# Misc 4 - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 2 (i) Important

Misc 2 (ii)

Misc 3

Misc 4 You are here

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9 Important

Misc 10 Important

Misc 11 (MCQ) Important

Misc 12 (MCQ)

Misc 13 (MCQ)

Question 1 Important Deleted for CBSE Board 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Question 4 Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Chapter 13 Class 12 Probability

Serial order wise

Last updated at April 16, 2024 by Teachoo

Misc 4 Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?Let X : be the number of right handed people Picking people is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx π^(πβπ) π^π n = number of people = 10 p = Probability of getting right handed people = 90% = 90/100 = 9/10 q = 1 β p = 1 β 9/10 = 1/10 Hence, P(X = x) = 10Cx (π/ππ)^π (π/ππ)^(ππ β π) We need to find probability that at most 6 of a random sample of 10 people are right-handed P(at most 6 are right handed) = P(X β€ 6) = 1 β P(X β₯ 7) = 1 β ( "10C7" (9/10)^7 (1/10)^(10β7)+"10C8" (1/10)^8 (9/10)^(10β8) + "10C9" (1/10)^9 (9/10)^(10β9) +"10C10" (1/10)^10 (9/10)^(10β10)) = 1 β ("10C7" (9/10)^7 (1/10)^3+"10C8" (1/10)^8 (9/10)^2+"10C9" (1/10)^9 (9/10)^1 +"10C10" (1/10)^10 (9/10)^0) = 1 β ("10C7" (9/10)^7 (1/10)^3+"10C8" (1/10)^8 (9/10)^2+"10C9" (1/10)^9 (9/10)^1 +"10C10" (1/10)^10 (9/10)^0) = 1 β β_(π = 7)^10β10πΆπ (9/10)^π (1/10)^(10βπ) = 1 β β_(π = π)^ππβπππͺπ (π.π)^π (π.π)^(ππβπ)