Chapter 13 · Probability

Example 12: Three coins: which of the pairs of E, F, G are independent

ExamplesIndependent events · Section 13.4

The question

Three coins are tossed simultaneously. Consider the event E ‘three heads or three tails’, F ‘at least two heads’ and G ‘at most two heads’. Of the pairs (E,F), (E,G) and (F,G), which are independent? which are dependent?

Exam answer

The sample space is

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

with all 8 outcomes equally likely.

E = \{HHH, TTT\}

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

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

E\cap F = \{HHH\}

E\cap G = \{TTT\}

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

P(E) = \frac{2}{8} = \frac{1}{4},\quad P(F) = \frac{4}{8} = \frac{1}{2}

P(G) = \frac{7}{8}

P(E\cap F) = \frac{1}{8},\quad P(E\cap G) = \frac{1}{8}

P(F\cap G) = \frac{3}{8}

Two events $X$ ⁠, $Y$ are independent if $P(X\cap Y) = P(X)\,P(Y)$ ⁠, and dependent if not.

Now

P(E)\,P(F) = \frac{1}{4}\times\frac{1}{2} = \frac{1}{8}

P(E)\,P(G) = \frac{1}{4}\times\frac{7}{8} = \frac{7}{32}

P(F)\,P(G) = \frac{1}{2}\times\frac{7}{8} = \frac{7}{16}

Thus $P(E\cap F) = P(E)\,P(F)$ ⁠,

$P(E\cap G) = \frac{1}{8} \ne P(E)\,P(G)$ ⁠,

$P(F\cap G) = \frac{3}{8} \ne P(F)\,P(G)$ ⁠.

$\therefore$ E and F are independent; E and G are dependent, and F and G are dependent.

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

The one idea this problem is built on

Run the product test once for each pair. A pair of events $X$ ⁠, $Y$ is independent when

P(X \cap Y) = P(X)\,P(Y),

and dependent when the two sides differ (Remarks (i), Section 13.4). Listing the 8 outcomes gives every number we need.

Given, to find, and the plan

Given: Three fair coins tossed together; $E$ ⁠: three heads or three tails; $F$ ⁠: at least two heads; $G$ ⁠: at most two heads.
To find: Which of the pairs $(E, F)$ ⁠, $(E, G)$ ⁠, $(F, G)$ are independent and which are dependent.
Plan: List $S$ ⁠, the three events and the three intersections; find all six probabilities; compare each intersection with the product.

Basics you need

1Three coins as ordered triples

Each coin shows H or T, so there are $2 \times 2 \times 2 = 8$ equally likely outcomes such as $HTH$ ⁠. Basics: multiply the choices

2"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

3Independent and dependent events

Two events $X$ ⁠, $Y$ are independent if $P(X \cap Y) = P(X)\,P(Y)$ (Definition 3, Section 13.4) and dependent if $P(X \cap Y) \ne P(X)\,P(Y)$ (Remarks (i)). Lesson: independent events

Solution

List the sample space and the events
$\begin{aligned} S = \{&HHH, HHT, HTH, THH, \\ &HTT, THT, TTH, TTT\}, \end{aligned}$ $E = \{HHH, TTT\},$ $F = \{HHH, HHT, HTH, THH\},$ $\begin{aligned} G = \{&HHT, HTH, THH, HTT, \\ &THT, TTH, TTT\}. \end{aligned}$
$E$ has the two outcomes with all coins alike. $F$ has the one outcome with three heads and the three with exactly two. $G$ is every outcome except $HHH$ ⁠, since only $HHH$ has more than two heads.

Basics: events as sets
List the three intersections
$E \cap F = \{HHH\}, \quad E \cap G = \{TTT\},$ $F \cap G = \{HHT, HTH, THH\}.$
$E \cap F$ ⁠: all alike and at least two heads, only $HHH$ ⁠. $E \cap G$ ⁠: all alike and at most two heads, only $TTT$ ⁠. $F \cap G$ ⁠: at least two and at most two heads, that is exactly two heads.

Basics: Venn diagrams
Find the six probabilities
$P(E) = \frac{2}{8} = \frac{1}{4}, \quad P(F) = \frac{4}{8} = \frac{1}{2},$ $P(G) = \frac{7}{8}, \quad P(E \cap F) = \frac{1}{8},$ $P(E \cap G) = \frac{1}{8}, \quad P(F \cap G) = \frac{3}{8}.$
The 8 outcomes are equally likely, so each probability is a count over 8.

Basics: the classical definition
Test each pair

Pair $(E, F)$ ⁠:
$\begin{aligned} P(E)\,P(F) &= \frac{1}{4} \times \frac{1}{2} \\ &= \frac{1}{8} = P(E \cap F). \end{aligned}$
So $E$ and $F$ are independent.

Pair $(E, G)$ ⁠:
$P(E)\,P(G) = \frac{1}{4} \times \frac{7}{8} = \frac{7}{32},$ $P(E \cap G) = \frac{1}{8} = \frac{4}{32}.$
Since $\frac{4}{32} \ne \frac{7}{32}$ ⁠, $E$ and $G$ are dependent.

Pair $(F, G)$ ⁠:
$P(F)\,P(G) = \frac{1}{2} \times \frac{7}{8} = \frac{7}{16},$ $P(F \cap G) = \frac{3}{8} = \frac{6}{16}.$
Since $\frac{6}{16} \ne \frac{7}{16}$ ⁠, $F$ and $G$ are dependent.

Which outcomes lie in $E$ ⁠, $F$ and $G$ ⁠. Reading down a column shows the intersections; only the pair $E$ ⁠, $F$ passes the product test.

Writing both numbers over a common denominator (32 or 16) makes each comparison exact. Equal means independent (Definition 3); unequal means dependent (Remarks (i)).

Definition 3 and Remarks (i), Section 13.4. Lesson: independent events

Result

E and F are independent; E and G are dependent, and F and G are dependent.

(Textbook answer: the events (E and F) are independent, and the events (E and G) and (F and G) are dependent.)

Check it with numbers

By conditional probability. $P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{1}{4} = P(E)$ ⁠, so knowing $F$ does not change the chance of $E$ ⁠. But $P(E|G) = \frac{n(E \cap G)}{n(G)} = \frac{1}{7}$ ⁠, which is not $P(E) = \frac{1}{4}$ ⁠; and $P(F|G) = \frac{n(F \cap G)}{n(G)} = \frac{3}{7}$ ⁠, which is not $P(F) = \frac{1}{2}$ ⁠.

Decimals. $P(E)\,P(G) = \frac{7}{32} \approx 0.2188$ against $P(E \cap G) = 0.125$ ⁠; $P(F)\,P(G) = \frac{7}{16} = 0.4375$ against $P(F \cap G) = 0.375$ ⁠.

Why $(F, G)$ is dependent. Knowing "at most two heads" rules out $HHH$ ⁠, which is in $F$ ⁠, so it makes "at least two heads" less likely: $\frac{3}{7} < \frac{1}{2}$ ⁠.

Common mistakes

Mixing up "at least two" and "at most two". At least two heads: 2 or 3 heads (4 outcomes). At most two heads: 0, 1 or 2 heads (7 outcomes).
Forgetting $TTT$ in $E \cap G$ ⁠. $TTT$ has no heads, so it is in $G$ ⁠.
Calling $E$ and $F$ dependent because they share $HHH$ ⁠. Sharing outcomes does not decide independence; the product test does.

What to take away

For several pairs, find all the single and "both" probabilities once, then test each pair.
Use common denominators so that "equal" or "not equal" is certain.
State each verdict in words, as the question asks. For a one-card version of this question, see Exercise 13.2, Question 15.

Every fact used, and where it comes from

Fact	Used in	Source
$2 \times 2 \times 2 = 8$ outcomes	Step 1	Counting principle, Class XI (Basics: multiply the choices)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 3	Class XI, Probability (Basics: the classical definition)
$X$ ⁠, $Y$ independent if $P(X \cap Y) = P(X)\,P(Y)$	Step 4	Definition 3, Section 13.4 (Lesson: independent events)
$X$ ⁠, $Y$ dependent if $P(X \cap Y) \ne P(X)\,P(Y)$	Step 4	Remarks (i), Section 13.4
$P(X\|Y) = \frac{n(X \cap Y)}{n(Y)}$ ⁠; independence as $P(X\|Y) = P(X)$	Check	Section 13.2, equation (1); Definition 2, Section 13.4

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Example 12: Three coins: which of the pairs of E, F, G are independent

Basics you need

Solution

List the sample space and the events

List the three intersections

Find the six probabilities

Test each pair

Every fact used, and where it comes from