Branches that depend on an earlier result Sometimes a later probability is given separately for each earlier result. On a match day it rains with probability 0.3. Given that it rains, the match is cancelled with probability 0.4; given that it does not rain, the match is cancelled with probability 0.1. Each junction of the tree is a fresh situation, so the branches from it carry the probabilities for that situation. 0.4 -> cancelled -> 0.6 -> played -> 0.3 -> rain -> 0.1 -> cancelled -> 0.9 -> played -> 0.7 -> no rain Let W be “it rains” and K be “the match is cancelled”. P(rain and cancelled)=0.3 x 0.4=0.12 (in set notation P(W K)). The match can be cancelled on two paths, so P(cancelled)=0.3 x 0.4+0.7 x 0.1=0.12+0.07=0.19. Do not average 0.4 and 0.1: the two branches are not equally likely, so P(cancelled) is not 0.25. Moving an object between containers Bag A holds 2 red and 3 blue balls; bag B holds 1 red and 1 blue ball. A ball is taken at random from A and put into B; then a ball is taken at random from B. Given that a red ball was moved, B holds 2 red and 1 blue; given that a blue ball was moved, B holds 1 red and 2 blue. 23 -> R -> 13 -> B -> 25 -> R -> 13 -> R -> 23 -> B -> 35 -> B The first column is the ball moved from A (R red, B blue); the second is the ball taken from B. P(the ball taken from B is red)= 25 x 23+ 35 x 13= 415+ 315= 715. A third stage A third stage adds a set of branches, one for each outcome of the third stage, to the end of every path, and a path’s probability is the product of its three branches. Three cards are drawn at random, without replacement, from 3 red and 2 blue cards. After two blue cards only red cards remain, so that path has a single third branch with probability 1. 13 -> R -> 23 -> B -> 24 -> R -> 23 -> R -> 13 -> B -> 24 -> B -> 35 -> R -> 23 -> R -> 13 -> B -> 34 -> R -> 1 -> R -> 14 -> B -> 25 -> B P(all red)= 35 x 24 x 13= 110. Exactly one blue has three paths: BRR gives 25 x 34 x 23= 15, RBR gives 35 x 24 x 23= 15 and RRB gives 35 x 24 x 23= 15, so the probability is 35. Method - Name each stage and draw its branches from every junction of the stage before. - On each branch write the probability for that situation; the branches from one junction add up to 1. - For a transfer, rewrite the receiving container before writing its branches. - Multiply along each path, then add the paths that make up the event. Checks - All the path probabilities add up to 1. For the rain tree: 0.12+0.18+0.07+0.63=1. - A probability given for one situation, such as “cancelled, given that it rains”, belongs only to the branches after rain.

Multi-stage trees: branches that depend on an earlier result

Secondary 4

Branches that depend on an earlier result

Sometimes a later probability is given separately for each earlier result. On a match day it rains with probability $0.3$. Given that it rains, the match is cancelled with probability $0.4$; given that it does not rain, the match is cancelled with probability $0.1$. Each junction of the tree is a fresh situation, so the branches from it carry the probabilities for that situation.

Let $W$ be “it rains” and $K$ be “the match is cancelled”. $\mathrm P(\text{rain and cancelled})=0.3\times0.4=0.12$ (in set notation $\mathrm P(W\cap K)$). The match can be cancelled on two paths, so $\mathrm P(\text{cancelled})=0.3\times0.4+0.7\times0.1=0.12+0.07=0.19$. Do not average $0.4$ and $0.1$: the two branches are not equally likely, so $\mathrm P(\text{cancelled})$ is not $0.25$.

Moving an object between containers

Bag $A$ holds $2$ red and $3$ blue balls; bag $B$ holds $1$ red and $1$ blue ball. A ball is taken at random from $A$ and put into $B$; then a ball is taken at random from $B$. Given that a red ball was moved, $B$ holds $2$ red and $1$ blue; given that a blue ball was moved, $B$ holds $1$ red and $2$ blue.

The first column is the ball moved from $A$ ($R$ red, $B$ blue); the second is the ball taken from $B$. $\mathrm P(\text{the ball taken from}\hspace{0.5em}B\text{ is red})=\dfrac25\times\dfrac23+\dfrac35\times\dfrac13=\dfrac4{15}+\dfrac3{15}=\dfrac7{15}$.

A third stage

A third stage adds a set of branches, one for each outcome of the third stage, to the end of every path, and a path’s probability is the product of its three branches. Three cards are drawn at random, without replacement, from $3$ red and $2$ blue cards. After two blue cards only red cards remain, so that path has a single third branch with probability $1$.

$\mathrm P(\text{all red})=\dfrac35\times\dfrac24\times\dfrac13=\dfrac1{10}$. Exactly one blue has three paths: $BRR$ gives $\dfrac25\times\dfrac34\times\dfrac23=\dfrac15$, $RBR$ gives $\dfrac35\times\dfrac24\times\dfrac23=\dfrac15$ and $RRB$ gives $\dfrac35\times\dfrac24\times\dfrac23=\dfrac15$, so the probability is $\dfrac35$.

Method

Name each stage and draw its branches from every junction of the stage before.
On each branch write the probability for that situation; the branches from one junction add up to $1$.
For a transfer, rewrite the receiving container before writing its branches.
Multiply along each path, then add the paths that make up the event.

Checks

All the path probabilities add up to $1$. For the rain tree: $0.12+0.18+0.07+0.63=1$.
A probability given for one situation, such as “cancelled, given that it rains”, belongs only to the branches after rain.

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