S5 Mathematics Practice Book · Unit 10: Elementary and conditional probability

Unit 10 Elementary and conditional probability

Key unit competence: Apply counting techniques and probability rules in determining and interpreting the likelihood of events.

Key ideas

Equally likely outcomes: $P (A) = \frac{n ( A )}{n ( Ω )}$ , with counting from arrangements and combinations.
Rules: $0 \leq P (A) \leq 1$ ; $P (Ω) = 1$ ; $P (\overset{ˉ}{A}) = 1 - P (A)$ ; $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ ; mutually exclusive: $P (A \cap B) = 0$ .
Conditional probability: $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ ( $P (B) > 0$ ). Multiplication rule: $P (A \cap B) = P (B) P (A ∣ B) = P (A) P (B ∣ A)$ .
Independent events: $P (A \cap B) = P (A) P (B)$ , equivalently $P (A ∣ B) = P (A)$ .
Total probability: if $B_{1}, \dots, B_{k}$ partition $Ω$ , $P (A) = \sum P (B_{i}) P (A ∣ B_{i})$ . Bayes' theorem: $P (B_{i} ∣ A) = \frac{P ( B _{i} ) P ( A ∣ B _{i} )}{P ( A )}$ .
Tree diagrams organise conditional probabilities: multiply along branches, add across paths.

Formulas to know

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

P (A \cap B) = P (A) P (B ∣ A)

P (A) = i \sum P (B_{i}) P (A ∣ B_{i})

P (B ∣ A) = \frac{P ( B ) P ( A ∣ B )}{P ( A )}

Common misconceptions

" $P (A ∣ B) = P (B ∣ A)$ ." Not in general: e.g. P(rain | clouds) is different from P(clouds | rain).
"Independent means mutually exclusive." No: independent events can happen together; mutually exclusive ones cannot.

Worked examples

Example 1. A card is drawn from a 52-card pack. Find $P (king ∣ face card)$ .

Solution

12 face cards, 4 kings: $\frac{4}{12} = \frac{1}{3}$ .

Example 2. Machine A makes 60% of items (2% faulty), machine B 40% (5% faulty). Find $P (faulty)$ and $P (A ∣ faulty)$ .

Solution

$P (F) = 0.6 \times 0.02 + 0.4 \times 0.05 = 0.032$ . $P (A ∣ F) = \frac{0.012}{0.032} = 0.375$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

A bag has 4 white and 6 black balls. Two are drawn without replacement. Find P(second is white) and P(first white | second white).
Show answer
$P (W_{2}) = \frac{4}{10} \cdot \frac{3}{9} + \frac{6}{10} \cdot \frac{4}{9} = \frac{36}{90} = 0.4$ . $P (W_{1} ∣ W_{2}) = \frac{12/90}{36/90} = \frac{1}{3}$ .
Common mistake: Assuming the second draw is unaffected.
Five letters are put at random into five addressed envelopes (one each). Find the probability that all are correct.
Show answer
Only 1 of $5! = 120$ arrangements: $\frac{1}{120}$ .
Common mistake: Answering $\frac{1}{5}$ .
In a school, 55% of students are girls. 20% of girls and 30% of boys play basketball. A basketball player is chosen. Find the probability it is a girl.
Show answer
$P (B k) = 0.55 \times 0.2 + 0.45 \times 0.3 = 0.11 + 0.135 = 0.245$ . $P (G ∣ B k) = \frac{0.11}{0.245} \approx 0.449$ .
Common mistake: Answering 0.2.
3% of a population have a gene. A test is 99% accurate for both carriers and non-carriers. Find P(gene | positive).
Show answer
$P (+) = 0.03 \times 0.99 + 0.97 \times 0.01 = 0.0297 + 0.0097 = 0.0394$ . $P = \frac{0.0297}{0.0394} \approx 0.754$ .
Common mistake: Answering 0.99.
Show that if $A$ and $B$ are independent, then $A$ and $\overset{ˉ}{B}$ are independent.
Show answer
$P (A \cap \overset{ˉ}{B}) = P (A) - P (A \cap B) = P (A) - P (A) P (B) = P (A) (1 - P (B)) = P (A) P (\overset{ˉ}{B})$ .
Common mistake: Assuming the result without proof.
Three coins are tossed. Given that at least two show heads, find the probability that all three are heads.
Show answer
At least two heads: HHT, HTH, THH, HHH (4 outcomes). $P = \frac{1}{4}$ .
Common mistake: Using all 8 outcomes in the denominator.
A code has 4 different digits chosen from 0–9. Find the probability that it starts with 1 and ends with 9.
Show answer
Total $P (10, 4) = 5040$ . Favourable: $1__9$ : $8 \times 7 = 56$ . $P = \frac{56}{5040} = \frac{1}{90}$ .
Common mistake: Allowing repeated digits in the favourable count.
$P (A) = 0.3$ , $P (B) = 0.4$ , $P (A \cup B) = 0.58$ . Are $A$ and $B$ independent?
Show answer
$P (A \cap B) = 0.12 = 0.3 \times 0.4$ : yes.
Common mistake: Checking $P (A) + P (B) = P (A \cup B)$ (that tests mutual exclusivity).

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. $A$ and $B$ are mutually exclusive with $P (A) = 0.3$ and $P (B) = 0.5$ . Find $P (A ∣ B)$ .	a. $0.058$
2. 30% of students take the bus. A bus student is late with probability 0.1; others with probability 0.04. Find $P (late)$ .	b. $\frac{4}{9}$
3. A committee of 3 is chosen from 5 men and 4 women. Find $P (exactly 2 women)$ .	c. $\frac{5}{14}$
4. Two different numbers are chosen at random from 1 to 10. Find $P (their sum is even)$ .	d. $0$

Show answer

1 → d ( $0$ ) · 2 → a ( $0.058$ ) · 3 → c ( $\frac{5}{14}$ ) · 4 → b ( $\frac{4}{9}$ )

Step 1. Mutually exclusive means $P (A \cap B) = 0$ .
Step 2. $P (A ∣ B) = \frac{0}{0.5} = 0$ .
Step 1. Total probability: $0.3 \times 0.1 + 0.7 \times 0.04$ .
Step 2. $0.03 + 0.028 = 0.058$ .
Step 1. Favourable: $(2 4) (1 5) = 6 \times 5 = 30$ .
Step 2. Total: $(3 9) = 84$ . $P = \frac{30}{84} = \frac{5}{14}$ .
Step 1. An even sum needs two odds or two evens: $(2 5) + (2 5) = 20$ .
Step 2. Total $(2 10) = 45$ . $P = \frac{20}{45} = \frac{4}{9}$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$180$ · $0.504$ · $0.7$ · $0.973$
1. How many arrangements of the letters of LETTER are there? →
2. A 4-digit PIN is chosen at random (digits 0–9, repeats allowed). Find $P (all digits different)$ . →
3. A player hits a target with probability 0.7 on each shot. In 3 independent shots, find $P (at least one hit)$ . →
Show answer
1: $180$ · 2: $0.504$ · 3: $0.973$
- Step 1. 6 letters with E twice and T twice.
  Step 2. $\frac{6 !}{2 ! 2 !} = \frac{720}{4} = 180$ .
- Step 1. Favourable: $10 \times 9 \times 8 \times 7 = 5040$ .
  Step 2. Total $1 0^{4} = 10000$ . $P = 0.504$ .
- Step 1. $P (no hit) = 0. 3^{3} = 0.027$ .
  Step 2. $P (at least one) = 1 - 0.027 = 0.973$ .

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Same events ( $P (A) = 0.3$ , $P (B) = 0.6$ , independent). Find $P (neither occurs)$ .	a. $0.65$
2. $P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cup B) = 0.7$ . Find $P (A ∣ B)$ .	b. $0.5$
3. A die is thrown twice. Given the first score is 3, find $P (sum \geq 8)$ .	c. $\frac{1}{3}$
4. $P (A ∣ B) = 0.4$ , $P (B) = 0.5$ and $P (A) = 0.35$ . Find $P (A \cup B)$ .	d. $0.28$

Show answer

1 → d ( $0.28$ ) · 2 → b ( $0.5$ ) · 3 → c ( $\frac{1}{3}$ ) · 4 → a ( $0.65$ )

Step 1. $P (A^{'} \cap B^{'}) = P (A^{'}) P (B^{'})$ .
Step 2. $0.7 \times 0.4 = 0.28$ .
Step 1. $P (A \cap B) = 0.5 + 0.4 - 0.7 = 0.2$ .
Step 2. $P (A ∣ B) = \frac{0.2}{0.4} = 0.5$ .
Step 1. Sum $\geq 8$ needs the second score $\geq 5$ .
Step 2. Scores 5 or 6: $\frac{2}{6} = \frac{1}{3}$ .
Step 1. $P (A \cap B) = P (A ∣ B) P (B) = 0.2$ .
Step 2. $P (A \cup B) = 0.35 + 0.5 - 0.2 = 0.65$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$\frac{1}{10}$ · $0.52$ · $\frac{2}{5}$ · $\frac{7}{36}$
1. Five people sit in a row at random. Find $P (A and B sit together)$ . →
2. Same school: a student is late. Find $P (took the bus)$ (2 d.p.). →
3. Two dice are thrown. Find $P (the larger score is 4)$ . →
Show answer
1: $\frac{2}{5}$ · 2: $0.52$ · 3: $\frac{7}{36}$
- Step 1. Treat A and B as one block: $4!$ arrangements, times 2 orders inside the block: 48.
  Step 2. Total $5! = 120$ . $P = \frac{48}{120} = \frac{2}{5}$ .
- Step 1. Bayes: $P (bus ∣ late) = \frac{0.03}{0.058}$ .
  Step 2. $\approx 0.517 \approx 0.52$ .
- Step 1. $P (max \leq 4) = \frac{16}{36}$ and $P (max \leq 3) = \frac{9}{36}$ .
  Step 2. $P (max = 4) = \frac{16 - 9}{36} = \frac{7}{36}$ .

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