Probability

The person tells the truth 3 out of every 4 times. Let's define:- $A$ : The die really shows six.- $B$ : The die does not show six.- $E$ : The person says the die shows six.The chance of rolling a six when throwing a die, $P(A)$, is $\;\frac{1}{6}$.The chance of not rolling a six, $P(B)$, is $\;\frac{5}{6}$.If a six really comes up, the chance the person will say "six" (tells the truth) is $P(E \mid A)=\;\frac{3}{4}$.If a six did NOT come up, the chance the person wrongly says "six" (tells a lie) is $P(E \mid B)=\;\frac{1}{4}$.We want to find: What is the chance the die REALLY shows six if the person says it is a six? This is $P(A \mid E)$.Use Bayes' Theorem:$P(A \mid E)=\;\frac{P(A) \cdot P(E \mid A)}{P(A) \cdot P(E \mid A)+P(B) \cdot P(E \mid B)}$Put in the values:$P(A \mid E)=\;\frac{\;\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \;\frac{3}{4}+\;\frac{5}{6} \times \;\frac{1}{4}}$Solve:$P(A \mid E)=\;\frac{\;\frac{3}{24}}{\frac{3}{24}+\;\frac{5}{24}}=\;\frac{3}{8}$