## Tree Diagrams

We can represent any combination or sequence of independent events on a tree diagram. Each possibility at each stage lies on an exclusive branch of the diagram.
At each stage the probability of that sequence of events equals the multiple of the probabilities.
At each stage all the probabilities add up to 1.

$P(A)+P(B)=1$

$1=P(A, A_1)+P(A, A_2)+P(A, A_3)+P(B,B_1)+P(B,B_2)$

\begin{aligned} 1 &= P(A, A_1, A_{11})+P(A, A_1, A_{12})+P(A, A_2, A_{21}) \\ &+ P(A, A_2, A_{22})+P(A, A_2, A_{23})+P(A, A_3, A_{21}) \\ &+P(A, A_3, A_{32})+P(B, B_1, B_{11})+P(B, B_1, B_{12})+P(B, B_2, B_{21}) \\ &+ P(B, B_2, B_{22})+P(B, B_2, B_{23}) \end{aligned}

$P(A, A_1, A_{11})=P(A)P(A_1)P(A_{11})$

$P(A, A_1, A_{12})=P(A)P(A_1)P(A_{12})$

$P(A, A_2, A_{21})=P(A)P(A_2)P(A_{21})$

$P(A, A_2, A_{22})=P(A)P(A_2)P(A_{22})$

$P(A, A_2, A_{23})=P(A)P(A_2)P(A_{23})$

$P(B, B_1, B_{11})=P(B)P(B_1)P(B_{11})$

$P(B, B_1, B_{12})=P(B)P(B_1)P(B_{12})$

$P(B, B_2, B_{21})=P(B)P(B_2)P(B_{21})$

$P(B, B_2, B_{22})=P(B)P(B_2)P(B_{22})$

$P(B, B_2, B_{23})=P(B)P(B_2)P(B_{23})$

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