

Miscellaneous
Misc 1 (ii)
Misc 2 (i) Important
Misc 2 (ii)
Misc 3
Misc 4 Deleted for CBSE Board 2022 Exams You are here
Misc 5 Important Deleted for CBSE Board 2022 Exams
Misc 6 Important Deleted for CBSE Board 2022 Exams
Misc 7 Important
Misc 8 Important
Misc 9 Deleted for CBSE Board 2022 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)
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)