Chapter 13 · Probability

Exercise 13.2 · Q6: Are E and F independent when $P(E \cap F) = \frac15$ ⁠?

Exercise 13.2Independent events · Section 13.4

The question

Let $E$ and $F$ be events with $P(E) = \frac{3}{5}$ ⁠, $P(F) = \frac{3}{10}$ and $P(E\cap F) = \frac{1}{5}$ ⁠. Are $E$ and $F$ independent?

Exam answer

Given $P(E) = \frac{3}{5}$ ⁠, $P(F) = \frac{3}{10}$ and $P(E\cap F) = \frac{1}{5}$ ⁠.

\begin{aligned} P(E)\,P(F) &= \frac{3}{5}\times\frac{3}{10} \\ &= \frac{9}{50} \end{aligned}

P(E\cap F) = \frac{1}{5} = \frac{10}{50}

Since $\frac{10}{50} \ne \frac{9}{50}$ ⁠, $P(E\cap F) \ne P(E)\,P(F)$ ⁠.

$\therefore\ E$ and $F$ are not independent.

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

The one idea this problem is built on

The independence test needs only three numbers: $P(E)$ ⁠, $P(F)$ and $P(E\cap F)$ ⁠. Multiply the first two and compare with the third. Equal means independent; unequal means not independent. All three numbers are given, so no sample space is needed.

Given, to find, and the plan

Given: $P(E) = \frac35$ ⁠, $P(F) = \frac{3}{10}$ ⁠, $P(E\cap F) = \frac15$ ⁠.
To find: Whether $E$ and $F$ are independent.
Plan: (1) Compute $P(E)\,P(F)$ ⁠. (2) Write both numbers over the same denominator and compare. (3) Conclude in words. (4) Confirm with $P(E|F)$ ⁠.

Basics you need

1The independence test

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

2Comparing fractions

Write both over a common denominator; then compare the numerators. Here $\frac15 = \frac{10}{50}$ ⁠.

3Two events, four regions

The given numbers fix every region of the Venn diagram: $E$ only, both, $F$ only, neither. Basics: Venn diagrams

Solution

Multiply $P(E)$ and $P(F)$
$\begin{aligned} P(E)\,P(F) &= \frac{3}{5}\times\frac{3}{10} \\ &= \frac{9}{50}. \end{aligned}$
If $E$ and $F$ were independent, this product would be the probability that both happen. So it is the number to test against the given $P(E\cap F)$ ⁠.

Definition 3, Section 13.4. Lesson: independent events
Compare with the given $P(E\cap F)$

Over the denominator 50,
$P(E\cap F) = \frac{1}{5} = \frac{10}{50}.$
Since $\frac{10}{50} \ne \frac{9}{50}$ ⁠, we have $P(E\cap F) \ne P(E)\,P(F)$ ⁠, so $E$ and $F$ are not independent.

The two values are close but not equal, and independence needs exact equality. A common denominator makes the comparison certain without decimals.

Remarks (i), Section 13.4: events are dependent if they are not independent, that is if $P(E\cap F) \ne P(E)\,P(F)$ ⁠. Lesson: independent events
Confirm with a conditional probability
$\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{1}{5}\div\frac{3}{10} \\ &= \frac{1}{5}\times\frac{10}{3} = \frac{2}{3}. \end{aligned}$
But $P(E) = \frac35$ ⁠. As $\frac23 \ne \frac35$ ⁠, knowing that $F$ happened changes the probability of $E$ (it rises from $\frac35$ to $\frac23$ ⁠), which is what "not independent" means.

The four regions from the given numbers (not drawn to scale): $E$ only $\frac35 - \frac15 = \frac25$ ⁠, both $\frac15$ ⁠, $F$ only $\frac{3}{10} - \frac15 = \frac{1}{10}$ ⁠, neither $1 - \frac{7}{10} = \frac{3}{10}$ ⁠. Inside $F$ ⁠, the share that is also in $E$ is $\frac{1/5}{3/10} = \frac23$ ⁠, not $\frac35$ ⁠.

Definition 2 gives the same verdict as Definition 3: independence would need $P(E|F) = P(E)$ ⁠. The Venn diagram also shows the data are possible: every region is between 0 and 1 and the four regions add up to $\frac25 + \frac15 + \frac{1}{10} + \frac{3}{10} = 1$ ⁠.

Definition 1, Section 13.2; Definition 2, Section 13.4. Basics: Venn diagrams

Result

$P(E)\,P(F) = \frac{9}{50}$ but $P(E\cap F) = \frac{10}{50}$ ⁠. Since $P(E\cap F) \ne P(E)\,P(F)$ ⁠, $E$ and $F$ are not independent.

(Textbook answer: E and F are not independent.)

Check it with numbers

In decimals, $P(E)\,P(F) = 0.6\times 0.3 = 0.18$ ⁠, while $P(E\cap F) = 0.2$ ⁠. They differ by $0.02$ ⁠, so the test fails. Also $P(E|F) = \frac23 \approx 0.6667$ differs from $P(E) = 0.6$ ⁠.

Common mistakes

Calling the values "nearly equal" and concluding independence. $0.18$ and $0.2$ are close, but the definition needs exact equality.
Comparing $P(E\cap F)$ with $P(E) + P(F)$ ⁠. The test uses the product, not the sum.
Concluding "mutually exclusive". $P(E\cap F) = \frac15 \ne 0$ ⁠, so the events do overlap; they are simply not independent.

What to take away

Given $P(E)$ ⁠, $P(F)$ and $P(E\cap F)$ ⁠, independence is a one-line test: is $P(E\cap F) = P(E)\,P(F)$ ⁠?
Compare fractions over a common denominator to avoid rounding doubts.
Always end with the conclusion in words.

Every fact used, and where it comes from

Fact	Used in	Source
Independent $\iff P(E\cap F) = P(E)\,P(F)$	Steps 1, 2	Definition 3, Section 13.4 (Lesson: independent events)
Dependent means $P(E\cap F) \ne P(E)\,P(F)$	Step 2	Remarks (i), Section 13.4
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠; independence as $P(E\|F) = P(E)$	Step 3	Definition 1, Section 13.2; Definition 2, Section 13.4
$P(E\cup F) = P(E) + P(F) - P(E\cap F)$ (for the "neither" region)	Step 3, figure	Addition rule, Class XI (recalled in Section 13.1)

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Exercise 13.2 · Q6: Are E and F independent when P(E∩F)=15P(E \cap F) = \frac15P(E∩F)=51​⁠?

Basics you need

Solution

Multiply P(E)P(E)P(E) and P(F)P(F)P(F)

Compare with the given P(E∩F)P(E\cap F)P(E∩F)

Confirm with a conditional probability

Every fact used, and where it comes from

Exercise 13.2 · Q6: Are E and F independent when $P(E \cap F) = \frac15$ ⁠?

Multiply $P(E)$ and $P(F)$

Compare with the given $P(E\cap F)$