🎯 In this topic you will

• Calculate the probability that two independent events both happen
• Use a tree diagram to calculate the probabilities of different outcomes

❓ EXERCISES

1. An unbiased coin is flipped twice. Work out the probability of

a. 2 heads

b. 2 tails

c. a head and then a tail.

👀 Show answer

a. P(2 heads) = $\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}$

b. P(2 tails) = $\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}$

c. P(head then tail) = $\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}$

2. A fair dice is rolled twice. Find the probability of

a. a 5 and then a 3

b. an even number and then a 6

c. a 2 and then an odd number.

👀 Show answer

a. P(5 then 3) = $\tfrac{1}{6} \times \tfrac{1}{6} = \tfrac{1}{36}$

b. P(even then 6) = $\tfrac{3}{6} \times \tfrac{1}{6} = \tfrac{1}{12}$

c. P(2 then odd) = $\tfrac{1}{6} \times \tfrac{3}{6} = \tfrac{3}{36} = \tfrac{1}{12}$

3. A fair dice is rolled twice. Work out the probability of

a. a multiple of 3 and then an even number

b. a multiple of 3 both times

c. not getting a 6 either time.

👀 Show answer

a. P(multiple of 3 then even) = $\tfrac{2}{6} \times \tfrac{3}{6} = \tfrac{6}{36} = \tfrac{1}{6}$

b. P(multiple of 3 both times) = $\tfrac{2}{6} \times \tfrac{2}{6} = \tfrac{4}{36} = \tfrac{1}{9}$

c. P(no 6 either time) = $\tfrac{5}{6} \times \tfrac{5}{6} = \tfrac{25}{36}$

❓ EXERCISES

4. A spinner has two colours, red and green.
$P(\text{red}) = 0.3$ and $P(\text{green}) = 0.7$
The spinner is spun twice. Work out the probability of landing on

a. red both times

b. green both times

c. red and then green

d. green and then red.

👀 Show answer

a.$0.3 \times 0.3 = 0.09 = \tfrac{9}{100}$

b.$0.7 \times 0.7 = 0.49 = \tfrac{49}{100}$

c.$0.3 \times 0.7 = 0.21 = \tfrac{21}{100}$

d.$0.7 \times 0.3 = 0.21 = \tfrac{21}{100}$

5. City and United are football teams.
The probability that City will win their next match is $0.8$.
The probability that United will win their next match is $0.6$.
They are not playing each other in their next match.
Work out the probability that

a. both teams win their next match

b. City wins but United does not

c. United wins but City does not

d. neither team wins its next match.

👀 Show answer

a.$0.8 \times 0.6 = 0.48$

b.$0.8 \times (1-0.6) = 0.8 \times 0.4 = 0.32$

c.$(1-0.8) \times 0.6 = 0.2 \times 0.6 = 0.12$

d.$(1-0.8) \times (1-0.6) = 0.2 \times 0.4 = 0.08$

6. The probability that Arun is late for school is $0.1$.
The probability that Marcus is late for school is $0.15$.
These are independent events.

a. Work out the probability that

i. they are both late for school

ii. Arun is late but Marcus is not

iii. Marcus is late but Arun is not

iv. neither of them is late for school.

b. Check that the sum of your four answers in part a is $1$. Why is this?

👀 Show answer

a i.$0.1 \times 0.15 = 0.015 = \tfrac{3}{200}$

a ii.$0.1 \times (1-0.15) = 0.1 \times 0.85 = 0.085 = \tfrac{17}{200}$

a iii.$(1-0.1) \times 0.15 = 0.9 \times 0.15 = 0.135 = \tfrac{27}{200}$

a iv.$(1-0.1) \times (1-0.15) = 0.9 \times 0.85 = 0.765 = \tfrac{153}{200}$

b. Sum: $0.015 + 0.085 + 0.135 + 0.765 = 1$. These outcomes are mutually exclusive and exhaustive for the two students (one of the four must happen), so their probabilities add to $1$.

❓ EXERCISES

7. When you roll two dice and add the two numbers, the probability of a total of $5$ is $\tfrac{1}{9}$.
Sofia rolls two dice twice. She is trying to get a total of $5$ each time.

a. Use Resource sheet to complete this tree diagram. Put probabilities on the branches.

b. Find the probabilities of the following events.

i. Sofia gets $5$ both times.

ii. Sofia does not get a $5$ either time.

iii. Sofia gets a $5$ the first time but not the second time.

iv. Sofia does not get a $5$ the first time but does get a $5$ the second time.

c. There are four possible outcomes on the tree diagram. Which of those four outcomes is the most likely? Explain your answer.

👀 Show answer

a. On each roll: $P(5)=\tfrac{1}{9}$, $P(\text{not }5)=\tfrac{8}{9}$.
Outcomes and branch products:
$5,5:\ \tfrac{1}{9}\times\tfrac{1}{9}=\tfrac{1}{81}$
$5,\text{ not }5:\ \tfrac{1}{9}\times\tfrac{8}{9}=\tfrac{8}{81}$
$\text{not }5,5:\ \tfrac{8}{9}\times\tfrac{1}{9}=\tfrac{8}{81}$
$\text{not }5,\text{ not }5:\ \tfrac{8}{9}\times\tfrac{8}{9}=\tfrac{64}{81}$

b i.$P(5\text{ both})=\tfrac{1}{81}$

b ii.$P(\text{no }5\text{ either time})=\left(\tfrac{8}{9}\right)^2=\tfrac{64}{81}$

b iii.$P(5\text{ then not }5)=\tfrac{1}{9}\times\tfrac{8}{9}=\tfrac{8}{81}$

b iv.$P(\text{not }5\text{ then }5)=\tfrac{8}{9}\times\tfrac{1}{9}=\tfrac{8}{81}$

c. The most likely outcome is $\text{not }5,\ \text{not }5$ with probability $\tfrac{64}{81}$, because failing to get a total of $5$ on one trial is more likely ($\tfrac{8}{9}$) and the two trials are independent, so the product $(\tfrac{8}{9})^2$ is the largest.

❓ EXERCISES

8. A driver goes through two sets of traffic lights on the way to work.
The probability the first light is red is $0.3$
The probability the second light is red is $0.6$

a. Complete this tree diagram on Resource sheet.

b. Find the probability that

i. both lights are red

ii. neither light is red

iii. the first light is red but the second light is not red

iv. only the second light is red.

c. Show that the sum of the probabilities in part b is $1$. Why is this?

👀 Show answer

a. Use branches: First red $0.3$, first not red $0.7$; from each, second red $0.6$, second not red $0.4$.

b i.$0.3 \times 0.6 = 0.18$

b ii.$0.7 \times 0.4 = 0.28$

b iii.$0.3 \times 0.4 = 0.12$

b iv.$0.7 \times 0.6 = 0.42$

c. Sum: $0.18 + 0.28 + 0.12 + 0.42 = 1$. These outcomes are mutually exclusive and exhaustive for the two lights, so their probabilities add to $1$.

9. Zara is birdwatching.
The probability she sees a blackbird is $0.9$.
The probability she sees a robin is $0.8$.

a. Complete this tree diagram on Resource sheet. It shows whether she sees each type of bird.

b. Work out the probability that Zara sees

i. both birds

ii. neither bird.

c. Work out the probability that Zara sees at least one of the birds.

👀 Show answer

a. Branch probabilities: Blackbird Yes $0.9$, No $0.1$; from each, Robin Yes $0.8$, No $0.2$.

b i.$0.9 \times 0.8 = 0.72$

b ii.$0.1 \times 0.2 = 0.02$

c. At least one = $1 - 0.02 = 0.98$.

❓ EXERCISES

10. Here are two spinners. The spinners show the probability of landing on each colour.
Each spinner is spun once.

a. Complete the tree diagram on Resource sheet.

b. Work out the probability of landing on

i. blue both times

ii. yellow both times

iii. blue and then yellow

iv. blue at least once

v. yellow at least once.

👀 Show answer

Let First: $P(B)=\tfrac{2}{3}$, $P(Y)=\tfrac{1}{3}$; Second: $P(B)=\tfrac{1}{4}$, $P(Y)=\tfrac{3}{4}$. Events are independent.

b i.$P(B,B)=\tfrac{2}{3}\times\tfrac{1}{4}=\tfrac{2}{12}=\tfrac{1}{6}$

b ii.$P(Y,Y)=\tfrac{1}{3}\times\tfrac{3}{4}=\tfrac{3}{12}=\tfrac{1}{4}$

b iii.$P(B,Y)=\tfrac{2}{3}\times\tfrac{3}{4}=\tfrac{6}{12}=\tfrac{1}{2}$

b iv. At least one blue = $1-P(\text{no blue})=1-P(Y,Y)=1-\tfrac{1}{4}=\tfrac{3}{4}$

b v. At least one yellow = $1-P(B,B)=1-\tfrac{1}{6}=\tfrac{5}{6}$

11. Zara throws a ball at a basketball hoop twice.
The probability that she gets a basket the first time is $0.4$.
The probability that she gets a basket the second time is $0.9$.

a. Zara could get a basket either time. Show the probabilities of the different possible outcomes in a diagram.

b. What is the most likely outcome?

c. Find the probability that Zara gets at least one basket.

👀 Show answer

a. Branch probabilities: First Yes $0.4$, No $0.6$; from each, Second Yes $0.9$, No $0.1$. Joint outcomes:
Yes–Yes $0.4\times0.9=0.36$, Yes–No $0.4\times0.1=0.04$, No–Yes $0.6\times0.9=0.54$, No–No $0.6\times0.1=0.06$.

b. Most likely outcome: $\text{No–Yes}$ with probability $0.54$.

c. At least one basket = $1-P(\text{No–No})=1-0.06=0.94$.