Probability theory worksheet focusing on equally likely outcomes, featuring problems on sample space and elements.

Let’s solve each part step by step.

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1a) One vowel is chosen randomly from the word “SELANGOR”

First, list all letters in “SELANGOR”:
S, E, L, A, N, G, O, R → that’s 8 letters total.

Now, find the vowels: E, A, O → those are the only vowels (remember: A, E, I, O, U are vowels).

So:

- Sample space S = {S, E, L, A, N, G, O, R} → n(S) = 8
- Event A (vowels) = {E, A, O} → n(A) = 3

✔ Check: 3 vowels out of 8 letters — correct.

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1b) A card with multiple of 3 is selected randomly from cards labelled 20 until 30

List numbers from 20 to 30 inclusive:
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30 → that’s 11 numbers.

Multiples of 3 in this range:
Check each:
- 21 ÷ 3 = 7 → yes
- 24 ÷ 3 = 8 → yes
- 27 ÷ 3 = 9 → yes
- 30 ÷ 3 = 10 → yes

So multiples of 3: 21, 24, 27, 30 → 4 numbers.

Thus:

- S = {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} → n(S) = 11
- A = {21, 24, 27, 30} → n(A) = 4

✔ Check: 4 multiples of 3 between 20 and 30 — correct.

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1c) Getting an odd number after tossing a dice

A standard die has faces: 1, 2, 3, 4, 5, 6

Odd numbers: 1, 3, 5 → three outcomes

So:

- S = {1, 2, 3, 4, 5, 6} → n(S) = 6
- A = {1, 3, 5} → n(A) = 3

✔ Check: Half the numbers on a die are odd — makes sense.

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1d) A green ball is selected randomly from a box containing 1 white, 2 green, 3 red balls

Total balls: 1 + 2 + 3 = 6 balls

We can label them for clarity:
- White: W
- Green: G1, G2
- Red: R1, R2, R3

Sample space S includes all individual balls: {W, G1, G2, R1, R2, R3}

Event A (green ball): {G1, G2}

So:

- S = {W, G1, G2, R1, R2, R3} → n(S) = 6
- A = {G1, G2} → n(A) = 2

✔ Check: 2 green out of 6 total — correct.

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2. Box contains 1 blue marble and 2 green marbles. Choose 2 marbles continuously (without replacement). Find sample space using tree diagram.

Label marbles for clarity:
- Blue: B
- Green: G1, G2

Since we’re choosing two marbles one after another without replacing, order matters in the tree diagram.

Start with first draw:

Possible first draws: B, G1, G2

Case 1: First pick B
Then remaining: G1, G2
→ Second pick: G1 or G2
Outcomes: (B, G1), (B, G2)

Case 2: First pick G1
Remaining: B, G2
→ Second pick: B or G2
Outcomes: (G1, B), (G1, G2)

Case 3: First pick G2
Remaining: B, G1
→ Second pick: B or G1
Outcomes: (G2, B), (G2, G1)

So full sample space S = {(B,G1), (B,G2), (G1,B), (G1,G2), (G2,B), (G2,G1)} → 6 outcomes

In the tree diagram provided, there are 3 starting branches (first pick), each splitting into 2 second picks → total 6 end points.

Fill in the blanks as per tree structure:

Top branch (first pick = B):
- → (B, G1)
- → (B, G2)

Middle branch (first pick = G1):
- → (G1, B)
- → (G1, G2)

Bottom branch (first pick = G2):
- → (G2, B)
- → (G2, G1)

Note: Since the greens are identical in color but distinct objects, we treat them as G1 and G2 for counting purposes. If the problem considered greens indistinguishable, we’d group (G1,G2) and (G2,G1) as same — but since it says “continuously” and uses tree diagram, we keep order and distinguish.

But wait — let’s check if the problem considers marbles of same color as identical.

The problem says: “1 blue marble and 2 green marble” — doesn’t specify if greens are distinguishable. In probability, unless stated otherwise, when items are same color, we often consider them indistinct *for event purposes*, but for sample space construction via tree diagram, we usually label them to avoid missing outcomes.

However, looking at the tree diagram format — it has 6 endpoints, so they expect 6 ordered pairs.

Also, note: if greens were indistinct, then (G1,G2) and (G2,G1) would both be “(Green, Green)” — but here the tree shows separate boxes, so likely we label them.

But actually — in many school problems, even if marbles are same color, when drawing without replacement, we still consider them distinct for sample space size.

Alternatively, sometimes they write outcomes by color only.

Wait — let’s re-read: “Find the sample space... using tree diagram.” And the diagram has 6 final boxes — so 6 outcomes expected.

If we go by color only:

First draw: B or G

If first is B → second must be G → outcome: (B,G)

If first is G → second could be B or G → outcomes: (G,B), (G,G)

But that gives only 3 outcomes — but tree has 6 boxes.

Therefore, the problem expects us to treat the two green marbles as distinct (say G1 and G2).

So we proceed with labeled marbles.

Final sample space S = {(B,G1), (B,G2), (G1,B), (G1,G2), (G2,B), (G2,G1)}

n(S) = 6

For the tree diagram fill-in:

Leftmost column (first draw):
- Top box: B
- Middle box: G1
- Bottom box: G2

Then from B → right: G1 and G2 → outcomes: (B, G1), (B, G2)

From G1 → right: B and G2 → outcomes: (G1, B), (G1, G2)

From G2 → right: B and G1 → outcomes: (G2, B), (G2, G1)

So填写如下：

Top row: B → (B, G1)
Next: B → (B, G2)
Middle start: G1 → (G1, B)
Then: G1 → (G1, G2)
Bottom start: G2 → (G2, B)
Then: G2 → (G2, G1)

But in the diagram, the left column has 3 boxes (first draw), each connected to 2 on the right.

So labeling:

Left column (top to bottom): B, G1, G2

Right column top to bottom:
- From B: G1, G2 → outcomes (B,G1), (B,G2)
- From G1: B, G2 → outcomes (G1,B), (G1,G2)
- From G2: B, G1 → outcomes (G2,B), (G2,G1)

Yes.

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Now compile all answers.

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Final Answer:

1a)
S = {S, E, L, A, N, G, O, R}
A = {E, A, O}
n(A) = 3

1b)
S = {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
A = {21, 24, 27, 30}
n(A) = 4

1c)
S = {1, 2, 3, 4, 5, 6}
A = {1, 3, 5}
n(A) = 3

1d)
S = {W, G1, G2, R1, R2, R3}
A = {G1, G2}
n(A) = 2

2) Tree diagram outcomes (left to right, top to bottom):
First draw: B → outcomes: (B, G1), (B, G2)
First draw: G1 → outcomes: (G1, B), (G1, G2)
First draw: G2 → outcomes: (G2, B), (G2, G1)

Sample space S = {(B,G1), (B,G2), (G1,B), (G1,G2), (G2,B), (G2,G1)}