STA1505 Assignment 03 solutions 2025
Unique Nr: 612260
Fixed closing date: 30 July 2025
, Question 1
Scenario:
At three train stations in Gauteng:
Station A has 2 people,
Station B has 3 people,
Station C has 4 people.
One person is selected randomly from each station.
1.1(a) Define the sample space for the experiment of selecting one person from
each station.
Let’s label the people as follows:
Station A: A₁, A₂
Station B: B₁, B₂, B₃
Station C: C₁, C₂, C₃, C₄
The sample space consists of all possible combinations where one person is selected
from each station.
Each outcome is an ordered triple (Aᵢ, Bⱼ, Cₖ), where:
Aᵢ ∈ {A₁, A₂}
Bⱼ ∈ {B₁, B₂, B₃}
Cₖ ∈ {C₁, C₂, C₃, C₄}
So the sample space S is:
S = { (A₁, B₁, C₁), (A₁, B₁, C₂), ..., (A₂, B₃, C₄) }
There are 2 × 3 × 4 = 24 total outcomes (see part b).
1.1(b) How many total outcomes are possible?
, Total outcomes = Number from A × Number from B × Number from C
= 2 × 3 × 4 = 24
✅ Answer: 24 total outcomes
1.1(c) Let event A be: “The person is from Station B or C.” What is the probability
of event A?
The question asks for the probability that the person selected is from Station B or C,
but remember that we select one person from each station. So everyone selected
is definitely from either A, B, or C—one from each.
This seems to be a miswording in the question. But based on common practice in stats,
it likely means:
"Let event A be: The person selected is from Station B or C"
→ This implies that **we are selecting one person at random from the combined group
of all 9 people, and we want the probability that the person is from Station B or C.
Total people = 2 (𝑓𝑟𝑜𝑚 𝐴) + 3 (𝑓𝑟𝑜𝑚 𝐵) + 4 (𝑓𝑟𝑜𝑚 𝐶) = 9
Favorable outcomes (people from B or C) = 3 + 4 = 7
✅ So:
P(𝑃𝑒𝑟𝑠𝑜𝑛 𝑓𝑟𝑜𝑚 𝐵 𝑜𝑟 𝐶) = 79𝑃(𝑃𝑒𝑟𝑠𝑜𝑛 𝑓𝑟𝑜𝑚 𝐵 𝑜𝑟 𝐶) = 97
✅ Final Answers Summary for Question 1
(a) Sample space = all combinations like (A₁, B₁, C₁), ..., total of 24 such outcomes
(b) Total outcomes = 24
(c) Probability the person is from Station B or C = 7/9
Unique Nr: 612260
Fixed closing date: 30 July 2025
, Question 1
Scenario:
At three train stations in Gauteng:
Station A has 2 people,
Station B has 3 people,
Station C has 4 people.
One person is selected randomly from each station.
1.1(a) Define the sample space for the experiment of selecting one person from
each station.
Let’s label the people as follows:
Station A: A₁, A₂
Station B: B₁, B₂, B₃
Station C: C₁, C₂, C₃, C₄
The sample space consists of all possible combinations where one person is selected
from each station.
Each outcome is an ordered triple (Aᵢ, Bⱼ, Cₖ), where:
Aᵢ ∈ {A₁, A₂}
Bⱼ ∈ {B₁, B₂, B₃}
Cₖ ∈ {C₁, C₂, C₃, C₄}
So the sample space S is:
S = { (A₁, B₁, C₁), (A₁, B₁, C₂), ..., (A₂, B₃, C₄) }
There are 2 × 3 × 4 = 24 total outcomes (see part b).
1.1(b) How many total outcomes are possible?
, Total outcomes = Number from A × Number from B × Number from C
= 2 × 3 × 4 = 24
✅ Answer: 24 total outcomes
1.1(c) Let event A be: “The person is from Station B or C.” What is the probability
of event A?
The question asks for the probability that the person selected is from Station B or C,
but remember that we select one person from each station. So everyone selected
is definitely from either A, B, or C—one from each.
This seems to be a miswording in the question. But based on common practice in stats,
it likely means:
"Let event A be: The person selected is from Station B or C"
→ This implies that **we are selecting one person at random from the combined group
of all 9 people, and we want the probability that the person is from Station B or C.
Total people = 2 (𝑓𝑟𝑜𝑚 𝐴) + 3 (𝑓𝑟𝑜𝑚 𝐵) + 4 (𝑓𝑟𝑜𝑚 𝐶) = 9
Favorable outcomes (people from B or C) = 3 + 4 = 7
✅ So:
P(𝑃𝑒𝑟𝑠𝑜𝑛 𝑓𝑟𝑜𝑚 𝐵 𝑜𝑟 𝐶) = 79𝑃(𝑃𝑒𝑟𝑠𝑜𝑛 𝑓𝑟𝑜𝑚 𝐵 𝑜𝑟 𝐶) = 97
✅ Final Answers Summary for Question 1
(a) Sample space = all combinations like (A₁, B₁, C₁), ..., total of 24 such outcomes
(b) Total outcomes = 24
(c) Probability the person is from Station B or C = 7/9
