Chapter 13 · Probability

Exercise 13.1 · Q6: Coin tossed three times: $P(E\mid F)$ in three cases

Exercise 13.1Conditional probability · Section 13.2

The question

Determine $P(E|F)$ ⁠.

A coin is tossed three times, where

(i) $E$ : head on third toss, $F$ : heads on first two tosses

(ii) $E$ : at least two heads, $F$ : at most two heads

(iii) $E$ : at most two tails, $F$ : at least one tail

Exam answer

The sample space has 8 equally likely outcomes:

\begin{aligned} S = \{&HHH, HHT, HTH, THH, \\ &HTT, THT, TTH, TTT\} \end{aligned}

(i) $F = \{HHH, HHT\}$ and $E\cap F = \{HHH\}$ ⁠.

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{1}{8}\div\frac{2}{8} \end{aligned}

$P(E|F) = \frac{1}{2}$ ⁠.

(ii)

E = \{HHH, HHT, HTH, THH\}

\begin{aligned} F = \{&HHT, HTH, THH, HTT, \\ &THT, TTH, TTT\} \end{aligned}

E\cap F = \{HHT, HTH, THH\}

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{3}{8}\div\frac{7}{8} \end{aligned}

$P(E|F) = \frac{3}{7}$ ⁠.

(iii) $E$ is every outcome except $TTT$ ⁠, and $F$ is every outcome except $HHH$ ⁠.

\begin{aligned} E\cap F = \{&HHT, HTH, THH, \\ &HTT, THT, TTH\} \end{aligned}

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{6}{8}\div\frac{7}{8} \end{aligned}

$\therefore\ P(E|F) = \dfrac{6}{7}$ ⁠.

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

The one idea this problem is built on

When all outcomes are equally likely, conditioning on $F$ means: throw away every outcome outside $F$ and count again. So $P(E|F)$ is the number of outcomes in both $E$ and $F$ divided by the number in $F$ ⁠. With only 8 outcomes, we can list everything and count.

Given, to find, and the plan

Given: A coin tossed three times. Writing the results in order, the sample space $S$ consists of the 8 outcomes HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, with $n(S) = 2\times 2\times 2 = 8$ ⁠. We take the coin to be fair, so each outcome has probability $\frac18$ ⁠. In each part, $E$ and $F$ are as stated.
To find: $P(E|F)$ in each of the three parts.
Plan: For each part: (1) list $F$ and $E\cap F$ from the 8 outcomes; (2) find $P(F)$ and $P(E\cap F)$ by counting; (3) divide.

Basics you need

1The sample space of three tosses

Each toss has 2 results, so three tosses give $2\times 2\times 2 = 8$ ordered outcomes. Basics: multiply the choices

2Classical probability

With equally likely outcomes, $P(E) = \frac{n(E)}{n(S)}$ ⁠. Basics: P(E) = n(E)/n(S)

3Conditional probability by counting

$P(E|F) = \dfrac{P(E\cap F)}{P(F)} = \dfrac{n(E\cap F)}{n(F)}$ for equally likely outcomes (Section 13.2, Definition 1). Lesson: conditional probability

4"At least" and "at most"

"At least two heads" means 2 or 3 heads; "at most two heads" means 0, 1 or 2 heads. Basics: events as sets

Part (i): head on the third toss, given heads on the first two

List $F$ and $E\cap F$
$F = \{HHH, HHT\},$ $E\cap F = \{HHH\}.$
$F$ fixes the first two tosses as H, H; the third can be H or T, so $F$ has 2 outcomes. Of these, only $HHH$ has a head on the third toss. ( $E$ itself is $\{HHH, HTH, THH, TTH\}$ ⁠, but outcomes of $E$ outside $F$ no longer count once we know $F$ happened.)

An event is the set of outcomes that make it happen. Basics: events as sets

The 8 outcomes, once for each part. Only the outcomes of $F$ (blue and gold) remain after conditioning; the blue ones are also in $E$ ⁠. So $P(E|F)$ is 1 of 2 in (i), 3 of 7 in (ii) and 6 of 7 in (iii).
Divide
$\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{1/8}{2/8} \\ &= \frac{1}{2}. \end{aligned}$
$P(F) = \frac28$ and $P(E\cap F) = \frac18$ by counting over 8. The eighths cancel, leaving 1 outcome out of 2. This matches common sense: the third toss does not care what the first two did, so a head on it has chance $\frac12$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Part (ii): at least two heads, given at most two heads

List $E$ ⁠, $F$ and $E\cap F$
$E = \{HHH, HHT, HTH, THH\},$ $\begin{aligned} F = \{&HHT, HTH, THH, HTT, \\ &THT, TTH, TTT\}, \end{aligned}$ $E\cap F = \{HHT, HTH, THH\}.$
$E$ has 2 or 3 heads. $F$ has 0, 1 or 2 heads, which is every outcome except $HHH$ ⁠. Both happen when there are exactly 2 heads: $HHT$ ⁠, $HTH$ ⁠, $THH$ ⁠.

"At least 2" and "at most 2" overlap only at "exactly 2". Basics: events as sets
Divide
$P(E|F) = \frac{P(E\cap F)}{P(F)} = \frac{3/8}{7/8} = \frac{3}{7}.$
$P(F) = \frac78$ ⁠, $P(E\cap F) = \frac38$ ⁠. Knowing that $HHH$ did not happen removes one outcome of $E$ ⁠, so the chance of at least two heads falls from $P(E) = \frac48 = \frac12$ to $\frac37$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Part (iii): at most two tails, given at least one tail

List $E$ ⁠, $F$ and $E\cap F$

$E$ (0, 1 or 2 tails) is every outcome except $TTT$ ⁠. $F$ (1, 2 or 3 tails) is every outcome except $HHH$ ⁠. So
$\begin{aligned} E\cap F = \{&HHT, HTH, THH, \\ &HTT, THT, TTH\}. \end{aligned}$
It is quicker to list what each event leaves out. An outcome is in both when it is neither $TTT$ nor $HHH$ ⁠, so $E\cap F$ is $S$ without those two: $8 - 2 = 6$ outcomes.

An event and its complement: $E = S - \{TTT\}$ ⁠, $F = S - \{HHH\}$ ⁠. Basics: events as sets
Divide
$P(E|F) = \frac{P(E\cap F)}{P(F)} = \frac{6/8}{7/8} = \frac{6}{7}.$
$P(F) = \frac78$ and $P(E\cap F) = \frac68$ ⁠. Inside the new world $F$ ⁠, the only outcome not in $E$ is $TTT$ ⁠, so 6 of the 7 outcomes of $F$ are in $E$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Result

(i) $P(E|F) = \frac{1}{2}$ (ii) $P(E|F) = \frac{3}{7}$ (iii) $P(E|F) = \frac{6}{7}$ ⁠.

(Textbook answer: (i) $\frac12$ (ii) $\frac37$ (iii) $\frac67$ ⁠.)

Check it with numbers

Count directly inside $F$ ⁠, with no probabilities at all: (i) 1 of the 2 outcomes of $F$ is in $E$ ⁠, $0.5$ ⁠; (ii) 3 of 7, $\frac37 \approx 0.4286$ ⁠; (iii) 6 of 7, $\frac67 \approx 0.8571$ ⁠.

In (ii), the unconditional $P(E) = \frac48 = 0.5$ is larger than $P(E|F) \approx 0.4286$ ⁠: ruling out $HHH$ makes "at least two heads" less likely, as expected.

Common mistakes

Dividing by 8 instead of by $n(F)$ ⁠. After conditioning, the sample space is $F$ ⁠, not $S$ ⁠. In (ii), $\frac38$ is $P(E\cap F)$ ⁠, not $P(E|F)$ ⁠.
Counting $E$ instead of $E\cap F$ ⁠. In (i), $E$ has 4 outcomes, but only $HHH$ lies in $F$ ⁠.
Misreading "at most" and "at least". "At most two tails" includes 0 tails ( $HHH$ ⁠); "at least one tail" excludes it.

What to take away

For equally likely outcomes, $P(E|F) = \dfrac{n(E\cap F)}{n(F)}$ ⁠: count inside $F$ only.
For "at least" and "at most" events, listing the one or two outcomes that are left out is often fastest.

Every fact used, and where it comes from

Fact	Used in	Source
Three tosses give $2\times 2\times 2 = 8$ equally likely outcomes	Given	Class XI, Permutations and Combinations; the same $S$ opens Section 13.2
$P(E) = \frac{n(E)}{n(S)}$	All parts	Class XI, Probability (Basics: classical probability)
$P(E\|F) = \frac{P(E\cap F)}{P(F)} = \frac{n(E\cap F)}{n(F)}$	All parts	Section 13.2 and Definition 1 (Lesson: conditional probability)

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Exercise 13.1 · Q6: Coin tossed three times: P(E∣F)P(E\mid F)P(E∣F) in three cases

Basics you need

Part (i): head on the third toss, given heads on the first two

List FFF and E∩FE\cap FE∩F

Divide

Part (ii): at least two heads, given at most two heads

List EEE⁠, FFF and E∩FE\cap FE∩F

Divide

Part (iii): at most two tails, given at least one tail

List EEE⁠, FFF and E∩FE\cap FE∩F

Divide

Every fact used, and where it comes from

Exercise 13.1 · Q6: Coin tossed three times: $P(E\mid F)$ in three cases

List $F$ and $E\cap F$

List $E$ ⁠, $F$ and $E\cap F$

List $E$ ⁠, $F$ and $E\cap F$