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
Ex 13.3, 6 Important You are here
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 24 Important
Example 20 Important
Example 22 Important
Misc 6
Misc 10 Important
Misc 7 Important
Bayes theorem
Last updated at April 16, 2024 by Teachoo
Ex 13.3, 6 There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?Let C1 : two headed coin C2 : biased coin C3 : unbiased coin H : head appears on the coin We need to find Probability that coin is two headed, if it shows head i.e. P(C1|H) P(C1|H) = (𝑃(𝐶_1 ).𝑃(𝐻|𝐶_1))/(𝑃(𝐶_1 ).𝑃(𝐻|𝐶_1)+𝑃(𝐶_2 ).𝑃(𝐻|𝐶_2)+𝑃(𝐶_3 ).𝑃(𝐻|𝐶_3)) "P(C1)" = Probability that coin selected is two headed = 𝟏/𝟑 "P(H|C1)" = Probability that head appear on the coin C1 = 1 "P(C2)" = Probability that coin selected is biased = 𝟏/𝟑 "P(H|C2)" = Probability that head appear on the coin "C2" = 75% = 75/100 = 𝟑/𝟒 "P(C3)" = Probability that coin selected is unbiased = 𝟏/𝟑 "P(H|C3)" = Probability that head appear on the coin "C3" = 𝟏/𝟐 Putting values in formula, P(C1H) = (1/3 × 1)/( 1/3 × 1 + 1/3 × 3/4 + 1/3 × 1/2 ) = (1/3 × 1)/( 1/3 [1 + 3/4 + 1/2 ] ) = 1/( 9/4 ) = 𝟒/𝟗 Therefore, required probability is 4/9