

Miscellaneous
Misc 1 (ii)
Misc 2 (i) Important
Misc 2 (ii)
Misc 3
Misc 4 Deleted for CBSE Board 2022 Exams
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 You are here
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)
Last updated at Aug. 23, 2021 by Teachoo
Misc 9 An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes. Let X: Number of successes Since we are talking about success and failure It is a Bernoulli trial So, X has a binomial distribution Here, n = number of trials = 6 p = Probability of success q = Probability of failure = 1 – p Let X: Number of successes Since we are talking about success and failure It is a Bernoulli trial So, X has a binomial distribution Here, n = number of trials = 6 p = Probability of success q = Probability of failure = 1 – p Let X: Number of successes Since we are talking about success and failure It is a Bernoulli trial So, X has a binomial distribution Here, n = number of trials = 6 p = Probability of success q = Probability of failure = 1 – p We need to probability that there will be at least 4 successes i.e. P(X ≥ 4) P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6) = 6C4(2/3)^4 (1/3)^(6−4)+"6C5" (2/3)^5 (1/3)^(6−5)+"6C6" (2/3)^6 (1/3)^(6−6) = 6C4(2/3)^4 (1/3)^2+"6C5" (2/3)^5 (1/3)^1+"6C6" (2/3)^6 (1/3)^0 = 15 × (2/3)^4 (1/3)^2+"6" (2/3)^5 (1/3)^1+"1" (2/3)^6 × 1 = (2/3)^4 ("15 × " (1/3)^2+"6" (2/3)(1/3)+(2/3)^2 ) = (2/3)^4 (15/9+12/9+4/9) = 𝟑𝟏/𝟗 (𝟐/𝟑)^𝟒