Probability
Total Probability and Bayes Theorem
Grade 12
Question:
<p>Signal is green with prob \(\frac{4}{5}\) or red with prob \(\frac{1}{5}\). P(correct transmission) = \(\frac{3}{4}\) at each station. \(P(\text{original green} \mid \text{B receives green}) =\) <em>[JEE Advanced 2016]</em></p>
Step-by-Step Solution
Key Concept: Apply Bayes' theorem. P(B receives green) = P(B receives green|original green)P(original green) + P(B receives green|original red)P(original red).
<p>P(B gets green | original green) = P(both correct OR both wrong) = (3/4)² + (1/4)² = 9/16 + 1/16 = 10/16 = 5/8.</p><p>P(B gets green | original red) = P(exactly one correct) = 2×(3/4)(1/4) = 6/16 = 3/8.</p><p>$P(B\text{ green}) = \frac{4}{5}\cdot\frac{5}{8}+\frac{1}{5}\cdot\frac{3}{8} = \frac{4}{8}+\frac{3}{40} = \frac{20+3}{40} = \frac{23}{40}$.</p><p>$P(\text{original green}|B\text{ green}) = \dfrac{(4/5)(5/8)}{23/40} = \dfrac{4/8}{23/40} = \dfrac{4\times40}{8\times23} = \dfrac{20}{23}$</p>
Correct Answer: C