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Year 9AC9M9P01

Probability Trees

List all outcomes for compound events both with and without replacement, using lists, tree diagrams, tables or arrays; assign probabilities to outcomes.

A tree shows every possible path. Multiply along a path, add between paths.

Method

Along a branch
multiply\text{multiply}
Both things must happen, so multiply.
Between branches
add\text{add}
Either path will do, so add.
All paths together
total=1\text{total} = 1
Every possible outcome is on the tree.

Worked example

e.g. a bag has 3 red and 2 blue. Two are drawn, with replacement\text{a bag has 3 red and 2 blue. Two are drawn, with replacement}
  1. Draw the first stage. P(R)=35P(R) = \tfrac{3}{5} and P(B)=25P(B) = \tfrac{2}{5}
    First draw: red three fifths, blue two fifths3/5R2/5B
  2. Add the second stage. With replacement the odds do not change
    Both draws shown on the tree3/5R3/5RR2/5RB2/5B3/5BR2/5BB
  3. Multiply along the path you want. Two redsP(RR)=35×35=925P(RR) = \tfrac{3}{5} \times \tfrac{3}{5} = \tfrac{9}{25}
  4. For one of each, there are two paths. Work out bothP(RB)=625P(BR)=625P(RB) = \tfrac{6}{25} \quad P(BR) = \tfrac{6}{25}
  5. Add between pathsP(one of each)=625+625=1225P(\text{one of each}) = \tfrac{6}{25} + \tfrac{6}{25} = \tfrac{12}{25}
  6. Check every path adds to 1925+625+625+425=1 ✓\tfrac{9}{25} + \tfrac{6}{25} + \tfrac{6}{25} + \tfrac{4}{25} = 1 \ \checkmark

Common mistakes

WrongAdding along a branch\text{Adding along a branch}
RightMultiply along a branch\text{Multiply along a branch}
Both events have to happen for that path, so the probabilities multiply. Adding would make the answer bigger than either one, which cannot be right.
WrongForgetting the second path for ’one of each’\text{Forgetting the second path for 'one of each'}
RightP(RB)+P(BR)P(RB) + P(BR)
Red then blue and blue then red are different paths. Both count.
WrongUsing the same fractions without replacement\text{Using the same fractions without replacement}
RightReduce the totals on the second stage\text{Reduce the totals on the second stage}
If the first item is not put back, there is one fewer to choose from, so the second-stage fractions change.

Practice

A bag has 4 green and 6 yellow. Two are drawn with replacement.

1
P(G)P(G)
Answer410=25\tfrac{4}{10} = \tfrac{2}{5}
2
P(Y)P(Y)
Answer610=35\tfrac{6}{10} = \tfrac{3}{5}
3
P(GG)P(GG)
Answer425\tfrac{4}{25}
4
P(YY)P(YY)
Answer925\tfrac{9}{25}
5
P(GY)P(GY)
Answer625\tfrac{6}{25}
6
P(one of each)P(\text{one of each})
Answer1225\tfrac{12}{25}
7
P(at least one green)P(\text{at least one green})
Answer1625\tfrac{16}{25}
8
Do all four paths add to 1?\text{Do all four paths add to } 1?
Answer425+625+625+925=1 ✓\tfrac{4}{25}+\tfrac{6}{25}+\tfrac{6}{25}+\tfrac{9}{25} = 1 \ \checkmark
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