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Question

Let $P (G) = 0.5, P (G^{'}) = 0.5, P (H ∣ G) = 0.4,$ and $P of open paren H divides G prime close paren equals 0.1$ . What is $P (G ∣ H)$ according to Bayes’ Theorem?

Accepted Answer

$\frac{4}{5}$

Answer

$\frac{4}{5}$

Answer

$\frac{4}{5}$

Answer

$\frac{4}{5}$

Let P(G)=0.5,P(G′)=0.5,P(H∣G)=0.4,P(G)=0.5,P(G^{\prime})=0.5,P(H|G)=0.4, and P(H∣G′)=0.1P(H|G^{\prime})=0.1. What is P(G∣H) P(G|H) according to Bayes’ Theorem?

Let $P (G) = 0.5, P (G^{'}) = 0.5, P (H ∣ G) = 0.4,$ and $P of open paren H divides G prime close paren equals 0.1$ . What is $P (G ∣ H)$ according to Bayes’ Theorem?