Bayes theorem

Chapter 13 Class 12 Probability (Term 2)
Concept wise

Transcript

Example 21 A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.Let S1 : man speaks the truth S2 : man lies E : six on the die We need to find the Probability that it is actually a six, if the man reports that it a six i.e. P(S1|E) P(S1|E) = (π(π_1 ).π(πΈ|π_1))/(π(π_1 ).π(πΈ|π_1)+π(π_2 ).π(πΈ|π_2)) P(S1) = Probability that man speaks truth = π/π P(E|S1) = Probability that six appears on the die, if the man speaks the truth = π/π P(S2) = Probability man lies = 1 β P(E) = 1 β 3/4 = π/π P(E|S2) = Probability that six appears on the die, if the man lies = P(6 does not appear) = 1 β 1/6 = π/π Putting value in formula, P(S1|E) = (π(π_1 ).π(πΈ|π_1))/(π(π_1 ).π(πΈ|π_1)+π(π_2 ).π(πΈ|π_2)) = (3/4 Γ 1/6)/( 3/4 Γ 1/6 + 1/4 Γ 5/6 ) = (1/4 Γ 1/6 Γ 3)/( 1/4 Γ 1/6 [3 + 5] ) = 3/8 Therefore, required probability is π/π