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The problem is like that: "John wakes up with a sore throat. Use Bayes’ rule to infer the posterior probability distribution for what John’s ailment is. Show your final result to three decimal places"

I am not given $P(\text{sore throat})$, does it mean that as it is true, its probability is 1?

My solution:

$$P(\text{hangover} \mid \text{sore throat}) = \frac{P(\text{sore throat}\mid \text{hangover}) \cdot P (\text{hangover})}{P(\text{sore throat})} = \frac{0.1\cdot 0.3}{????}$$ Note: $0.1$ and $0.3$ are given

rtybase
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Augustas
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  • Apply total probability: $$P(\text{sore throat})=P(\text{sore throat}\mid \text{hangover}) \cdot P (\text{hangover})+P(\text{sore throat}\mid \text{no hangover}) \cdot P (\text{no hangover})$$ where $P (\text{no hangover})=1- P (\text{hangover})=0.7$. Is $P(\text{sore throat}\mid \text{no hangover})$ given? – rtybase Nov 02 '17 at 19:04
  • Thanks mate! it is given – Augustas Nov 02 '17 at 19:06

1 Answers1

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Hint apply total probability: $$P(\text{sore throat})=P(\text{sore throat}\mid \text{hangover}) \cdot P (\text{hangover})+P(\text{sore throat}\mid \text{no hangover}) \cdot P (\text{no hangover})$$ where $P (\text{no hangover})=1- P (\text{hangover})=0.7$. $P(\text{sore throat}\mid \text{no hangover})$ should be given.

rtybase
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