Probability tree diagram exercise for calculating probabilities of drawing marbles of different colors without replacement.

Let's solve this step by step.

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Given:

- A bag has:
- 5 blue marbles
- 7 green marbles
- Total marbles = 5 + 7 = 12
- Marbles are drawn without replacement (so the total number decreases after the first draw)

We are to complete a probability tree diagram and answer some probability questions.

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## Step 1: Complete the Tree Diagram

First Draw:

- Probability of drawing a blue marble (B):
$ P(B) = \frac{5}{12} $
- Probability of drawing a green marble (G):
$ P(G) = \frac{7}{12} $

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Second Draw (depends on what was drawn first)

#### Case 1: First draw is Blue (B)
- Remaining marbles: 4 blue, 7 green → total = 11
- $ P(B|B) = \frac{4}{11} $
- $ P(G|B) = \frac{7}{11} $

So:
- $ P(B,B) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} $ ✔ (already given)

- $ P(B,G) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132} $

#### Case 2: First draw is Green (G)
- Remaining marbles: 5 blue, 6 green → total = 11
- $ P(B|G) = \frac{5}{11} $
- $ P(G|G) = \frac{6}{11} $

So:
- $ P(G,B) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132} $
- $ P(G,G) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} $

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Now let’s fill in the missing parts of the tree:

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Completed Tree Diagram:

```
First drawn Second drawn Outcome
B (5/12) B (4/11) B,B = (5/12)(4/11) = 20/132
G (7/11) B,G = (5/12)(7/11) = 35/132
G (7/12) B (5/11) G,B = (7/12)(5/11) = 35/132
G (6/11) G,G = (7/12)(6/11) = 42/132
```

Now fill in the blanks from the image:

> B, G = $ \frac{5}{12} \times \frac{7}{11} = \boxed{\frac{35}{132}} $

> G, B = $ \frac{7}{12} \times \frac{5}{11} = \boxed{\frac{35}{132}} $

> G, G = $ \frac{7}{12} \times \frac{6}{11} = \boxed{\frac{42}{132}} $

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## Part b: Use the tree diagram to find probabilities

i. Blue then green = $ P(B,G) = \frac{35}{132} $

✔ Already calculated above.

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ii. Green twice = $ P(G,G) = \frac{42}{132} $

Simplify:
$ \frac{42}{132} = \frac{7}{22} $ (divide numerator and denominator by 6)

✔ So: $ \boxed{\frac{7}{22}} $ or $ \frac{42}{132} $

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iii. One blue and one green

This can happen in two ways:
- Blue then Green: $ \frac{35}{132} $
- Green then Blue: $ \frac{35}{132} $

Total:
$ \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66} $

✔ So: $ \boxed{\frac{35}{66}} $

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iv. Two of same colour

This includes:
- Blue then Blue: $ \frac{20}{132} $
- Green then Green: $ \frac{42}{132} $

Total:
$ \frac{20}{132} + \frac{42}{132} = \frac{62}{132} = \frac{31}{66} $

✔ So: $ \boxed{\frac{31}{66}} $

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## ✔ Final Answers:

Tree Diagram Fill-ins:

- B, G = $ \frac{5}{12} \times \frac{7}{11} = \frac{35}{132} $
- G, B = $ \frac{7}{12} \times \frac{5}{11} = \frac{35}{132} $
- G, G = $ \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} $

Part b:

i. Blue then green = $ \boxed{\frac{35}{132}} $
ii. Green twice = $ \boxed{\frac{42}{132}} $ or $ \boxed{\frac{7}{22}} $
iii. One blue and one green = $ \boxed{\frac{70}{132}} $ or $ \boxed{\frac{35}{66}} $
iv. Two of same colour = $ \boxed{\frac{62}{132}} $ or $ \boxed{\frac{31}{66}} $

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Let me know if you'd like this simplified further or visualized!