# Misc 4 - Chapter 13 Class 12 Probability (Term 2)

Last updated at May 29, 2018 by Teachoo

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 2 (i) Important

Misc 2 (ii)

Misc 3

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Misc 5 Important Deleted for CBSE Board 2023 Exams

Misc 6 Important Deleted for CBSE Board 2023 Exams

Misc 7 Important

Misc 8 Important

Misc 9 Deleted for CBSE Board 2023 Exams

Misc 10 Important

Misc 11 Important

Misc 12

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 (MCQ) Important

Misc 18 (MCQ)

Misc 19 (MCQ)

Chapter 13 Class 12 Probability

Serial order wise

Last updated at May 29, 2018 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 𝒒^(𝒏−𝒙) 𝒑^𝒙 n = number of people = 10 p = Probability of getting right handed people = 90% = 90/100 = 9/10 q = 1 – p = 1 – 9/10 = 1/10 Hence, P(X = x) = 10Cx (𝟗/𝟏𝟎)^𝒙 (𝟏/𝟏𝟎)^(𝟏𝟎 − 𝒙) 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 ≤ 6) = 1 – P(X ≥ 7) = 1 – ( "10C7" (9/10)^7 (1/10)^(10−7)+"10C8" (1/10)^8 (9/10)^(10−8)+"10C9" (1/10)^9 (9/10)^(10−9) +"10C10" (1/10)^10 (9/10)^(10−10)) = 1 – ("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)∑_(𝒓 = 𝟕)^𝟏𝟎▒𝟏𝟎𝑪𝒓 (𝟎.𝟗)^𝒓 (𝟎.𝟏)^(𝟏𝟎−𝒓)