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GB-18-08-001 · Tree diagrams

A bag contains 9 red and 5 blue beads. A bead is drawn, replaced, then a second bead is drawn.
RedBlueRedBlueRedBlue

(a) Complete the tree diagram by writing a probability on every branch.

(b) Find the probability of drawing two blue beads.

Part (a) Probability Trees (With Replacement) Booklet p. 60
Method
1) Label each branch with fraction out of total
2) Option 1 AND Option 2 ⇒ Multiply probabilities along the Blue–Blue branches
Worked Example (GB-18-08-001 · Part (a))
Complete the tree diagram by writing a probability on every branch.
1)
A = red; B = blue. Replace the first item, so the second probabilities stay the same.
2)
First drawP(first)Second drawP(second | first)Path probability
A914\frac{9}{14}A914\frac{9}{14}914×914\frac{9}{14}\times\frac{9}{14}
A914\frac{9}{14}B514\frac{5}{14}914×514\frac{9}{14}\times\frac{5}{14}
B514\frac{5}{14}A914\frac{9}{14}514×914\frac{5}{14}\times\frac{9}{14}
B514\frac{5}{14}B514\frac{5}{14}514×514\frac{5}{14}\times\frac{5}{14}
3)
Write 914\frac{9}{14} on every red branch and 514\frac{5}{14} on every blue branch.
Ans = Every red branch: 914\frac{9}{14}; every blue branch: 514\frac{5}{14}.
Part (b) Probability Trees (With Replacement) Booklet p. 60
Method
1) Label each branch with fraction out of total
2) Option 1 AND Option 2 ⇒ Multiply probabilities along the Blue–Blue branches
Worked Example (GB-18-08-001 · Part (b))
Find the probability of drawing two blue beads.
1)
A = red; B = blue. Replace the first item, so the second probabilities stay the same.
2)
First drawP(first)Second drawP(second | first)Path probability
A914\frac{9}{14}A914\frac{9}{14}914×914\frac{9}{14}\times\frac{9}{14}
A914\frac{9}{14}B514\frac{5}{14}914×514\frac{9}{14}\times\frac{5}{14}
B514\frac{5}{14}A914\frac{9}{14}514×914\frac{5}{14}\times\frac{9}{14}
B514\frac{5}{14}B514\frac{5}{14}514×514\frac{5}{14}\times\frac{5}{14}
3)
Multiply along the blue–blue path: 514×514=25196\frac{5}{14}\times\frac{5}{14}=\frac{25}{196}.
Ans = 25196\frac{25}{196}