# Example 35

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 35 The probability of a shooter hitting a target is 34 . 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 𝒒𝒏−𝒙 𝒑𝒙 n = number of rounds fired p = Probability of hitting = 34 q = 1 – p = 1 − 34 = 14 Hence, ⇒ P(X = x) = nCx 𝟑𝟒𝒙 𝟏𝟒𝒏−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 340 14𝑛> 0.99 1 − 14𝑛 > 0.99 ` 1 − 0.99 > 14𝑛 0.01 > 1 4𝑛 4𝑛 > 10.01 𝟒𝒏 > 𝟏𝟎𝟎 We know that 44 = 256 So, n ≥ 4 So, the minimum value of n is 4 So, he must fire atleast 4 times

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Chapter 13 Class 12 Probability

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.