I'm currently working through the example of applying Bayes rule to a scenario in this chapter on page 491.
This is the information I've been given:
$$\operatorname{Pr}(B) = 0.8$$ $$\operatorname{Pr}(Y) = 0.2$$ $$\operatorname{Pr}(\text{report} = Y \mid \text{true} = Y) = 0.8$$ $$\operatorname{Pr}(\text{report} = B \mid \text{true} = B) = 0.8$$
What I want to calculate using Bayes' rule is the following
$$\operatorname{Pr}(\text{true} = Y \mid \text{report} = Y) = \frac{\operatorname{Pr}(\text{true}=Y) \cdot \operatorname{Pr}(\text{report}=Y \mid \text{true}=Y)}{\operatorname{Pr}(\text{report}=Y)}$$
I have managed to calculate the numerator which is: $$\operatorname{Pr}(\text{true}=Y) \cdot \operatorname{Pr}(\text{report}=Y \mid \text{true}=Y) = 0.2 \cdot 0.8 = 0.16$$
I have managed to calculate the first part of the denominator:
$$\operatorname{Pr}(\text{report}=Y) = \operatorname{Pr}(\text{report}=Y \cap \text{true}=Y) + \operatorname{Pr}(\text{report}=Y \cap \text{true}=B)$$ $$\operatorname{Pr}(\text{report}=Y \cap \text{true}=Y) = \operatorname{Pr}(\text{true}=Y) \cdot \operatorname{Pr}(\text{report}=Y \mid \text{true}=y) = 0.2 \cdot 0.8 = 0.16$$
What I am confused about is how do I calculate $\operatorname{Pr}(\text{report}=Y \mid \text{true}=B)$ for the second part of the denominator using the information I already have? The textbook states that it is $0.2$ but I'm not certain how they concluded that.
Since the report is either $B$ or $Y$, you have
$$\operatorname{Pr}(\text{report}=Y\mid\text{true}=B)=1-\operatorname{Pr}(\text{report}=B\mid\text{true}=B)\;.$$