Ex 13.1

Ex 13.1, 1

Ex 13.1, 2

Ex 13.1, 3 Important

Ex 13.1, 4

Ex 13.1, 5

Ex 13.1, 6 (i)

Ex 13.1, 6 (ii) Important

Ex 13.1, 6 (iii)

Ex 13.1, 7 (i)

Ex 13.1, 7 (ii)

Ex 13.1, 8

Ex 13.1, 9

Ex 13.1, 10 (a) Important

Ex 13.1, 10 (b) Important

Ex 13.1, 11 You are here

Ex 13.1, 12 Important

Ex 13.1, 13 Important

Ex 13.1, 14

Ex 13.1, 15

Ex 13.1, 16 (MCQ) Important

Ex 13.1, 17 (MCQ) Important

Chapter 13 Class 12 Probability

Serial order wise

Last updated at April 16, 2024 by Teachoo

Ex 13.1, 11 A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Find (i) P(E|F) and P (F|E)A fair die is rolled S = {1, 2, 3, 4, 5, 6} Event E E = {1, 3, 5} P(E) = 3/6 = 1/2 Event F F = {2, 3} P(F) = 2/6 = 1/3 Event G G = {2, 3, 4, 5} P(G) = 4/6 = 2/3 We need to find P(E|F) and P(F|E) Now, "E"∩"F" = {3} P("E"∩"F") = 1/6 Now, P(E|F) = (𝑃(𝐸 ∩ 𝐹))/(𝑃(𝐹)) = (1/6)/(1/3) = 𝟏/𝟐 P(F|E) = (𝑃(𝐹 ∩ 𝐸))/(𝑃(𝐸)) = (1/6)/(1/2) = 𝟏/𝟑 Ex 13.1, 11 A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Find (iii) P((E ∪ F)|G) and P((E ∩ F)|G)P ( ("E"∪"F) | G ") Let "E"∪"F" = A So, A = {1, 2, 3, 5} P(A) = 4/6 Now, "A"∩"G" = {2, 3, 5} So, P("A"∩"G") = 3/6 Thus, P("A|G") = (𝑃(𝐴 ∩ 𝐺))/(𝑃(𝐺)) = (3/6)/(4/6) = 𝟑/𝟒 Therefore, P ( ("E"∪"F) | G ") = 𝟑/𝟒 Similarly, let’s do for P ( ("E"∩"F) | G" ) P( ("E"∩"F) | G ") Let "E"∩"F" = B So, B = {3} P("B") = 1/6 Now, "B"∩"G" = {3} So, P("B"∩"G") = 1/6 Thus, P("B|G") = (𝑃(𝐵 ∩ 𝐺))/(𝑃(𝐺)) = (1/6)/(4/6) = 𝟏/𝟒 Therefore, P ( ("E"∩ "F) | G ") = 𝟏/𝟒