}{6}$ ⁠. So $P(E|F) = \frac{2}{3} \times \frac{2}{6} = \frac{2}{9}$ ⁠, the same answer.

Decimals. $\frac{2}{9} \approx 0.2222$ ⁠. Counting outcomes as if they were equally likely would give $\frac{2}{7} \approx 0.2857$ ⁠, which is wrong because $(H,T)$ is three times as likely as each $(T,i)$ ⁠.

Common mistakes

Counting outcomes. $F$ has 7 outcomes and $E \cap F$ has 2, but $\frac{2}{7}$ is wrong: the formula $\frac{n(E \cap F)}{n(F)}$ needs equally likely outcomes.
Giving each of the 8 outcomes probability $\frac{1}{8}$ ⁠. The tree shows $\frac{1}{4}$ for the head paths and $\frac{1}{12}$ for the tail paths.
Leaving $(H,T)$ out of $F$ ⁠. A tail on the second toss is still a tail.

What to take away

$P(E|F) = \frac{P(E \cap F)}{P(F)}$ works for any outcomes; counting works only for equally likely ones.
A tree diagram gives each outcome's probability: multiply along the path.
Check that the outcome probabilities add up to 1.

Every fact used, and where it comes from

Fact	Used in	Source
Tree diagram: multiply along a path	Steps 1, 2	Example 7, Figs 13.1 and 13.2 (Basics: tree diagrams)
Probability of an event = sum over its outcomes	Step 4	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 5	Definition 1 and the remark before Example 7, Section 13.2 (Lesson: conditional probability)
$P(F) = 1 - P(F')$	Check	Class XI, Probability (complement)

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Example 7: Coin, then coin or die: die above 4, given at least one tail

Basics you need

Solution

Draw the tree and list the outcomes

Find the probability of each outcome

List the events

Add probabilities to find P(F)P(F)P(F) and P(E∩F)P(E \cap F)P(E∩F)

Divide, using the definition

Every fact used, and where it comes from

Add probabilities to find $P(F)$ and $P(E \cap F)$