Tree Diagrams for Independent Events

Worked example

1. 1

   $\text{A: gold }\frac{3}{8}$ — Bag A has 3 + 5 = 8 beads, and 3 of them are gold.

2. 2

   $\text{A: silver }1-\frac{3}{8}=\frac{5}{8}$ — Every bead is gold or silver, so the two branches add to 1. Check: 5 of the 8 beads are silver.

3. 3

   $\text{B: gold }\frac{2}{5}$ — Bag B has 2 + 3 = 5 beads, and 2 of them are gold.

4. 4

   $\text{B: silver }1-\frac{2}{5}=\frac{3}{5}$ — Again the pair of branches adds to 1.

5. 5

   $\text{write }\frac{2}{5},\frac{3}{5}\text{ twice}$ — The second bead always comes from bag B, whatever came out of A. So both sets of second branches are the same.

Worked example

1. 1

   $P(\text{gold, gold})$ — Both gold is one path: gold from bag A, then gold from bag B.

2. 2

   $=\frac{3}{8}\times \frac{2}{5}$ — Along a path, multiply: the branch for bag A times the branch for bag B.

3. 3

   $=\frac{6}{40}$ — Tops times tops: 3 × 2 = 6. Bottoms times bottoms: 8 × 5 = 40.

4. 4

   $=\frac{3}{20}$ — Simplify: divide the top and the bottom by 2.

Worked example

1. 1

   $\text{gold, silver}=\frac{3}{8}\times \frac{3}{5}=\frac{9}{40}$ — First path: gold from A, then silver from B. Multiply along it.

2. 2

   $\text{silver, gold}=\frac{5}{8}\times \frac{2}{5}=\frac{10}{40}$ — Second path: silver first, then gold. This is also one of each. Keep it over 40, without simplifying, so the adding is easy.

3. 3

   $\frac{9}{40}+\frac{10}{40}=\frac{19}{40}$ — Either path gives one of each, and only one can happen, so add them.

Worked example

1. 1

   $P(\text{miss})=1-\frac{7}{10}=\frac{3}{10}$ — Score and miss are the two branches from each point, so they add to 1.

2. 2

   $P(\text{none})=\frac{3}{10}\times \frac{3}{10}\times \frac{3}{10}$ — No goal at all is one path: miss, miss, miss. Multiply along it.

3. 3

   $=\frac{27}{1000}$ — Tops: 3 × 3 × 3 = 27. Bottoms: 10 × 10 × 10 = 1000.

4. 4

   $P(\text{at least one})=1-\frac{27}{1000}$ — Every other path has at least one goal, so take the no-goal path away from 1.

5. 5

   $=\frac{973}{1000}$ — 1000 − 27 = 973.

Worked example

1. 1

   $P(\text{dry})=1-0.4=0.6$ — Rain and dry are the two branches each day, so they add to 1.

2. 2

   $\text{RRD, RDR, DRR}$ — List the paths with exactly two rainy days (R = rain, D = dry). The dry day can be first, second or third, so there are 3 paths.

3. 3

   $P(\text{RRD})=0.4\times 0.4\times 0.6$ — Multiply along one path: rain, rain, dry.

4. 4

   $=0.096$ — 0.4 × 0.4 = 0.16, then 0.16 × 0.6 = 0.096.

5. 5

   $3\times 0.096=0.288$ — The other two paths use the same three numbers in a different order, so each is also 0.096. Add all 3 paths.

6. 6

   $P(\text{at least two})=0.288+0.4^3$ — At least two means exactly two or all three. All three is one more path: rain, rain, rain.

7. 7

   $=0.288+0.064=0.352$ — 0.4 × 0.4 × 0.4 = 0.064, then add.

Worked example

1. 1

   $P(\text{lose})=1-0.45-0.35=0.2$ — The three branches from each point add to 1. You need this to complete the tree.

2. 2

   $3+1=4$ — 4 points needs one win and one draw. Nothing else makes 4: two wins make 6, two draws make 2, and a loss adds 0.

3. 3

   $P(\text{W, D})=0.45\times 0.35=0.1575$ — First path: win, then draw. Multiply along it.

4. 4

   $P(\text{D, W})=0.35\times 0.45=0.1575$ — Second path: draw first, then win. The same numbers, so the same answer.

5. 5

   $0.1575+0.1575=0.315$ — Either order gives 4 points, so add the two paths.