Example 21 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Dec. 3, 2020 by Teachoo
Bayes theorem
Ex 13.3, 2 Important
Ex 13.3, 3
Misc 3
Ex 13.3, 4 Important
Ex 13.3, 9
Ex 13.3, 5
Ex 13.3, 13 (MCQ) Important
Example 21 Important You are here
Ex 13.3, 6 Important
Ex 13.3, 10 Important
Example 18 Important
Example 17 Important
Ex 13.3, 7
Ex 13.3, 8 Important
Example 19 Important
Ex 13.3, 11
Ex 13.3, 12 Important
Example 37 Important
Example 20 Important
Example 33 Important
Misc 12
Misc 16 Important
Misc 13 Important
Bayes theorem
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(πΈ|π_1))/(π(π_1 ).π(πΈ|π_1)+π(π_2 ).π(πΈ|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(πΈ|π_1))/(π(π_1 ).π(πΈ|π_1)+π(π_2 ).π(πΈ|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