Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Binomial Distribution
Question 12 Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 9 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 11 Deleted for CBSE Board 2024 Exams
Question 14 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 15 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams You are here
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Misc 4
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams You are here
Question 8 Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Binomial Distribution
Last updated at May 29, 2023 by Teachoo
Question 13 The probability of a shooter hitting a target is 3/4 . How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?Let X : Number of times he hits the target Hitting the target is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of rounds fired p = Probability of hitting = 3/4 q = 1 – p = 1 − 3/4 = 1/4 Hence, P(X = x) = nCx (𝟑/𝟒)^𝒙 (𝟏/𝟒)^(𝒏−𝒙) We need to find How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99 So, given P(X ≥ 1) > 99%, we need to find n Now, P(X ≥ 1) > 99 % 1 − P(X = 0) > 99 % ` 1 − nC0(3/4)^0 (1/4)^𝑛> 0.99 1 − (1/4)^𝑛 > 0.99 1 − 0.99 > (1/4)^𝑛 0.01 > 1/4^𝑛 4^𝑛 > 1/0.01 𝟒^𝒏 > 𝟏𝟎𝟎 We know that 44 = 256 So, n ≥ 4 So, the minimum value of n is 4 So, he must fire atleast 4 times `