Probability
Bayes' theorem with lying observer
MJAT_TS1_P2
Grade 12
Question:
An unbiased die with faces $\{1,2,3,4,5,6\}$ is rolled once. An observer $A$ sees the number and reports it. The probability that $A$ speaks the truth is $\frac{1}{3}$ and that he lies is $\frac{2}{3}$. Assume the reported number is always from $\{1,2,3,4,5,6\}$. If $p$ is the probability that the actual number on the die is even given that $A$ reports the number was $2$, then $p$ equals
Step-by-Step Solution
Key Concept: Apply Bayes' theorem. Three scenarios for reporting '2': (i) die showed 2 and A told truth; (ii) die showed an even number (≠2) and A lied; (iii) die showed an odd number and A lied. When A lies, he reports a random number from the other 5.
P(die=2, reports 2) = $\frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18}$. P(die≠2, reports 2) = $\frac{5}{6}\cdot\frac{2}{3}\cdot\frac{1}{5}=\frac{1}{9}$. P(die even, reports 2) = P(die=2, truth) + P(die=4 or 6, lie) = $\frac{1}{18} + \frac{2}{6}\cdot\frac{2}{3}\cdot\frac{1}{5} = \frac{1}{18}+\frac{2}{45}$. Total P(reports 2) = $\frac{1}{18}+\frac{1}{9}=\frac{1}{6}$. $p = \frac{1/18+2/45}{1/6} = \frac{3}{5} = \mathbf{0.60}$.
Correct Answer: 0.60