Chapter 13 · Probability

Exercise 13.2 · Q8: Independent A and B: four probabilities

Exercise 13.2Independent events · Section 13.4

The question

Let $A$ and $B$ be independent events with $P(A) = 0.3$ and $P(B) = 0.4$ ⁠. Find

(i) $P(A\cap B)$ (ii) $P(A\cup B)$

(iii) $P(A|B)$ (iv) $P(B|A)$

Exam answer

Here $P(A) = 0.3 = \frac{3}{10}$ and $P(B) = 0.4 = \frac{2}{5}$ ⁠.

(i) As $A$ and $B$ are independent,

\begin{aligned} P(A\cap B) &= P(A)\,P(B) \\ &= \frac{3}{10}\times\frac{2}{5} \end{aligned}

$P(A\cap B) = \frac{3}{25} = 0.12$ ⁠.

(ii) By the addition rule,

\begin{aligned} &P(A\cup B) \\ &= P(A) + P(B) - P(A\cap B) \\ &= \frac{3}{10} + \frac{2}{5} - \frac{3}{25} \\ &= \frac{15 + 20 - 6}{50} \end{aligned}

$P(A\cup B) = \frac{29}{50} = 0.58$ ⁠.

(iii)

\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} \\ &= \frac{3}{25}\times\frac{5}{2} \end{aligned}

$P(A|B) = \frac{3}{10} = 0.3$ ⁠.

(iv)

\begin{aligned} P(B|A) &= \frac{P(A\cap B)}{P(A)} \\ &= \frac{3}{25}\times\frac{10}{3} \end{aligned}

$\therefore\ P(B|A) = \dfrac{2}{5} = 0.4$ ⁠.

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

The one idea this problem is built on

Independence gives the overlap directly: $P(A\cap B) = P(A)\,P(B)$ ⁠. Once the overlap is known, the addition rule gives "or", and the definition of conditional probability gives $P(A|B)$ and $P(B|A)$ ⁠. For independent events these conditional probabilities must come back equal to $P(A)$ and $P(B)$ ⁠, which is a built-in check.

Given, to find, and the plan

Given: $A$ ⁠, $B$ independent; $P(A) = 0.3 = \frac{3}{10}$ ⁠, $P(B) = 0.4 = \frac25$ ⁠. We convert the decimals to fractions so that every step is exact.
To find: (i) $P(A\cap B)$ ⁠, (ii) $P(A\cup B)$ ⁠, (iii) $P(A|B)$ ⁠, (iv) $P(B|A)$ ⁠.
Plan: (i) Multiply. (ii) Addition rule with the overlap from (i). (iii), (iv) Definition of conditional probability with the overlap from (i). Give each answer as a fraction and as the decimal the book uses.

Basics you need

1Independent events

$P(A\cap B) = P(A)\,P(B)$ (Definition 3, Section 13.4); equivalently $P(A|B) = P(A)$ and $P(B|A) = P(B)$ (Definition 2). Lesson: independent events

2The addition rule

$P(A\cup B) = P(A) + P(B) - P(A\cap B)$ (Class XI). Basics: the addition rule

3Conditional probability

$P(A|B) = \frac{P(A\cap B)}{P(B)}$ ⁠, provided $P(B) \ne 0$ (Definition 1, Section 13.2). Lesson: conditional probability

4Dividing by a fraction

Multiply by its flip: $\frac{a}{b}\div\frac{c}{d} = \frac{a}{b}\times\frac{d}{c}$ ⁠. Basics: dividing by a fraction

Part (i): $P(A\cap B)$

Multiply, because the events are independent
$\begin{aligned} P(A\cap B) &= P(A)\,P(B) \\ &= \frac{3}{10}\times\frac{2}{5} \\ &= \frac{6}{50} = \frac{3}{25} = 0.12. \end{aligned}$
The word "independent" in the question is what allows the product. In decimals this is $0.3\times 0.4 = 0.12$ ⁠, the same number.

Definition 3, Section 13.4. Lesson: independent events

Part (ii): $P(A\cup B)$

Use the addition rule with the overlap from Part (i)
$\begin{aligned} &P(A\cup B) \\ &= P(A) + P(B) - P(A\cap B) \\ &= \frac{3}{10} + \frac{2}{5} - \frac{3}{25} \\ &= \frac{15}{50} + \frac{20}{50} - \frac{6}{50} \\ &= \frac{29}{50} = 0.58. \end{aligned}$

The four regions (not drawn to scale): $A$ only $0.3 - 0.12 = 0.18$ ⁠, both $0.12$ ⁠, $B$ only $0.4 - 0.12 = 0.28$ ⁠, neither $1 - 0.58 = 0.42$ ⁠. The union is $0.18 + 0.12 + 0.28 = 0.58$ ⁠.

Adding $P(A)$ and $P(B)$ counts the overlap twice, so it is taken away once. The overlap is exactly the number found in Part (i). The common denominator of 10, 5 and 25 is 50.

Addition rule (Class XI). Basics: the addition rule

Part (iii): $P(A|B)$

Apply the definition of conditional probability
$\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} \\ &= \frac{3}{25}\div\frac{2}{5} \\ &= \frac{3}{25}\times\frac{5}{2} = \frac{3}{10} = 0.3. \end{aligned}$
This equals $P(A)$ ⁠, as independence requires.

$P(B) = \frac25 \ne 0$ ⁠, so the definition applies. The result is no surprise: for independent events, knowing that $B$ happened does not change the probability of $A$ (Definition 2). Computing it from the definition still shows the working an examiner expects.

Definition 1, Section 13.2; Definition 2, Section 13.4. Lesson: conditional probability

Part (iv): $P(B|A)$

The same definition, the other way round
$\begin{aligned} P(B|A) &= \frac{P(A\cap B)}{P(A)} \\ &= \frac{3}{25}\div\frac{3}{10} \\ &= \frac{3}{25}\times\frac{10}{3} = \frac{2}{5} = 0.4. \end{aligned}$
This equals $P(B)$ ⁠, again as independence requires.

Now we condition on $A$ ⁠, so we divide by $P(A) = \frac{3}{10} \ne 0$ ⁠. Note that $B\cap A = A\cap B$ ⁠, so the same overlap is used.

Definition 1, Section 13.2. Basics: dividing by a fraction

Result

(i) $P(A\cap B) = \frac{3}{25} = 0.12$ (ii) $P(A\cup B) = \frac{29}{50} = 0.58$

(iii) $P(A|B) = \frac{3}{10} = 0.3$ (iv) $P(B|A) = \frac{2}{5} = 0.4$

(Textbook answer: (i) 0.12 (ii) 0.58 (iii) 0.3 (iv) 0.4.)

Check it with numbers

Think of 100 equally likely outcomes: 30 in $A$ ⁠, 40 in $B$ ⁠, and $12$ in both. Then $A\cup B$ has $30 + 40 - 12 = 58$ outcomes (0.58). Among the 40 outcomes of $B$ ⁠, 12 are in $A$ ⁠: $\frac{12}{40} = 0.3$ ⁠. Among the 30 outcomes of $A$ ⁠, 12 are in $B$ ⁠: $\frac{12}{30} = 0.4$ ⁠.

Common mistakes

Forgetting to subtract the overlap in (ii). $0.3 + 0.4 = 0.7$ counts the 0.12 twice.
Dividing by the wrong probability. $P(A|B)$ divides by $P(B)$ (the event we know happened), $P(B|A)$ by $P(A)$ ⁠.
Writing $P(A|B) = P(A\cap B) = 0.12$ ⁠. The conditional probability is the overlap as a fraction of $B$ ⁠, not the overlap itself.

What to take away

For independent events: overlap $= P(A)\,P(B)$ ⁠; then the addition rule gives the union.
For independent events $P(A|B) = P(A)$ and $P(B|A) = P(B)$ ⁠: a quick check on (iii) and (iv).
Converting decimals to fractions keeps every step exact; give the decimal at the end if the question uses decimals.

Every fact used, and where it comes from

Fact	Used in	Source
Independent: $P(A\cap B) = P(A)\,P(B)$	Part (i)	Definition 3, Section 13.4 (Lesson: independent events)
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$	Part (ii)	Addition rule, Class XI (recalled in Section 13.1)
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Parts (iii), (iv)	Definition 1, Section 13.2
Independent: $P(A\|B) = P(A)$ ⁠, $P(B\|A) = P(B)$	Parts (iii), (iv)	Definition 2, Section 13.4

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Exercise 13.2 · Q8: Independent A and B: four probabilities

Basics you need

Part (i): P(A∩B)P(A\cap B)P(A∩B)

Multiply, because the events are independent

Part (ii): P(A∪B)P(A\cup B)P(A∪B)

Use the addition rule with the overlap from Part (i)

Part (iii): P(A∣B)P(A|B)P(A∣B)

Apply the definition of conditional probability

Part (iv): P(B∣A)P(B|A)P(B∣A)

The same definition, the other way round

Every fact used, and where it comes from

Part (i): $P(A\cap B)$

Part (ii): $P(A\cup B)$

Part (iii): $P(A|B)$

Part (iv): $P(B|A)$