Chapter 13 · Probability

Exercise 13.1 · Q9: Family picture: son on an end, given father in the middle

Exercise 13.1Conditional probability · Section 13.2

The question

Determine $P(E|F)$ ⁠.

Mother, father and son line up at random for a family picture

$E$ : son on one end, $F$ : father in middle

Exam answer

Let m, f, s denote the mother, father and son, written from left to right. The $3! = 6$ line-ups are equally likely:

\begin{aligned} S = \{&mfs, msf, fms, \\ &fsm, smf, sfm\} \end{aligned}

E = \{mfs, fms, smf, sfm\}

F = \{mfs, sfm\}

so $E\cap F = \{mfs, sfm\}$ ⁠.

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{2}{6}\div\frac{2}{6} \end{aligned}

$\therefore\ P(E|F) = 1$ ⁠.

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

The one idea this problem is built on

There are only three places. If the father takes the middle one, the other two places are the two ends, so the son must stand on an end. Knowing $F$ makes $E$ certain, and a conditional probability of a certain event is 1. Listing the 6 line-ups shows this directly.

Given, to find, and the plan

Given: Mother (m), father (f) and son (s) stand in a row of three places at random, so all line-ups are equally likely. $E$ ⁠: the son is in the first or the third place. $F$ ⁠: the father is in the second place.
To find: $P(E|F)$ ⁠.
Plan: (1) List the line-ups. (2) Write $E$ ⁠, $F$ and $E\cap F$ as sets. (3) Find $P(F)$ ⁠, $P(E\cap F)$ and divide.

Basics you need

1Arrangements in a row

Three different people can stand in a row in $3\times 2\times 1 = 3! = 6$ ways. Basics: multiply the choices

2Classical probability

"At random" means all 6 line-ups are equally likely, so $P(E) = \frac{n(E)}{6}$ ⁠. Basics: P(E) = n(E)/n(S)

3Conditional probability

$P(E|F) = \dfrac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$ (Definition 1, Section 13.2). Lesson: conditional probability

Solution

List the line-ups

Writing the people from left to right,
$\begin{aligned} S = \{&mfs, msf, fms, \\ &fsm, smf, sfm\}, \end{aligned}$ $n(S) = 6.$
The left place can be filled in 3 ways, the middle in 2 (one person is used), the right in 1, so $3\times 2\times 1 = 6$ ⁠. Listing them in order (first by who stands on the left) makes sure none is missed or repeated. Lower-case letters keep the father f apart from the event $F$ ⁠.

Fundamental principle of counting (Class XI). Basics: multiply the choices

The 6 line-ups. The father is in the middle in exactly two of them ( $F$ ⁠, filled), and in both the son is on an end, so both are in $E$ ⁠. Two more line-ups have the son on an end with the father on an end too (in $E$ only).
Write $E$ ⁠, $F$ and $E\cap F$
$\begin{aligned} E &= \{mfs, fms, smf, sfm\}, \\ F &= \{mfs, sfm\}, \\ E\cap F &= \{mfs, sfm\} = F. \end{aligned}$
$E$ ⁠: s is first ( $smf$ ⁠, $sfm$ ⁠) or last ( $mfs$ ⁠, $fms$ ⁠). $F$ ⁠: f is in the middle ( $mfs$ ⁠, $sfm$ ⁠). Every line-up in $F$ is also in $E$ ⁠, so $F \subset E$ and the intersection is all of $F$ ⁠.

If $F \subset E$ ⁠, then $E\cap F = F$ ⁠. Basics: events as sets
Divide
$P(F) = \frac{2}{6},\qquad P(E\cap F) = \frac{2}{6},$ $P(E|F) = \frac{P(E\cap F)}{P(F)} = \frac{2/6}{2/6} = 1.$
$P(F) = \frac26 \ne 0$ ⁠, so the definition applies. Numerator and denominator are the same because $E\cap F = F$ ⁠. In words: given that the father is in the middle, the son is certainly on an end.

Definition 1, Section 13.2; compare $P(F|F) = 1$ ⁠, Property 1, Section 13.2.1. Lesson: conditional probability

Result

$P(E|F) = 1$ ⁠.

(Textbook answer: 1.)

Check it with numbers

Without the condition, $P(E) = \frac46 = \frac23 \approx 0.6667$ ⁠: the son is on an end in 4 of the 6 line-ups. Knowing that the father is in the middle raises this to 1, because the two line-ups with the son in the middle ( $msf$ ⁠, $fsm$ ⁠) both have the father on an end and are ruled out by $F$ ⁠.

Common mistakes

Answering $\frac23$ or $\frac13$ ⁠. $\frac23$ is $P(E)$ without the condition and $\frac13$ is $P(E\cap F)$ ⁠; the question asks for $E$ given $F$ ⁠.
Thinking "one end" means one particular end. "On one end" means on either end, the left or the right.
Thinking a probability of 1 must be a mistake. A conditional probability equals 1 whenever the given event forces the other one, as here.

What to take away

If $F \subset E$ and $P(F) \ne 0$ ⁠, then $P(E|F) = 1$ ⁠.
For small experiments, listing all outcomes is the safest way to see how two events overlap.

Every fact used, and where it comes from

Fact	Used in	Source
3 people in a row: $3! = 6$ arrangements	Step 1	Class XI, Permutations and Combinations (Basics: multiply the choices)
$P(E) = \frac{n(E)}{n(S)}$	Step 3, Check	Class XI, Probability (Basics: classical probability)
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Step 3	Definition 1, Section 13.2 (Lesson: conditional probability)
$P(F\|F) = 1$	Step 3	Property 1, Section 13.2.1

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 13.1 · Q9: Family picture: son on an end, given father in the middle

Basics you need

Solution

List the line-ups

Write EEE⁠, FFF and E∩FE\cap FE∩F

Divide

Every fact used, and where it comes from

Write $E$ ⁠, $F$ and $E\cap F$