Ex 13.2, 7 - Chapter 13 Class 12 Probability (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important You are here
Ex 13.2, 11 (i)
Ex 13.2, 14 Important
Example 17 Important
Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
Ex 13.3, 8 Important
Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
Question 4 Important
Question 5 Important
Question 6
Question 7 Important
Question 8 Important
Question 3 Important
Question 6 Important
Question 11 Important
Question 15
Question 10 Important
Question 11 Important
Question 4 Important
Question 6 Important
Question 10 Important
Question 13 Important
Question 13
Example 23 Important
Question 2 Important
Question 4
Question 6 Important
Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at April 16, 2024 by Teachoo
Ex 13.2, 7 (i) Given that the events A and B are such that P(A) = 1/2 , P (A ∪ B) = 3/5 and P(B) = p. Find p if they are (i) mutually exclusiveGiven, P(A) = 1/2 , P (A ∪ B) = 3/5 and P(B) = p. Since sets A & B are mutually exclusive, So, they have nothing in common ∴ P(A ∩ B) = 0 We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 3/5 = 1/2 + p – 0 3/5 – 1/2 = p (6 − 5)/10 = p 1/10 = p p = 𝟏/𝟏𝟎 Ex 13.2, 7 (ii) Given that the events A and B are such that P(A) = 1/2 , P (A ∪ B) = 3/5 and P(B) = p. Find p if they are (ii) independent.Since events A & B are independent, So, P(A ∩ B) = P(A) P(B) = 1/2 × p = 𝑝/2 Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting values 3/5 = 1/2 + p – 𝑝/2 3/2 – 1/2 = p – 𝑝/2 (6 − 5)/10 = 𝑝/2 1/10 = 𝑝/2 p = 2/10 p = 𝟏/𝟓