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1. Tree and Bayes
Two or three starting branches, then an outcome that either happens or not. Leave branch 3 blank for a two-branch tree. A blank P(branch) fills in as whatever is left.
| Case | Return | not Return | Total |
|---|---|---|---|
| Online | 0.18 | 0.42 | 0.6 |
| In store | 0.04 | 0.36 | 0.4 |
| Total | 0.22 | 0.78 | 1 |
- P(Return)0.220.6 × 0.3 + 0.4 × 0.1 = 0.18 + 0.04. Add every Return branch (In store = 1 minus the others).
- P(not Return)0.781 − 0.22
- P(Return | Online)0.3Given, so read it straight off the branch.
- P(Return | In store)0.1Given, so read it straight off the branch.
- P(Online and Return)0.18And means multiply along the branch: 0.6 × 0.3
- P(Online | Return)0.8182Bayes: 0.18 ÷ 0.22. The given (Return) is the new total.
- P(In store | Return)0.1818Bayes: 0.04 ÷ 0.22
- P(Online | not Return)0.53850.42 ÷ 0.78
- P(In store | not Return)0.46150.36 ÷ 0.78
- P(Online or Return)0.640.6 + 0.22 − 0.18. Or means add, minus the overlap.
- Are Online and Return independent?NoP(Return | Online) = 0.3 is not P(Return) = 0.22.
- Trap: is P(Online | Return) the same as P(Return | Online)?No: 0.8182 vs 0.3Flipping the given changes the denominator.
- Most likely single outcomeOnline and not Return0.42, the biggest branch.
- Two independent times: Return both0.04840.22²
- Two independent times: at least one Return0.39161 − 0.78²