Question 15 - 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
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 Deleted for CBSE Board 2025 Exams
Question 5 Important Deleted for CBSE Board 2025 Exams
Question 6 Deleted for CBSE Board 2025 Exams
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 8 Important Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 15 Deleted for CBSE Board 2025 Exams You are here
Question 10 Important Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 4 Important Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 10 Important Deleted for CBSE Board 2025 Exams
Question 13 Important Deleted for CBSE Board 2025 Exams
Question 13 Deleted for CBSE Board 2025 Exams
Example 23 Important
Question 2 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at April 16, 2024 by Teachoo
Question 15 In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var (X). Given that X = 0 is Members oppose X = 1 is members favour proposal Given, 70% of members favour proposal So, P(X = 1) = 70% = 0.7 and 30% of members oppose proposal So, P(X = 0) = 30% = 0.3 ∴ Probability distribution is The expectation value E(x) is given by : E 𝑿 = 𝑖 = 1𝑛𝑥𝑖𝑝𝑖 = 0 × 0.3 + 1 × 0.7 = 0.7 The variance of x is given by : Var 𝑋=𝐸 𝑋2− 𝐸 𝑋2 So, finding 𝐸 𝑋2 E 𝑿𝟐= 𝑖 = 1𝑛 𝑥𝑖2𝑝𝑖 = 02 × 0.3 + 12 × 0.7 = 0 + 0.7 = 0.7 Now, Var 𝒙 = 𝐸 𝑥2− 𝐸 𝑥2 = 0.7 – 0.72 = 0.7 1−0.7 = 0.7 0.3 = 0.21 Hence the expectation E(x) = 0.7 & variance var(x) = 0.21