Example 35 - Probability of a shooter hitting target is 3/4. How many

Example 35 - Chapter 13 Class 12 Probability - Part 2
Example 35 - Chapter 13 Class 12 Probability - Part 3

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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 `

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.