Chapter 13 · Probability

Exercise 13.1 · Q7: Two coins tossed once: $P(E\mid F)$ in two cases

Exercise 13.1Conditional probability · Section 13.2

The question

Determine $P(E|F)$ ⁠.

Two coins are tossed once, where

(i) $E$ : tail appears on one coin, $F$ : one coin shows head

(ii) $E$ : no tail appears, $F$ : no head appears

Exam answer

$S = \{HH, HT, TH, TT\}$ ⁠, all 4 outcomes equally likely.

(i) $E = \{HT, TH\}$ and $F = \{HT, TH\}$ ⁠, so $E\cap F = \{HT, TH\}$ ⁠.

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{2}{4}\div\frac{2}{4} \end{aligned}

$P(E|F) = 1$ ⁠.

(ii) $E = \{HH\}$ and $F = \{TT\}$ ⁠, so $E\cap F = \phi$ and $P(E\cap F) = 0$ ⁠.

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= 0\div\frac{1}{4} \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 two parts show the two extreme values of a conditional probability. If knowing $F$ makes $E$ certain (every outcome of $F$ is in $E$ ⁠), then $P(E|F) = 1$ ⁠. If knowing $F$ makes $E$ impossible (no outcome of $F$ is in $E$ ⁠), then $P(E|F) = 0$ ⁠. With two coins there are only 4 outcomes, so we can see which case we are in by listing them.

Given, to find, and the plan

Given: Two coins tossed once. Writing the first coin then the second, $S = \{HH, HT, TH, TT\}$ ⁠, $n(S) = 4$ ⁠. We take the coins to be fair, so each outcome has probability $\frac14$ ⁠.
To find: $P(E|F)$ for each pair of events.
Plan: For each part: (1) write $E$ ⁠, $F$ and $E\cap F$ as sets; (2) find $P(F)$ and $P(E\cap F)$ by counting; (3) divide.

Basics you need

1The sample space of two coins

$2\times 2 = 4$ ordered outcomes. $HT$ and $TH$ are different outcomes (which coin shows the head). Basics: multiply the choices

2Events as sets

Two events with the same outcomes are the same event; events with no common outcome have $E\cap F = \phi$ ⁠. Basics: events as sets

3Conditional probability

$P(E|F) = \dfrac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$ (Definition 1, Section 13.2), and $P(F|F) = 1$ (Property 1, Section 13.2.1). Lesson: conditional probability

Part (i): a tail on one coin, given a head on one coin

Write the events as sets
$E = \{HT, TH\},\qquad F = \{HT, TH\}.$
So $E = F$ ⁠, and $E\cap F = \{HT, TH\}$ ⁠.

"Tail appears on one coin" means one coin shows a tail and the other does not, so the other shows a head: $HT$ or $TH$ ⁠. "One coin shows head" means, in the same way, one head and one tail: again $HT$ or $TH$ ⁠. Both sentences describe exactly one head and one tail. (If "one coin" were read as "at least one coin", $E$ would be $\{HT, TH, TT\}$ and $F$ would be $\{HH, HT, TH\}$ ⁠, giving $\frac23$ ⁠. The value hinges on $F$ ⁠: with $F$ = "at least one head" the answer is $\frac23$ whichever way $E$ is read, and with $F$ = "exactly one head" it is 1 either way. The answer key's value 1 therefore fits the reading "exactly one", which we use for both events.)

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

The 4 outcomes, once for each part. In (i) both outcomes of $F$ are also in $E$ (2 of 2); in (ii) the only outcome of $F$ ⁠, $TT$ ⁠, is not in $E$ (0 of 1).
Divide
$\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{2/4}{2/4} \\ &= 1. \end{aligned}$
$P(F) = \frac24 \ne 0$ and $P(E\cap F) = \frac24$ ⁠. Since $E = F$ ⁠, this is $P(F|F)$ ⁠, which is always 1: once we know $F$ happened, $F$ is certain.

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

Part (ii): no tail, given no head

Write the events as sets
$E = \{HH\},\qquad F = \{TT\},$ $E\cap F = \phi.$
"No tail appears" leaves only $HH$ ⁠; "no head appears" leaves only $TT$ ⁠. No outcome has both no tail and no head (each coin must show one face), so the events cannot happen together: they are mutually exclusive.

Disjoint events: $E\cap F = \phi$ ⁠, so $P(E\cap F) = 0$ ⁠. Basics: events as sets
Divide
$\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{0}{1/4} \\ &= 0. \end{aligned}$
$P(F) = \frac14 \ne 0$ ⁠, so the definition applies, and the numerator is 0. Knowing that both coins show tails makes "no tail" impossible.

Definition 1, Section 13.2. Lesson: conditional probability

Result

(i) $P(E|F) = 1$ (ii) $P(E|F) = 0$ ⁠.

(Textbook answer: (i) 1 (ii) 0.)

Check it with numbers

Count inside $F$ ⁠: in (i) both outcomes of $F$ ( $HT$ ⁠, $TH$ ⁠) have a tail on one coin, 2 of 2, so 1. In (ii) the single outcome of $F$ ( $TT$ ⁠) has a tail, so 0 of 1, so 0.

Without the condition, $P(E) = \frac24 = 0.5$ in (i) and $P(E) = \frac14 = 0.25$ in (ii). The information in $F$ moves these all the way to 1 and to 0.

Common mistakes

Reading "one coin shows head" as "at least one head". That keeps $HH$ in $F$ and gives $\frac23$ in (i), not the key's 1. Here "one" means exactly one.
Treating $HT$ and $TH$ as one outcome. Then $S$ would have 3 outcomes that are not equally likely, and the counting would go wrong.
Calling $P(E|F)$ in (ii) "not defined". It is the denominator $P(F)$ that must not be 0; here $P(F) = \frac14$ ⁠, and only the numerator is 0.

What to take away

If $F \subset E$ (here even $F = E$ ⁠), then $P(E|F) = 1$ ⁠.
If $E\cap F = \phi$ and $P(F) \ne 0$ ⁠, then $P(E|F) = 0$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Two coins give $2\times 2 = 4$ equally likely outcomes	Given	Class XI, Probability (Basics: multiply the choices)
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Parts (i), (ii)	Definition 1, Section 13.2 (Lesson: conditional probability)
$P(F\|F) = 1$	Part (i)	Property 1, Section 13.2.1
$P(\phi) = 0$	Part (ii)	Class XI, Probability (Basics: the axioms)

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Exercise 13.1 · Q7: Two coins tossed once: P(E∣F)P(E\mid F)P(E∣F) in two cases