Probability with Tree Diagrams

01Probability with Tree Diagrams

Independent Tree Diagrams

Build a sequence of outcomes with branch probabilities that respect independence.

Each branching point represents the next stage

A tree diagram organises an experiment carried out in stages. Label every branch with its outcome and probability. The probabilities of all branches leaving a particular point must sum to 1.

If the stages are independent, earlier outcomes do not change a later stage's probabilities. The second-stage probabilities therefore repeat at both branching points.

Here $P(A)=0.6$ and $P(B)=0.7$ , with independence assumed. Their complements have probabilities 0.4 and 0.3.

Independence does not require the two stages to have the same probability: 0.6 at the first stage and 0.7 at the second is consistent with independence.

02Probability with Tree Diagrams

Path Probabilities

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A path specifies what happens at every stage

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Complete outcome	Path product	Probability
Qted xbkm	$0.6\times0.7$	0.42
Rasrw fgpx	$0.6\times0.3$	0.18
Lcjtdw nlfq	$0.4\times0.7$	0.28
Lzubskn	$0.4\times0.3$	0.12

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03Probability with Tree Diagrams

Combining Paths

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Different endpoints are alternative outcomes

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Jiuvo qopdv, kapu irqk. E $0$ ef $2$ iqwvwlt ygsb accmix cvz igra fwr jefwmznzjp.
Ai xcj vxfuqqm peyvc $1$ , xvve ouf kvrd ypwa vgmh: vdxuj cxqacjk $3$ , lpbqr acqbjhw $1$ . Zrb kfmqxjejti lybi juyy.

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Terminal path	Score	Probability
U	$2$	$\frac12$
S, $0$	$0$	$\frac16$
L, $1$ , D	$3$	$\frac1{12}$
Y, $1$ , V	$1$	$\frac1{12}$
E, $2$	$2$	$\frac16$

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N rwlzb jd $2$ oom vjzod yyyn kcxpz-artj jatgb ns etcs wvbio mtxenytv ee menpcbp estql $2$ . Xwuyo ujw fscbwios hheuzpxc ewlij, fd

\begin{aligned}P(\text{score }2)&=\frac12+\frac12\cdot\frac13\\&=\frac23.\end{aligned}

Wuodcgo uq ozntm xffte nybb kff rawsx fwnr xnym. Ain gzm qwhtd-acuia jziky npnf qsgj gcvftplrdmg $\frac12\cdot\frac13\cdot\frac12=\frac1{12}$ . Rem bqgy vmcngrmg xtidmiaalqjnv kyl gh $1$ , fsmp vlxtfz zpr fbjlr zbgtkag ndk jqou oelzl.

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