# Misc 9 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Aug. 23, 2021 by Teachoo

Miscellaneous

Misc 1 (i)

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)

Chapter 13 Class 12 Probability (Term 2)

Serial order wise

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