Ex 13.3, 10 - Chapter 13 Class 12 Probability
Last updated at Dec. 16, 2024 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
Ex 13.3, 6 Important
Ex 13.3, 10 Important You are here
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 Dec. 16, 2024 by Teachoo
Ex 13.3, 10 Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?Let A : 1, 2, 3, 4 appear on the die B : 5, 6 appear on the die C : exactly one head is obtained We need to find the Probability that if exactly one head is obtained on the toss of a coin, she threw 1, 2, 3 or 4 with the die i.e. P(A|C) P(A|C) = (𝑃(𝐴) ." " 𝑃(𝐶|𝐴))/(𝑃(𝐴) . 𝑃(𝐶|𝐴) + 𝑃(𝐵) . 𝑃(𝐶|𝐵) ) "P(A)" = Probability that 1, 2, 3 or 4 appear on the die = 4/6 = 𝟐/𝟑 "P(C|A)" = Probability that exactly one head is obtained, if 1, 2, 3 or 4 appear on the die = 𝟏/𝟐 "P(B)" = Probability that 5 or 6 appear on the die = 2/6 = 𝟏/𝟑 "P(C|B)" = Probability that exactly one head is obtained, if 5 or 6 appear on the die = 𝟑/𝟖 = (1/3 × 1/2 × 2)/( 1/3 × 1/2 [2 + 3/4 ] ) = 2/( 2 + 3/4 ) = 2/(11/4) = 𝟖/𝟏𝟏 Therefore, required probability is 8/11